1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 1

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 2

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 3

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 4

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

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Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 6

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 7

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 8

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 9

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 10

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 11

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 12

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 13

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 14

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 15

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 16

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 17

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 18

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 19

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 20

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 21

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 22

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 23

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 24

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 25

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 26

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 27

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 28

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 29

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 30

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 31

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 32

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 33

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 34

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

eSolutions Manual - Powered by Cognero Page 35

Study Guide and Review - Chapter 2

1. The coefficient of the term with the greatest exponent of the variable is the (leading coefficient, degree) of the polynomial.

SOLUTION: leading coefficient

2. A (polynomial function, power function) is a function of the form f (x) = anxn + an-1x

n-1 + … + a1x + a0, where a1,

a2, …, an are real numbers and n is a natural number.

SOLUTION: polynomial function

3. A function that has multiple factors of (x – c) has (repeated zeros, turning points).

SOLUTION: repeated zeros

4. (Polynomial division, Synthetic division) is a short way to divide polynomials by linear factors.

SOLUTION: Synthetic division

5. The (Remainder Theorem, Factor Theorem) relates the linear factors of a polynomial with the zeros of its related function.

SOLUTION: Factor Theorem

6. Some of the possible zeros for a polynomial function can be listed using the (Factor, Rational Zeros) Theorem.

SOLUTION: Rational Zeros

7. (Vertical, Horizontal) asymptotes are determined by the zeros of the denominator of a rational function.

SOLUTION: Vertical

8. The zeros of the (denominator, numerator) determine the x-intercepts of the graph of a rational function.

SOLUTION: numerator

9. (Horizontal, Oblique) asymptotes occur when a rational function has a denominator with a degree greater than 0 anda numerator with degree one greater than its denominator.

SOLUTION: Oblique

10. A (quartic function, power function) is a function of the form f (x) = axn where a and n are nonzero constant real

numbers.

SOLUTION: power function

11. f (x) = –8x3

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an odd degree and a negative value for a.

D = (− , ), R = (− , ); intercept: 0; continuous for all real numbers;

decreasing: (− , )

X −3 −2 −1 0 1 2 3 f (x) 216 64 8 0 −8 −64 −216

12. f (x) = x –9

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (− , 0) (0, ); no intercepts; infinite discontinuity

at x = 0; decreasing: (− , 0) and (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) −0.026 −1 −512 512 1 0.026

13. f (x) = x –4

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since the power is negative, the function will be undefined at x = 0.

D = (− , 0) (0, ), R = (0, ); no intercepts; infinite discontinuity at x = 0;

increasing: (− , 0); decreasing: (0, )

X −1.5 −1 −0.5 0 0.5 1 1.5 f (x) 0.07 0.33 5.33 5.33 0.33 0.07

14. f (x) = – 11

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 5x − 6. Solve for x when the radicand is 0 to find the restriction on the domain.

To find the x-intercept, solve for x when f (x) = 0.

D = [1.2, ), R = [–11, ); intercept: (25.4, 0); = ; continuous on [1.2, ); increasing: (1.2, )

X 1.2 2 4 6 8 10 12 f (x) −11 −9 −7.3 −6.1 −5.2 −4.4 −3.7

15.

SOLUTION: Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

To find the x-intercepts, solve for x when f (x) = 0.

D = (– , ), R = (– , 2.75]; x-intercept: about –1.82 and 1.82, y-intercept: 2.75; = – and

= − ; continuous for all real numbers; increasing: (– , 0); decreasing: (0, )

X −6 −4 −2 0 2 4 6 f (x) −2.5 −1.4 −0.1 2.75 −0.1 −1.4 −2.5

Solve each equation.

16. 2x = 4 +

SOLUTION:

x = or x = 4.

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x =

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

17. + 1 = 4x

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 1

One solution checks and the other solution does not. Therefore, the solution is x = 1.

18. 4 = −

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 4

One solution checks and the other solution does not. Therefore, the solution is x = 4.

19.

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

X = 15

x = −15

The solutions are x = 15 and x = −15.

Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning using the leading term test.

20. F(x) = –4x4 + 7x

3 – 8x

2 + 12x – 6

SOLUTION:

f (x) = – ; f (x) = – ; The degree is even and the leading coefficient is negative.

21. f (x) = –3x5 + 7x

4 + 3x

3 – 11x – 5

SOLUTION:

f (x) = – ; f (x) = ; The degree is odd and the leading coefficient is negative.

22. f (x) = x2 – 8x – 3

SOLUTION:

f (x) = and f (x) = ; The degree is even and the leading coefficient is positive.

23. f (x) = x3(x – 5)(x + 7)

SOLUTION:

f (x) = and f (x) = – ; The degree is odd and the leading coefficient is positive.

State the number of possible real zeros and turning points of each function. Then find all of the real zeros by factoring.

24. F(x) = x3 – 7x

2 +12x

SOLUTION: The degree of f (x) is 3, so it will have at most three real zeros and two turning points.

So, the zeros are 0, 3, and 4.

25. f (x) = x5 + 8x

4 – 20x

3

SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

So, the zeros are −10, 0, and 2.

26. f (x) = x4 – 10x

2 + 9

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Let u = x2.

So, the zeros are –3, –1, 1, and 3.

27. f (x) = x4 – 25

SOLUTION: The degree of f (x) is 4, so it will have at most four real zeros and three turning points.

Because ± are not real zeros, f has two distinct real zeros, ± .

For each function, (a) apply the leading term test, (b) find the zeros and state the multiplicity of any repeated zeros, (c) find a few additional points, and then (d) graph the function.

28. f (x) = x3(x – 3)(x + 4)

2

SOLUTION: a. The degree is 6, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 0, 3, and −4. The zero −4 has multiplicity 2 since (x + 4) is a factor of the polynomial twice and the zero 0 has a multiplicity 3 since x is a factor of the polynomial three times. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X −1 0 1 2 f (x) 36 0 −50 −288

X −5 −4 −3 −2 3 4 5 6 f (x) 1000 0 162 160 0 4096 20,250 64,800

29. f (x) = (x – 5)2(x – 1)

2

SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,

b. The zeros are 5 and 1. Both zeros have multiplicity 2 because (x − 5) and (x − 1) are factors of the polynomial twice. c. Sample answer: Evaluate the function for a few x-values in its domain.

d. Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

X 2 3 4 5 f (x) 9 16 9 0

X −3 −2 −1 0 1 6 7 8 f (x) 1024 441 144 25 0 25 144 441

Divide using long division.

30. (x3 + 8x

2 – 5) ÷ (x – 2)

SOLUTION:

So, (x3 + 8x

2 – 5) ÷ (x – 2) = x

2 + 10x + 20 + .

31. (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4)

SOLUTION:

So, (–3x3 + 5x

2 – 22x + 5) ÷ (x2

+ 4) = .

32. (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1)

SOLUTION:

So, (2x5 + 5x

4 − 5x3 + x

2 − 18x + 10) ÷ (2x – 1) = x4 + 3x

3 − x2 − 9 + .

Divide using synthetic division.

33. (x3 – 8x

2 + 7x – 15) ÷ (x – 1)

SOLUTION:

Because x − 1, c = 1. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is .

34. (x4 – x

3 + 7x

2 – 9x – 18) ÷ (x – 2)

SOLUTION:

Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + x

2 + 9x + 9.

35. (2x4 + 3x

3 − 10x2 + 16x − 6 ) ÷ (2x – 1)

SOLUTION:

Rewrite the division expression so that the divisor is of the form x − c.

Because Set up the synthetic division as follows. Then follow the synthetic division procedure.

The quotient is x3 + 2x

2 − 4x + 6.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

36. f (x) = x3 + 3x

2 – 8x – 24; (x + 3)

SOLUTION: Use synthetic division to test the factor (x + 3).

Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x).

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

– 8).

37. f (x) =2x4 – 9x

3 + 2x

2 + 9x – 4; (x – 1), (x + 1)

SOLUTION:

Use synthetic division to test each factor, (x − 1) and (x + 1).

Because the remainder when f (x) is divided by (x − 1) is 0, (x − 1) is a factor. Test the second factor, (x + 1), with

the depressed polynomial 2x3 − 7x

2 − 5x + 4.

Because the remainder when the depressed polynomial is divided by (x + 1) is 0, (x + 1) is a factor of f (x).

Because (x − 1) and (x + 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x − 1)(x + 1)(2x2 − 9x + 4). Factoring the quadratic expression yields f (x) = (x – 1)(x + 1)(x – 4)(2x – 1).

38. f (x) = x4 – 2x

3 – 3x

2 + 4x + 4; (x + 1), (x – 2)

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x − 2).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x − 2), with

the depressed polynomial x3 − 3x

2 + 4.

Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x).

Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x − 2)(x2 − x − 2). Factoring the quadratic expression yields f (x) = (x – 2)

2(x + 1)

2.

39. f (x) = x3 – 14x − 15

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15.

Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (x + 3)(x2

−3x − 5). Because the factor (x2 − 3x − 5) yields no rational zeros, the rational zero of f is −3.

40. f (x) = x4 + 5x

2 + 4

SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros of f are ±1, ±2, and ±4. Using synthetic division, it does not appear that the polynomial has any rational zeros.

Factor x4 + 5x

2 + 4. Let u = x

2.

Since (x2 + 4) and (x

2 + 1) yield no real zeros, f has no rational zeros.

41. f (x) = 3x4 – 14x

3 – 2x

2 + 31x + 10

SOLUTION:

The leading coefficient is 3 and the constant term is 10. The possible rational zeros are or 1, 2,

5, 10, .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that is a rational zero.

Because (x − 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as

Since (3x2 − 9x − 15) yields no rational zeros, the rational zeros of f are

Solve each equation.

42. X4 – 9x3 + 29x

2 – 39x +18 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros

are 1, 2, 3, 6, 9 and 18. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x − 1) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 1)(x − 2)(x2 − 6x + 9). Factoring the quadratic expression yields 0 = (x − 1)(x − 2)(x − 3)

2. Thus, the solutions

are 1, 2, and 3 (multiplicity: 2).

43. 6x3 – 23x

2 + 26x – 8 = 0

SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the

constant term is −8. The possible rational zeros are or 1, 2, 4, 8,

.

By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as 0 = (x − 2)(6x2

− 11x + 4). Factoring the quadratic expression yields 0 = (x − 2)(3x − 4)(2x − 1).Thus, the solutions are 2, , and

.

44. x4 – 7x3 + 8x

2 + 28x = 48

SOLUTION:

The equation can be written as x4 – 7x

3 + 8x

2 + 28x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors

of the constant term −48. Therefore, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.

By using synthetic division, it can be determined that x = −2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x + 2) and (x − 2) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x + 2)(x − 2)(x2 − 7x + 12). Factoring the quadratic expression yields 0 = (x + 2)(x − 2)(x − 3)(x − 4). Thus, the

solutions are −2, 2, 3, and 4.

45. 2x4 – 11x

3 + 44x = –4x

2 + 48

SOLUTION:

The equation can be written as 2x4 – 11x

3 + 4x

2 + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible

rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros

are or ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± .

By using synthetic division, it can be determined that x = 2 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 4 is a rational zero.

Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a factored form as 0 =

(x − 2)(x − 4)(2x2 + x − 6). Factoring the quadratic expression yields 0 = (x − 2)(x − 4)(2x − 3)(x + 2). Thus, the

solutions are , –2, 2, and 4.

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

46. f (x) = x4 + x

3 – 41x

2 + x – 42; i

SOLUTION: Use synthetic substitution to verify that I is a zero of f (x).

Because x = I is a zero of f , x = −I is also a zero of f . Divide the depressed polynomial by −i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + i)(x − i)(x2 + x − 42).

Factoring the quadratic expression yields f (x) = (x + i)(x − i)(x + 7)(x − 6).Therefore, the zeros of f are −7, 6, I, and –i.

47. f (x) = x4 – 4x

3 + 7x

2 − 16x + 12; −2i

SOLUTION:

Use synthetic substitution to verify that −2i is a zero of f (x).

Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.

Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x2 − 4x + 3).

Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i,

and −2i.

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

48. F(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D

= x | x −4, x R. Check for vertical asymptotes.

Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the right.

Because x = −4 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, there does not appear to be a horizontal

asymptote.

X −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9 f (x) −158 −1508 −15,008 undef 14,992 1492 142

x −100 −50 −10 0 10 50 100 f (x) −104.2 −54.3 −16.5 −0.3 7.1 46.3 96.1

49. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 − 25. The real zeros of b(x) are 5 and −5.

Therefore, D = x | x 5, –5, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = −5 is a point of infinite discontinuity. Find the limit as x approaches −5 from the left and the right.

Because x = −5 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) −24 −249 −2499 undef 2501 251 26

X −5.1 −5.01 −5.001 −5 −4.999 −4.99 −4.9 f (x) 26 251 2500 undef −2499 −249 −24

x −100 −50 −10 0 10 50 100 f (x) 1.0025 1.0101 1.3333 0 1.3333 1.0101 1.0025

50. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x − 5)2(x + 3)

2. The real zeros of b(x) are 5

and −3. Therefore, D = x | x 5, –3, x R. Check for vertical asymptotes. Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the right.

Because x = 5 is a vertical asymptote of f .

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 0 is a horizontal asymptote of f .

X 4.9 4.99 4.999 5 5.001 5.01 5.1 f (x) 15 1555 156,180 undef 156,320 1570 16

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) 29 2819 281,320 undef 281,180 2806 27

x −100 −50 −10 0 10 50 100 f (x) 0.0001 0.0004 0.0026 0 0.0166 0.0004 0.0001

51. f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 9). The real zeros of b(x) are −3

and −9. Therefore, D = x | x −3, –9, x R. Check for vertical asymptotes.

Determine whether x = −3 is a point of infinite discontinuity. Find the limit as x approaches −3 from the left and the right.

Because x = −3 is a vertical asymptote of f .

Determine whether x = −9 is a point of infinite discontinuity. Find the limit as x approaches −9 from the left and the right.

Because x = −9 is a vertical asymptote of f .

Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore, you know that y = 1 is a horizontal asymptote of f .

X −3.1 −3.01 −3.001 −3 −2.999 −2.99 −2.9 f (x) −70 −670 −6670 undef 6663 663 63

X −9.1 −9.01 −9.001 −9 −8.999 −8.99 −8.9 f (x) 257 2567 25,667 undef −25,667 −2567 −257

x −1000 −100 −50 0 50 100 1000 f (x) 1.0192 1.2133 1.4842 03704 0.6908 0.8293 0.9812

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

52. F(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x 5, x R. There are vertical asymptotes at x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because f (0) =0. Graph the asymptotes and intercepts. Then find and plot points.

53. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x −4, x R. There are vertical asymptotes at x = −4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 2, the zero of the numerator. The y-intercept is because .

Graph the asymptotes and intercepts. Then find and plot points.

54. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –5, 6, x R. There are vertical asymptotes at x = −5 and x = 6. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −3 and 4, the zeros of the numerator. The y-intercept is because f (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

55. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –6, 3, x R. There are vertical asymptotes at x = −6 and x = 3. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are −7 and 0, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

56. f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = x | x –1, 1, x R. There are vertical asymptotes at x = −1 and x = 1. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is −2, the zero of the numerator. The y-intercept is −2 because f (0) = −2. Graph the asymptote and intercepts. Then find and plot points.

57. f (x) =

SOLUTION:

The function is undefined at b(x) = 0. b(x) = x3 − 6x

2 + 5x or x(x − 5)(x − 1). So, D = x | x 0, 1, 5, x R.

There are vertical asymptotes at x = 0, x = 1, and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are −4 and 4, the zeros of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptote and intercepts. Then find and plot points.

Solve each equation.

58. + x – 8 = 1

SOLUTION:

59.

SOLUTION:

60.

SOLUTION:

Because 0 ≠ 5, the equation has no solution.

61. = +

SOLUTION:

62. x2 – 6x – 16 > 0

SOLUTION:

Let f (x) = x2 – 6x − 16. A factored form of f (x) is f (x) = (x − 8)(x + 2). F(x) has real zeros at x = 8 and x = −2. Set

up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 – 6x – 16 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

63. x3 + 5x2 ≤ 0

SOLUTION:

Let f (x) = x3 + 5x

2. A factored form of f (x) is f (x) = x

2(x + 5). F(x) has real zeros at x = 0 and x = −5. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

3 + 5x

2 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see

that the solution set is .

64. 2x2 + 13x + 15 < 0

SOLUTION:

Let f (x) = 2x2 + 13x + 15. A factored form of f (x) is f (x) = (x + 5)(2x + 3). F(x) has real zeros at x = −5 and

. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f

(x) is positive or negative at that point.

The solutions of 2x2 + 13x + 15 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is .

65. x2 + 12x + 36 ≤ 0

SOLUTION:

Let f (x) = x2 + 12x + 36. A factored form of f (x) is f (x) = (x + 6)

2. f (x) has a real zero at x = −6. Set up a sign

chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x

2 + 12x + 36 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can

see that the solution set is –6.

66. x2 + 4 < 0

SOLUTION:

Let f (x) = x2 + 4. Use a graphing calculator to graph f (x).

f(x) has no real zeros, so there are no sign changes. The solutions of x2 + 4 < 0 are x-values such that f (x) is

negative. Because f (x) is positive for all real values of x, the solution set of x2 + 4 < 0 is .

67. x2 + 4x + 4 > 0

SOLUTION:

Let f (x) = x2 + 4x + 4. A factored form of f (x) is f (x) = (x + 2)

2. f (x) has a real zero at x = −2. Set up a sign chart.

Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.

The solutions of x2 + 4x + 4 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is (– , –2) (–2, ).

68. < 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = 5, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution

set is (0, 5).

69. ≥0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =

−1, and denominator, . Create a sign chart. Then choose and test x-values in each interval to

determine if f (x) is positive or negative.

The solutions of ≥0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you

can see that the solution set is . are not part of the solution set because

the original inequality is undefined at .

70. + > 0

SOLUTION:

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x = ,

and denominator, x = 3 and x = 4. Create a sign chart. Then choose and test x-values in each interval to determine iff(x) is positive or negative.

The solutions of + > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the

solution set is .

71. PUMPKIN LAUNCH Mr. Roberts technology class constructed a catapult to compete in the county’s annual pumpkin launch. The speed v in miles per hour of a launched pumpkin after t seconds is given in the table.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a pumpkin is traveling after 1.2 seconds. d. Use the function to predict the time at which the pumpkin’s speed is 47 miles per hour.

SOLUTION: a. Use a graphing calculator to plot the data.

b. Use the power regression function on the graphing calculator to find values for a and n

v(t) = 43.63t-1.12

c. Graph the regression equation using a graphing calculator. To predict the speed at which a pumpkin is traveling after 1.2 seconds, use the CALC function on the graphing calculator. Let x = 1.2.

The speed a pumpkin is traveling after 1.2 seconds is approximately 35.6 miles per hour. d. Graph the line y = 47. Use the CALC menu to find the intersection of y = 47 with v(t).

A pumpkin’s speed is 47 miles per hour at about 0.94 second.

72. AMUSEMENT PARKS The elevation above the ground for a rider on the Big Monster roller coaster is given in the table.

a. Create a scatter plot of the data and determine the type of polynomial function that could be used to represent the data. b. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth and state the correlation coefficient. c. Use the model to estimate a rider’s elevation at 17 seconds. d. Use the model to determine approximately the first time a rider is 50 feet above the ground.

SOLUTION: a.

; Given the position of the points, a cubic function could be a good representation of the data.

b. Sample answer: Use the cubic regression function on the graphing calculator.

f(x) = 0.032x3 – 1.171x

2 + 6.943x + 76; r

2 = 0.997

c. Sample answer: Graph the regression equation using a graphing calculator. To estimate a rider’s elevation at 17 seconds, use the CALC function on the graphing calculator. Let x = 17.

A rider’s elevation at 17 seconds is about 12.7 feet. d. Sample answer: Graph the line y = 50. Use the CALC menu to find the intersection of y = 50 with f (x).

A rider is 50 feet for the first time at about 11.4 seconds.

73. GARDENING Mark’s parents seeded their new lawn in 2001. From 2001 until 2011, the amount of crab grass

increased following the model f (x) = 0.021x3 – 0.336x

2 + 1.945x – 0.720, where x is the number of years since 2001

and f (x) is the number of square feet each year . Use synthetic division to find the number of square feet of crab grass in the lawn in 2011. Round to the nearest thousandth.

SOLUTION: To find the number of square feet of crab grass in the lawn in 2011, use synthetic substitution to evaluate f (x) for x =10.

The remainder is 6.13, so f (10) = 6.13. Therefore, rounded to the nearest thousand, there will be 6.13 square feet of crab grass in the lawn in 2011.

74. BUSINESS A used bookstore sells an average of 1000 books each month at an average price of $10 per book. Dueto rising costs the owner wants to raise the price of all books. She figures she will sell 50 fewer books for every $1 she raises the prices. a. Write a function for her total sales after raising the price of her books x dollars. b. How many dollars does she need to raise the price of her books so that the total amount of sales is $11,250? c. What is the maximum amount that she can increase prices and still achieve $10,000 in total sales? Explain.

SOLUTION: a. The price of the books will be 10 + x for every x dollars that the owner raises the price. The bookstore will sell

1000 − 50x books for every x dollars that the owner raises the price. The total sales can be found by multiplying the price of the books by the amount of books sold.

A function for the total sales is f (x) = -50x2 + 500x + 10,000.

b. Substitute f (x) = 11,250 into the function found in part a and solve for x.

The solution to the equation is x = 5. Thus, the owner will need to raise the price of her books by $5. c. Sample answer: Substitute f (x) = 10,000 into the function found in part a and solve for x.

The solutions to the equation are x = 0 and x = 10. Thus, the maximum amount that she can raise prices and still achieve total sales of $10,000 is $10. If she raises the price of her books more than $10, she will make less than $10,000.

75. AGRICULTURE A farmer wants to make a rectangular enclosure using one side of her barn and 80 meters of fence material. Determine the dimensions of the enclosure if the area of the enclosure is 750 square meters. Assume that the width of the enclosure w will not be greater than the side of the barn.

SOLUTION: The formula for the perimeter of a rectangle is P = 2l + 2w. Since the side of the barn is accounting for one of the

widths of the rectangular enclosure, the perimeter can be found by P = 2l + w or 80 = 2l + w. Solving for w, w = 80 −2l.

The area of the enclosure can be expressed as A = lw or 750 = lw. Substitute w = 80 − 2l and solve for l.

l = 25 or l = 15. When l = 25, w = 80 − 2(25) or 30. When l = 15, w = 80 − 2(15) or 50. Thus, the dimensions of the enclosure are 15 meters by 50 meters or 25 meters by 30 meters.

76. ENVIRONMENT A pond is known to contain 0.40% acid. The pond contains 50,000 gallons of water. a. How many gallons of acid are in the pond? b. Suppose x gallons of pure water was added to the pond. Write p (x), the percentage of acid in the pond after x gallons of pure water are added. c. Find the horizontal asymptote of p (x). d. Does the function have any vertical asymptotes? Explain.

SOLUTION: a. 0.40% = 0.004. Thus, there are 0.004 · 50,000 or 200 gallons of acid in the pond. b. The percent of acid p that is in the pond is the ratio of the total amount of gallons of acid to the total amount of

gallons of water in the pond, or p = . If x gallons of pure water is added to the pond, the total amount of

water in the pond is 50,000 + x gallons. Thus,

p(x) = .

c. Since the degree of the denominator is greater than the degree of the numerator, the function has a horizontal asymptote at y = 0.

d. No. Sample answer: The domain of function is [0, ∞) because a negative amount of water cannot be added to thepond. The function does have a vertical asymptote at x = −50,000, but this value is not in the relevant domain. Thus,

p(x) is defined for all numbers in the interval [0, ).

77. BUSINESS For selling x cakes, a baker will make b(x) = x2 – 5x − 150 hundreds of dollars in revenue. Determine

the minimum number of cakes the baker needs to sell in order to make a profit.

SOLUTION:

Solve for x for x2 – 5x − 150 > 0. Write b(x) as b(x) = (x − 15)(x + 10). b(x) has real zeros at x = 15 and x = −10.

Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if b(x) is positive or negative at that point.

The solutions of x

2 – 5x − 150 > 0 are x-values such that b(x) is positive. From the sign chart, you can see that the

solution set is (– , –10) (15, ). Since the baker cannot sell a negative number of cakes, the solution set

becomes (15, ). Also, the baker cannot sell a fraction of a cake, so he must sell 16 cakes in order to see a profit.

78. DANCE The junior class would like to organize a school dance as a fundraiser. A hall that the class wants to rent costs $3000 plus an additional charge of $5 per person. a. Write and solve an inequality to determine how many people need to attend the dance if the junior class would liketo keep the cost per person under $10. b. The hall will provide a DJ for an extra $1000. How many people would have to attend the dance to keep the cost under $10 per person?

SOLUTION: a. The total cost of the dance is $3000 + $5x, where x is the amount of people that attend. The average cost A per

person is the total cost divided by the total amount of people that attend or A = . Use this expression to

write an inequality to determine how many people need to attend to keep the average cost under $10.

Solve for x.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =600, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (600, ∞). Since a negative number of people cannot attend the dance, the solution set is (600, ∞). Thus, more than 600 people need to attend the dance. b. The new total cost of the dance is $3000 + $5x + $1000 or $4000 + $5x. The average cost per person is A =

+ 5. Solve for x for + 5 < 10.

Let f (x) = . The zeros and undefined points of the inequality are the zeros of the numerator, x =800, and

denominator, x = 0. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

The solutions of < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the

solution set is (–∞, 0) ∪ (800, ∞). Since a negative number of people cannot attend the dance, the solution set is (800, ∞). Thus, more than 800 people need to attend the dance.

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Study Guide and Review - Chapter 2