state whether each sentence is true or false . if false , replace the … · 2014-03-31 · state...

31
State whether each sentence is true or false . If false , replace the underlined term to make a true sentence. 1. The axis of symmetry of a quadratic function can be found by using the equation x = . SOLUTION: The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation . The statement is true. 2. The vertex is the maximum or minimum point of a parabola. SOLUTION: The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true. 3. The graph of a quadratic function is a straight line . SOLUTION: Quadratic functions are nonlinear and can be written in the form f (x) = ax 2 + bx + c, where a ≠ 0. So, the statement is false. The graph of a quadratic function is a parabola . 4. The graph of a quadratic function has a maximum if the coefficient of the x 2 -term is positive. SOLUTION: When a > 0, the graph of y = ax 2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the statement is false. The graph of a quadratic function has a minimum if the coefficient of the x 2 -term is positive. 5. A quadratic equation with a graph that has two x-intercepts has one real root. SOLUTION: The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots. 6. The expression b 2 4ac is called the discriminant . SOLUTION: In the Quadratic Formula, the expression under the radical sign, b 2 4ac, is called the discriminant. The statement is true. 7. The solutions of a quadratic equation are called roots . SOLUTION: The solutions of a quadratic equation are called roots. The statement is true. 2 eSolutions Manual - Powered by Cognero Page 1 Study Guide and Review - Chapter 9

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Page 1: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 1

Study Guide and Review - Chapter 9

Page 2: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 2

Study Guide and Review - Chapter 9

Page 3: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 3

Study Guide and Review - Chapter 9

Page 4: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 4

Study Guide and Review - Chapter 9

Page 5: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 5

Study Guide and Review - Chapter 9

Page 6: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 6

Study Guide and Review - Chapter 9

Page 7: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 7

Study Guide and Review - Chapter 9

Page 8: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 8

Study Guide and Review - Chapter 9

Page 9: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 9

Study Guide and Review - Chapter 9

Page 10: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 10

Study Guide and Review - Chapter 9

Page 11: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 11

Study Guide and Review - Chapter 9

Page 12: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 12

Study Guide and Review - Chapter 9

Page 13: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 13

Study Guide and Review - Chapter 9

Page 14: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 14

Study Guide and Review - Chapter 9

Page 15: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 15

Study Guide and Review - Chapter 9

Page 16: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 16

Study Guide and Review - Chapter 9

Page 17: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 17

Study Guide and Review - Chapter 9

Page 18: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 18

Study Guide and Review - Chapter 9

Page 19: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 19

Study Guide and Review - Chapter 9

Page 20: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 20

Study Guide and Review - Chapter 9

Page 21: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 21

Study Guide and Review - Chapter 9

Page 22: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 22

Study Guide and Review - Chapter 9

Page 23: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 23

Study Guide and Review - Chapter 9

Page 24: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 24

Study Guide and Review - Chapter 9

Page 25: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 25

Study Guide and Review - Chapter 9

Page 26: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 26

Study Guide and Review - Chapter 9

Page 27: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 27

Study Guide and Review - Chapter 9

Page 28: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 28

Study Guide and Review - Chapter 9

Page 29: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 29

Study Guide and Review - Chapter 9

Page 30: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

eSolutions Manual - Powered by Cognero Page 30

Study Guide and Review - Chapter 9

Page 31: State whether each sentence is true or false . If false , replace the … · 2014-03-31 · State whether each sentence is true or false . If false , replace the underlined term to

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The axis of symmetry of a quadratic function can be found by using the equation x = .

SOLUTION:  The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation

. The statement is true.

2. The vertex is the maximum or minimum point of a parabola.

SOLUTION:  The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a parabola. The statement is true.

3. The graph of a quadratic function is a straight line.

SOLUTION:  

Quadratic functions are nonlinear and can be written in the form f (x) = ax2 + bx + c, where a ≠ 0. So, the statement 

is false. The graph of a quadratic function is a parabola .

4. The graph of a quadratic function has a maximum if the coefficient of the x2-term is positive.

SOLUTION:  

When a > 0, the graph of y = ax2 + bx + c opens upward. The lowest point on the graph is the minimum. So, the

statement is false. The graph of a quadratic function has a minimum if the coefficient of the x2-term is positive.

5. A quadratic equation with a graph that has two x-intercepts has one real root.

SOLUTION:  The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So, the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.

6. The expression b2 − 4ac is called the discriminant.

SOLUTION:  

In the Quadratic Formula, the expression under the radical sign, b2 − 4ac, is called the discriminant. The statement is

true.

7. The solutions of a quadratic equation are called roots.

SOLUTION:  The solutions of a quadratic equation are called roots. The statement is true.

8. The graph of the parent function is translated down to form the graph of f (x) = x2 + 5.

SOLUTION:  

If c > 0 in f (x) = x2 + c, the graph of the parent function is translated  units up. So, the statement is false. The 

graph of the parent function is translated up 5 units to form the graph of f (x) = x2 + 5.

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

9. The range of the greatest integer function is the set of all real numbers.

SOLUTION:  The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all integers. 

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.

SOLUTION:  true

Consider each equation. a.  Determine whether the function has a maximum or minimum value.b.  State the maximum or minimum value. c.  What are the domain and range of the function?

11. y = x2 − 4x + 4

SOLUTION:  

a. For y = x2 – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a

minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. 

The minimum value is 0. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ 0}.

12. y = −x2 + 3x

SOLUTION:  

a. For y = −x2 + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has

a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2.25. c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y |y  ≤ 2.25}.

13. y = x2 − 2x − 3

SOLUTION:  

a. For y = x2 − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has

a minimum value.

b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The minimum value is –4. c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y |y ≥ –4}.

14. y = −x2 + 2

SOLUTION:  

a. For y = −x2 + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a

maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 2. c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y |y  ≤ 2}.

15. BASEBALL  A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t2 + 32t

gives the height of the rocket t seconds after it is launched. a.  Determine whether the function has a maximum or minimum value. b.  State the maximum or minimum value. c.  State a reasonable domain and range of this situation.

SOLUTION:  

a. For h = −16t2 + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function

has a maximum value.

b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is .

 

  The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate.   

  The maximum value is 16. c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet, and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}.  

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

16. x2 − 3x − 4 = 0

SOLUTION:  

Graph the related function f (x) = x2 − 3x − 4.

The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.CHECK: Check each solution in the original equation. 

     Therefore, the solutions are –1 and 4.

17. −x2 + 6x − 9 = 0

SOLUTION:  

Graph the related function f (x) = −x2 + 6x − 9.

The x-intercept of the graph appears to be only at 3, so the only solution is 3.CHECK: Check the solution in the original equation.

Therefore, the solution is 3.

18. x2 − x − 12 = 0

SOLUTION:  

Graph the related function f (x) = x2 − x − 12.

The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.CHECK: Check each solution in the original equation.

         Therefore, the solutions are –3 and 4.

19. x2 + 4x − 3 = 0

SOLUTION:  

Graph the related function f (x) = x2 + 4x − 3.

The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are approximately –4.6 and 0.6.

x –4.9 –4.8 –4.7 –4.6 y 1.41 0.84 0.29 –0.24 x –4.4 –4.3 –4.2 –4.1 y –1.24 –1.71 –2.16 –2.59 x 0.1 0.2 0.3 0.4 y –2.59 –2.16 –1.71 –1.24 x 0.6 0.7 0.8 0.9 y –0.24 0.29 0.84 1.41

20. x2 − 10x = −21

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = x2 − 10x + 21.

The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.CHECK: Check each solution in the original equation.

       Therefore, the solutions are 3 and 7.

21. 6x2 − 13x = 15

SOLUTION:  Rewrite the equation in standard form.

Graph the related function f (x) = 6x2 − 13x – 15.

The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3. CHECK: Check the solution in the original equation.

  To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look for a change in the signs of the function values. The function value that is closest to zero is the best approximation for a zero of the function.  

  The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –0.8. So, the solutions are about  –0.8 and 3.

x –0.9 –0.8 –0.7 y 1.56 –0.76 –2.96 x –0.6 –0.5 –0.4 y –5.04 –7 –8.84 x –0.3 –0.2 –0.1 x –10.56 –12.16 –13.64

22. NUMBER THEORY  Find two numbers that have a sum of 2 and a product of −15.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.   x + y = 2

x • y = –15   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

 

Graph the related function f (x) = x2 − 2x – 15.

The x-intercepts of the graph appear to be at –3 and 5, so the solutions  for x are –3 and 5. When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is –15. Therefore the two numbers are –3 and 5.   

Describe how the graph of each function is related to the graph of f (x) = x2.

23. f (x) = x2 + 8

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.

So, the graph is shifted up 8 units from the parent function.

24. f (x) = x2 − 3

SOLUTION:  

The graph of f (x) = x2 + c represents a translation up or down of the parent graph. Since c = –3, the translation is

down. So, the graph is shifted down 3 units from the parent function.

25. f (x) = 2x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x

2 is the

graph of y = x2 vertically stretched.

26. f (x) = 4x2 − 18

SOLUTION:  

The function can be written f (x) = ax2 + c, where a = 4 and c = –18. Since –18 < 0 and  > 1, the graph of y = 4x

2

− 18 is the graph of y = x2 vertically stretched and shifted down 18 units.

27. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. Since a = , the graph of y = x

2 is

the graph of y = x2 vertically compressed.

28. f (x) = x2

SOLUTION:  

The graph of f (x) = ax2 stretches or compresses the parent graph vertically. ince a = , the graph of y = x

2  is 

the graph of y = x2 vertically compressed.

29. Write an equation for the function shown in the graph.

SOLUTION:  Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have

the form y = ax2 + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this

point to find the value of a.

Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x2 – 3.

30. PHYSICS  A ball is dropped off a cliff that is 100 feet high. The function h = −16t2 + 100 models the height h of the

ball after t seconds. Compare the graph of this function to the graph of h = t2.

SOLUTION:  

Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at2 + c,

where a = –16 and c = 100. Since 100 > 0 and  > 1, the graph of h = −16t2 + 100 is the graph of y = x

2

vertically stretched and shifted up 100 units from the parent function.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

31. x2 + 6x + 9 = 16

SOLUTION:  

  The solutions are 1 and –7.

32. −a2 − 10a + 25 = 25

SOLUTION:  

  The solutions are 0 and –10.

33. y2 − 8y + 16 = 36

SOLUTION:  

The solutions are 10 and –2.

34. y2 − 6y + 2 = 0

SOLUTION:  

  Use a calculator to approximate each value of y .  

  The solutions are approximately 5.6 and 0.4.

35. n2 − 7n = 5

SOLUTION:  

  Use a calculator to approximate each value of n.  

  The solutions are approximately 7.7 and –0.7.

36. −3x2 + 4 = 0

SOLUTION:  

  Use a calculator to approximate each value of x.  

  The solutions are approximately 1.2 and –1.2.

37. NUMBER THEORY  Find two numbers that have a sum of −2 and a product of −48.

SOLUTION:  From the given information, write two equations. Let x and y represent the numbers.  x + y = –2

x • y = –48   Solve the first equation for y .  

  Substitute this value for y into the second equation.  

  Now solve by completing the square.  

  The solutions are 6 and –8.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

38. x2 − 8x = 20

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 1, b = –8, and c = –20.  

  The solutions are 10 and –2.

39. 21x2 + 5x − 7 = 0

SOLUTION:  For this equation, a = 21, b = 5, and c = –7.

The solutions are approximately 0.5 and –0.7.

40. d2 − 5d + 6 = 0

SOLUTION:  For this equation, a = 1, b = –5, and c = 6.

The solutions are 3 and 2.

41. 2f2 + 7f − 15 = 0

SOLUTION:  For this equation, a = 2, b = 7, and c = –15.

The solutions are 1.5 and –5.

42. 2h2 + 8h + 3 = 3

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 2, b = 8, and c = 0.

The solutions are 0 and –4.

43. 4x2 + 4x = 15

SOLUTION:  Rewrite the equation in standard form.  

  For this equation, a = 4, b = 4, and c = –15.

The solutions are 1.5 and –2.5.

44. GEOMETRY  The area of a square can be quadrupled by increasing the side length and width by 4 inches. What isthe side length?

SOLUTION:  

If x = the original side length, then the area of the square = x2. If x is increased by 4, the area of the square is

quadrupled. Write an equation to represent the new area of the square.  

  For this equation, a = –3, b = 8, and c = 16.

Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.

Look for a pattern in each table of values to determine which kind of model best describes the data. Then write an equation for the function that models the data.

45. 

SOLUTION:  

First differences:

Second differences:

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

3) to find the value of a.

An equation that models the data is y = 3x2.

46. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

  Because neither the first difference nor the second differences are equal, the table does not represent a linear or quadratic function. Compare the ratios of the y-values. Calculate the ratios.

The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential

function. The equation has the form y = abx. The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the

value of a.

An equation that models the data is y = 1 • 2x or y = 2

x.

47. 

SOLUTION:  

Calculate the first differences.

Calculate the second differences.

Since the second differences are equal, a quadratic function models the data.

Write an equation for the function that models the data. The equation has the form y = ax2. Use the ordered pair (1,

–1) to find the value of a.

An equation that models the data is y = –x2.

Graph each function. State the domain and range.

48.  

 

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

49. f (x) =

SOLUTION:  Make a table of values.  

 

Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers. The range is all integers.

x 2x f (x) 0 0 0

0.25 0.5 0 0.5 1 1 1 2 2

1.25 2.5 2 1.5 3 3 2 4 4

50. f (x) = |x|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,

so the range is {y | y  0}.

x −2 −1 0 1 2 f (x) 2 1 0 1 2

51. f (x) = |2x − 2|

SOLUTION:  Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.

Make a table of values. Be sure to include the domain values for which the function changes.  

 

The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,

so the range is {y | y  0}.

x −1 0 1 2 3 f (x) 4 2 0 2 4

52.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

  Notice that both functions are linear.  

The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between

−1 and 3. Thus, the range is {y | y < −1 or y  3}.

x −2 −1 0 1 2 f (x) −4 −3 −2 3 6

53.  

SOLUTION:  This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes.

Notice that both functions are linear.

The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values

between 1 and 3. Thus, the range is {y | y  1 or y > 3}.

x −1 0 1 2 3 f (x) −5 −3 −1 1 4

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Study Guide and Review - Chapter 9