polynomial functions: graphs, applications, and models

Polynomial Functions: Graphs, Applications, and Models

Graph functions (x) = axn

Graph General Polynomial Functions

Find Turning Points and End Behavior

Intermediate Value Theorems

Approximate Real Zeros Using GraphingCalculator

Use Graphing Calculator to Determine Polynomial Models and Curve Fitting

Objectives

Students will learn how to:

Example 1 GRAPHING FUNCTIONS OF THE FORM (x) = axn

Solution Choose several values for x, and find the corresponding values of (x), or y.

a.Graph the function.

3( )x xf

x (x)

– 2 – 8– 1 – 10 0

1 1

2 8

3( )x xf


Solution The graphs of (x) = x3 and g(x) = x5 are both symmetric with respect to the origin.

b.Graph the function.

5( )x xg

x g(x)

– 1.5 –7.6– 1 – 10 0

1 1

1.5 7.6

5( )x xg


Solution The graphs of (x) = x3 and g(x) = x5 are both symmetric with respect to the origin.

c.Graph the function.

4 6( ) ( )x x x x f , g

x g(x)

– 1.5 11.4– 1 10 0

1 1

1.5 11.4

6( )x xg

x (x)

– 2 16– 1 10 0

1 1

2 16

4( )x xf

Graphs of General Polynomial Functions

As with quadratic functions, the absolute value of a in (x) = axn determines the width of the graph. When a> 1, the graph is stretched vertically, making it narrower, while when 0 < a < 1, the graph is shrunk or compressed vertically, so the graph is broader. The graph of (x) = – axn is reflected across the x-axis compared to the graph of (x) = axn.

Graphs of General Polynomial Functions

Compared with the graph of the graph of is translated (shifted) k units up if k > 0 andkunits down if k < 0. Also, when compared with the graph of the graph of (x) = a(x – h)n is translated h units to the right if h > 0 and hunits to the left if h < 0.The graph of shows a combination of these translations. The effects here are the same as those we saw earlier with quadratic functions.

( ) ,nx axf

( ) nx ax k f

( ) nx ax k f

( ) ,nx axf

Example 2 EXAMINING VERTICAL AND HORIZONTAL TRANSLATIONS

Solution The graph will be the same as that of (x) = x5, but translated 2 units down.

a.Graph the function.

5( ) 2x x f


Solution In (x) = (x + 1)6, function has a graph like that of (x) = x6, but since x + 1 = x – (– 1), it is translated 1 unit to the left.

b.Graph the function.

6( ) ( 1)x x f


Solution The negative sign in – 2 causes the graph of the function to be reflected across the x-axis when compared with the graph of (x) = x3. Because – 2> 1, the graph is stretched vertically as compared to the graph of (x) = x3. It is also translated 1 unit to the right and 3 units up.

c.Graph the function.

3( ) 2( 1) 3x x f

Unless otherwise restricted, the domain of a polynomial function is the set of all real numbers. Polynomial functions are smooth, continuous curves on the interval (– , ). The range of a polynomial function of odd degree is also the set of all real numbers. Typical graphs of polynomial functions of odd degree are shown in next slide. These graphs suggest that for every polynomial function of odd degree there is at least one real value of x that makes (x) = 0 . The real zeros are the x-intercepts of the graph.

Odd Degree

A polynomial function of even degree has a range of the form (– , k] or [k, ) for some real number k. Here are two typical graphs of polynomial functions of even degree.

Even Degree

Recall that a zero c of a polynomial function has as multiplicity the exponent of the factor x – c. Determining the multiplicity of a zero aids in sketching the graph near that zero. If the zero has multiplicity one, the graph crosses the x-axis at the corresponding x-intercept as seen here.

If the zero has even multiplicity, the graph is tangent to the x-axis at the corresponding x-intercept (that is, it touches but does not cross the x-axis there).

If the zero has odd multiplicity greater than one, the graph crosses the x-axis and is tangent to the x-axis at the corresponding x-intercept. This causes a change in concavity, or shape, at the x-intercept and the graph wiggles there.

Turning Points and End Behavior

The previous graphs show that polynomial functions often have turning points where the function changes from increasing to decreasing or from decreasing to increasing.

Turning Points

A polynomial function of degree n has at most n – 1 turning points, with at least one turning point between each pair of successive zeros.

End Behavior

The end behavior of a polynomial graph is determined by the dominating term, that is, the term of greatest degree. A polynomial of the form

11 0( ) n n

n nx a x a x a f

has the same end behavior as . ( ) nnx a xf

End Behavior

For instance,3 2( ) 2 8 9x x x f

has the same end behavior as . It is large and positive for large positive values of x and large and negative for negative values of x with large absolute value.

3( ) 2x xf

End Behavior

The arrows at the ends of the graph look like those of the graph shown here; the right arrow points up and the left arrow points down.The graph shows that as x takes on larger and larger positive values, y does also. This is symbolized as , ,x y

read “as x approaches infinity, y approaches infinity.”

End Behavior

For the same graph, as x takes on negative values of larger and larger absolute value, y does also:

as , ,x y

End Behavior

For this graph, we have

as

and as

, ,x y

yx ,

End Behavior of Polynomials

Suppose that axn is the dominating term of a polynomial function of odd degree.1.If a > 0, then as and as Therefore, the end behavior of the graph is of the type that looks like the figure shown here.

We symbolize it as .

, ( ) ,x x f, ( ) .x x f


Suppose that axn is the dominating term of a polynomial function of odd degree.2. If a < 0, then as and as

Therefore, the end behavior of the graph looks like the graph shown here.


, ( ) ,x x f, ( ) .x x f


Suppose that axn is the dominating term of a polynomial function of even degree.1.If a > 0, then as Therefore, the end behavior of the graph looks like the graph shown here.


, ( ) .x x f


Suppose that is the dominating term of a polynomial function of even degree.2. If a < 0, then as Therefore, the end behavior of the graph looks like the graph shown here.


, ( ) .x x f

Example 3 DETERMINING END BEHAVIOR GIVEN THE DEFINING POLYNOMIAL

Match each function with its graph.4 2( ) 5 4x x x x f

Solution Because is of even degree with positive leading coefficient, its graph is C.

A. B. C. D.


Match each function with its graph.6 2( ) 3 4x x x x g

Solution Because g is of even degree with negative leading coefficient, its graph is A.

A. B. C. D.


Match each function with its graph.3 2( ) 3 2 4x x x x h

Solution Because function h has odd degree and the dominating term is positive, its graph is in B.

A. B. C. D.


Match each function with its graph.7( ) 4x x x k

Solution Because function k has odd degree and a negative dominating term, its graph is in D.

A. B. C. D.

Graphing Techniques

We have discussed several characteristics of the graphs of polynomial functions that are useful for graphing the function by hand. A comprehensive graph of a polynomial function will show the following characteristics:1. all x-intercepts (zeros)2. the y-intercept3. the sign of (x) within the intervals formed by the x-intercepts, and all turning points4. enough of the domain to show the end behavior.

In Example 4, we sketch the graph of a polynomial function by hand. While there are several ways to approach this, here are some guidelines.

Graphing a Polynomial Function

Let be a polynomial function of degree n. To sketch its graph, follow these steps.Step 1 Find the real zeros of . Plot them as x-intercepts.Step 2 Find (0). Plot this as the y-intercept.

11 1 0( ) , 0,n n

n n nx a x a x a x a a f


Step 3 Use test points within the intervals formed by the x-intercepts to determine the sign of (x) in the interval. This will determine whether the graph is above or below the x-axis in that interval.


Use end behavior, whether the graph crosses, bounces on, or wiggles through the x-axis at the x-intercepts, and selected points as necessary to complete the graph.

Synthetic DivisionSolve Equations of Higher Order Than Quadratics by Factoring

Use Synthetic Division to Divide Polynomials

Evaluate Polynomial Functions Using the Remainder Theorem (synthetic substitution)

Test Potential Zeros (Roots, Solutions)

Objectives


Solving Higher Degree Equations by Factoring

The equation x3 + 8 = 0 that follows is called a cubic equation because of the degree 3 term. Some higher-degree equations can be solved using factoring and/or the quadratic formula.

SOLVING A CUBIC EQUATION

Solve 3 8 0.x Solution 3 8 0x

22 2 4 0x x x Factor as a sum of cubes.

2 0x or 2 2 4 0x x Zero-factor property

2x or2( 2) ( 2) 4(1)(4)

2(1)x

Quadratic formula; a = 1, b = – 2, c = 4


Solve 3 8 0.x Solution

Simplify.2 122

x

2 2 32i

x Simplify the radical.

32

2

1 ix

Factor out 2 in the

numerator.


Solve 3 8 0.x Solution

Lowest terms1 3x i

The solution set is 2,1 3 .i

Solve 0482 23 xxx

Factor out common factor

0)482( 2 xxx

Factor completely if possible or use other methods.

0)6)(8( xxx

Set each factor equal to zero and solve


6

6066

06

8

8088

08

0

x

x

x

x

x

x

x

Solution set x={0,8,-6}


Division Algorithm

Let (x) and g(x) be polynomials with g(x) of lower degree than (x) and g(x) of degree one or more. There exists unique polynomials q(x) and r(x) such that

where either r(x) = 0 or the degree of r(x) is less than the degree of g(x).

,x x x x f g q r

If a polynomial is not easily factored than we must use division to determine the solutions

Synthetic Division

Synthetic division is a shortcut method of performing long division with polynomials.

It is used only when a polynomial is divided by a first-degree binomial of the form x – k, where the coefficient of x is 1.

Traditional Division

3 24 3 2 0 150x x x x 3 23 12x x

23x

210 0x x

10x

210 40x x40 150x

40

40 160x 10

Divide

by

23 10 4010

4x x

x

Quotient Remainder

Synthetic Division

4 3 2 0 150 12 40 160

103 10 40

23 10 4010

4x x

x

Quotient

Remainder

Caution To avoid errors, use 0 as the coefficient for any missing terms, including a missing constant, when setting up the division.

Example 1 USING SYNTHETIC DIVISION

Solution Express x + 2 in the form x – k by writing it as x – (– 2). Use this and the coefficients of the polynomial to obtain

Use synthetic division to divide 3 25 6 28 2

.2

x x xx

2 5 6 28 2. x + 2 leads to – 2

Coefficients


Solution Bring down the 5, and multiply: – 2(5) = – 10


.2

x x xx

2 25 6 28

510


Solution Add – 6 and – 10 to obtain – 16. Multiply – 2(– 16) = 32.


.2

x x xx

2 5 6 28 2

51016

32


Solution Add – 28 and 32 to obtain 4. Finally, – 2(4) = – 8.


.2

x x xx

2 5 6 28 2

51016

324

8

Add columns. Watch your

signs.


Solution Add – 2 and – 8 to obtain – 10.


.2

x x xx

2 5 6 28 2

51016

324

810 Remainder

Quotient


Since the divisor x – k has degree 1, the degree of the quotient will always be written one less than the degree of the polynomial to be divided. Thus,


.2

x x xx

22

3

51

16 45 6 28 2 0

2.

2x

xx

xx

xx

Remember to add remainder

.divisor

Special Case of the Division Algorithm

For any polynomial (x) and any complex number k, there exists a unique polynomial q(x) and number r such that

( ) ( ) ( ) .x x k q x r f

For Example

In the synthetic division in Example 1,

3 2 25 6 28 2 ( 2)(5 16 4) ( 10).x x x x x x

( )xf ( )x k ( )xq r

Here g(x) is the first-degree polynomial x – k.

Remainder Theorem

If the polynomial (x) is divided by x – k, the remainder is equal to (k).

Remainder Theorem

In Example 1, when (x) = 5x3 – 6x2 – 28x – 2 was divided by x + 2 or x –(– 2), the remainder was – 10. Substituting – 2 for x in (x) gives

3 2( ) 5( ) 6( ) 28(2 2)2 22 f

40 24 56 2 10 Use parentheses

around substituted values to avoid errors.

Remainder Theorem

A simpler way to find the value of a polynomial is often by using synthetic division. By the remainder theorem, instead of replacing x by – 2 to find (– 2), divide (x) by x + 2 using synthetic division as in Example 1. Then (– 2) is the remainder, – 10.

2 5 6 28 2

51016

324

810 (– 2)

Example 2 APPLYING THE REMAINDER THEOREM

Solution Use synthetic division with k = – 3.

Let (x) = – x4 + 3x2 – 4x – 5. Use the remainder theorem to find (– 3).

3 1 0 3 4 5 3 9 18 42

1 3 6 1 474 Remainder

By this result, (– 3) = – 47.

Testing Potential Zeros

A zero of a polynomial function is a number k such that (k) = 0. The real number zeros are the x-intercepts of the graph of the function.

The remainder theorem gives a quick way to decide if a number k is a zero of the polynomial function defined by (x). Use synthetic division to find (k); if the remainder is 0, then (k) = 0 and k is a zero of (x). A zero of (x) is called a root or solution of the equation (x) = 0.

Proposed zero

Example 3 DECIDING WHETHER A NUMBER IS A ZERO

Solution

a.

Decide whether the given number k is a zero of (x).

3 2( ) 4 9 6; 1x x x x k f

1 1 4 9 6 1 3 6

1 3 6 0

3 2( ) 4 9 6x x x x f

Remainder

Since the remainder is 0, (1) = 0, and 1 is a zero of the polynomial function defined by (x) = x3 – 4x2 + 9x – 6. An x-intercept of the graph (x) is 1, so the graph includes the point (1, 0).

Proposed zero


Solution Remember to use 0 as coefficient for the missing x3-term in the synthetic division.

b.


4 2( ) 3 1; 4x x x x k f

4 1 30 1 1 4 16 68 284

Remainder1 2854 17 71 The remainder is not 0, so – 4 is not a zero of (x) = x4 +x2 – 3x + 1. In fact, (– 4) = 285, indicating that (– 4, 285) is on the graph of (x).


Solution Use synthetic division and operations with complex numbers to determine whether 1 + 2i is a zero of (x) = x4 – 2x3 + 4x2 + 2x – 5.

c.


4 3 2( ) 2 4 2 5; 1 2x x x x x k i f

1 2 1 2 4 2 5i 1 2 5 1 2 5i i

Remainder

1 1 2 01 2 1i i 2

(1 2 )( 1 2 )

1 4

5

i i

i


Since the remainder is 0, 1 + 2i is a zero of the given polynomial function. Notice that 1 + 2i is not a real number zero. Therefore, it would not appear as an x-intercept in the graph of (x).

c.


4 3 2( ) 2 4 2 5; 1 2x x x x x k i f

1 2 1 2 4 2 5i 1 2 5 1 2 5i i

Remainder

1 1 2 01 2 1i i 2

(1 2 )( 1 2 )

1 4

5

i i

i

Zero of Polynomial Functions

Use the Factor Theorem to find Solutions

Use Rational Zeros Theorem to find Possible Solutions

Use Number of Zeros Theorum to determine the Maximum Number of Zeros

Use Conjugate Zeros Theorem to find additional solutions

Find All Zeros(solutions) of a Polynomial Function

Descartes’ Rule of Signs

Objectives


Factor Theorem

The polynomial x – k is a factor of the polynomial (x) if and only if (k) = 0.

Example 1 DECIDING WHETHER x – k IS A FACTOR OF (x)

Solution By the factor theorem, x – 1 will be a factor of (x) if and only if (1) = 0. Use synthetic division and the remainder theorem to decide.

a.Determine whether x – 1 is a factor of (x).

4 2( ) 2 3 5 7x x x x f

1 2 0 3 5 72 2 5 0

2 2 5 0 7Use a zero

coefficient for the missing

term.

(1) = 7

Since the remainder is 7 and not 0, x – 1 is not a factor of (x).


Solution

b.Determine whether x – 1 is a factor of (x).

5 4 3 2( ) 3 2 8 5 1x x x x x x f

1 3 2 1 8 5 1 3 1 2 6 1

3 1 2 6 1 0 (1) = 0

Because the remainder is 0, x – 1 is a factor. Additionally, we can determine from the coefficients in the bottom row that the other factor is


Solution

b.Determine whether x – k is a factor of (x).

5 4 3 2( ) 3 2 8 5 1x x x x x x f

1 3 2 1 8 5 1 3 1 2 6 1

3 1 2 6 1 0 (1) = 0

4 3 23 2 6 1.x x x x

Thus, 4 3 2( ) ( 1)(3 2 6 1).x x x x x x f

Example 2 FACTORING A POLYNOMIAL GIVEN A ZERO

Solution Since – 3 is a zero of , x – (– 3) = x + 3 is a factor.

Factor the following into linear factors if – 3 is a zero of . 3 2( ) 6 19 2 3x x x x f

3 6 19 2 3 Use synthetic division to divide (x) by x + 3.18 3 3

6 1 01

The quotient is 6x2 + x – 1.


Solution x – (– 3) = x + 3 is a factor.

Factor the following into linear factors if – 3 is a zero of . 3 2( ) 6 19 2 3x x x x f

Factor 6x2 + x – 1.

2( ) ( 3)(6 1)x x x x f

( ) ( 3)(2 1)(3 1).x x x x f

The quotient is 6x2 + x – 1, so

These factors are all linear.


Solve: ( ) ( 3)(2 1)(3 1).x x x x f

)13)(12)(3(0 xxx

Set:

013

012

03

x

x

x

3/1

2/1

3

x

x

x

Rational Zeros Theorem

If is a rational number written in lowest terms, and if is a zero of , a polynomial function with integer coefficients, then p is a factor of the constant term and q is a factor of the leading coefficient.

pq p

q

If no zeros are given than must use trial and error. The following theorum helps limit the possibilities.

Example 3 USING THE RATIONAL ZERO THEOREM

Solution For a rational number to be zero, p must be a factor of a0 = 2 and q must be a factor of a4 = 6. Thus, p can be 1 or 2, and q can be 1, 2, 3, or 6. The possible rational zeros, are,

a. List all possible rational zeros.

Do the following for the polynomial function defined by 4 3 2( ) 6 7 12 3 2.x x x x x f

pq

pq

1 1 1 21, 2, , , , .

2 3 6 3

qpossible

ppossible


Solution Use the remainder theorem to show that 1 is a zero.

b. Find all rational zeros and factor (x) into linear factors.


1 6 7 12 3 2 6 13 1 2

6 13 1 02

Use “trial and error” to find

zeros.(1) = 0

The 0 remainder shows that 1 is a zero. The quotient is 6x3 +13x2 + x – 4, so (x) = (x – 1)(6x3 +13x2 + x – 2).


Solution Now, use the quotient polynomial and synthetic division to find that – 2 is a zero.

b. Find all rational zeros and factor (x) into linear equations.


2 6 13 1 2 12 2 2

6 1 01 (– 2 ) = 0

The new quotient polynomial is 6x2 + x – 1. Therefore, (x) can now be factored.


Solution


Do the following for the polynomial function defined by

2( ) ( 1)( 2)(6 1)x x x x x f

(3 1)(2 1)( 1)( 2) .x xx x

4 3 2( ) 6 7 12 3 2.x x x x x f


Solution Setting 3x – 1 = 0 and 2x + 1 = 0 yields the zeros ⅓ and – ½. In summary the rational zeros are 1, – 2, ⅓, – ½, and the linear factorization of (x) is



4 3 2( ) 6 7 12 3 2x x x x x f

( 1)( 2)(3 1)(2 1).x x x x Check by

multiplying these factors.

Note In Example 3, once we obtained the quadratic factor of 6x2 + x – 1, we were able to complete the work by factoring it directly. Had it not been easily factorable, we could have used the quadratic formula to find the other two zeros (and factors).

Caution The rational zeros theorem gives only possible rational zeros; it does not tell us whether these rational numbers are actual zeros. We must rely on other methods to determine whether or not they are indeed zeros. Furthermore, the function must have integer coefficients. To apply the rational zeros theorem to a polynomial with fractional coefficients, multiply through by the least common denominator of all fractions. For example, any rational zeros of p(x) defined below will also be rational zeros of q(x).

4 3 21 2 1 1( )

6 3 6 3p x x x x x

4 3 2( ) 6 4 2q x x x x x Multiply the terms of p(x) by 6.

Fundamental Theorem of AlgebraEvery function defined by a polynomial of degree 1 or more has at least one complex zero.

Number of Zeros Theorem

A function defined by a polynomial of degree n has at most n distinct zeros.

Example 4 FINDING A POLYNOMIAL FUNCTION THAT SATISFIES GIVEN CONDITIONS (REAL ZEROS)

Solution These three zeros give x – (– 1) = x + 1, x – 2, and x – 4 as factors of (x). Since (x) is to be of degree 3, these are the only possible factors by the number of zeros theorem. Therefore, (x) has the form

for some real number a.

a. Zeros of – 1, 2, and 4; (1) = 3

Find a function defined by a polynomial of degree 3 that satisfies the given conditions.

( ) ( 1)( 2)( 4)x a x x x f


Solution To find a, use the fact that (1) = 3.

a. Zeros of – 1, 2, and 4; (1) = 3


( ) ( 1)(1 1 1 ) 4)12 (a f Let x = 1.

(2)( ( 3)3 1)a (1) = 3

3 6a12

a

Solve for a.


Solution Thus,

a. Zeros of – 1, 2, and 4; (1) = 3


Multiply.

( ) ( 1)( 2)( 4),12

x x x x f

or 3 21 5( ) 4.

2 2x x x x f

Note In Example 4a, we cannot clear the denominators in (x) by multiplying both sides by 2 because the result would equal 2 • (x), not (x).


Solution The polynomial function defined by (x) has the form

b. – 2 is a zero of multiplicity 3; (– 1) = 4


( ) ( 2)( 2)( 2)x a x x x f

3( 2) .a x


Solution Since (– 1) = 4,

b. – 2 is a zero of multiplicity 3; (– 1) = 4


3( 1) ( 1 2)a f34 (1)a

4,a and 3 3 2( ) 4( 2) 4 24 48 32.x x x x x f

Remember: (x + 2)3 ≠ x3 + 23

Conjugate Zeros Theorem

If (x) defines a polynomial function having only real coefficients and if z = a + bi is a zero of (x), where a and b are real numbers, Then its conjugate

is also a zero of ( ).z a bi x f

Example 5 FINDING A POLYNOMIAL FUNCTION THAT SATISFIES GIVEN CONDITIONS

(COMPLEX ZEROS)

Solution The complex number 2 – i must also be a zero, so the polynomial has at least three zeros, 3, 2 + i, and 2 – i. For the polynomial to be of least degree, these must be the only zeros. By the factor theorem there must be three factors, x – 3, x – (2 + i), and x – (2 – i), so

Find a polynomial function of least degree having only real coefficients and zeros 3 and 2 + i.

Example 5 FINDING A POLYNOMIAL FINCTION THAT SATISFIES GIVEN CONDITIONS

(COMPLEX ZEROS)

Solution


( ) ( 3) (2 ) (2 )x x x i x i f

( 3 2 ( ))( ) 2x i x ix 2( 5)3)( 4x xx

Remember:i2 = – 1

3 27 17 15.x x x

Example 5 FINDING A POLYNOMIAL FINCTION THAT SATISFIES GIVEN CONDITIONS

(COMPLEX ZEROS)

Solution Any nonzero multiple of x3 – 7x2 + 17x – 15 also satisfies the given

conditions on zeros. The information on zeros given in the problem is not enough to give a specific value for the leading coefficient.


Example 6FINDING ALL ZEROS OF A POLYNOMIAL FUNCTION GIVEN ONE ZERO

Solution Since the polynomial function has only real coefficients and since 1 – i is a zero, by the conjugate zeros theorem 1 + i is also a zero. To find the remaining zeros, first use synthetic division to divide the original polynomial by x – (1 – i).

Find all zeros of (x) = x4 – 7x3 + 18x2 – 22x + 12,

given that 1 – i is a zero.


Solution

1 1 7 18 22 12i 1 7 5 16 6 12i i i

1 6 11 5 6 6 0i i i




Solution By the factor theorem, since x = 1 – i is a zero of (x), x – (1 – i) is a factor, and (x) can be written as

3 2( ) (1 ) ( 6 ) (11 5 ) ( 6 6 ) .x x i x i x i x i f

We know that x = 1 + i is also a zero of (x), so

( ) (1 ) (1 ) ( ),x x i x i q x ffor some polynomial q(x).




Solution Thus,

3 2( 6 ) (11 5 ) ( 6 6 ) (1 ) ( ).x i x i x i x i q x

Use synthetic division to find q(x).

1 6 11 51 6 6i i ii 1 5 5 6 6i i i

1 5 6 0




Solution Since q(x) = x2 – 5x + 6, (x) can

be written as

Factoring x2 – 5x + 6 as (x – 2)(x – 3), we see that the remaining zeros are 2 and 3. The four zeros of (x) are 1 – i, 1 + i, 2, and 3.

2( ) (1 ) (1 ) ( 5 6).x x i x i x x f



Descartes’ Rule of Signs

Let (x) define a polynomial function with real coefficients and a nonzero constant term, with terms in descending powers of x.a.The number of positive real zeros of either equals the number of variations in sign occurring in the coefficients of (x), or is less than the number of variations by a positive even integer.b.The number of negative real zeros of either equals the number of variations in sign occurring in the coefficients of (– x), or is less than the number of variations by a positive even integer.

Example 7 APPLYING DESCARTES’ RULE OF SIGNS

Solution We first consider the possible number of positive zeros by observing that (x) has three variations in signs:

Determine the possible number of positive real zeros and negative real zeros of

4 3 2( ) 6 8 2 1.x x x x x f

4 3 26 8 2 1x x x x

1 2 3


Solution Thus, by Descartes’ rule of signs, has either 3 or 3 – 2 = 1 positive real zeros.For negative zeros, consider the variations in signs for (– x):


4 3 2( ) 6 8 2 1.x x x x x f

4 3 2( ) ( ) 6( ) 8( ) 2( ) 1x x x x x f4 3 26 8 2 1.x x x x

1


Solution


4 3 2( ) 6 8 2 1.x x x x x f

4 3 26 8 2 1.x x x x

1

Since there is only one variation in sign, (x) has only 1 negative real zero.

Example 4 GRAPHING A POLYNOMIAL FUNCTION

Graph 3( ) 2 5 6.x x x x f

Solution Step 1 The possible rational zeros are 1, 2, 3, 6, ½, and 3/2. Use synthetic division to show that 1 is a zero.

1 2 5 1 6 2 7 6

2 7 6 0 (1) = 0


Graph 3( ) 2 5 6.x x x x f

Solution Step 1 Thus,

Factor 2x2 + 7x + 6.

2( ) ( 1)(2 7 6)x x x x f

( 1)(2 3)( 2).x x x

Set each linear factor equal to 0, then solve for x to find real zeros. The three real zeros of are 1, – 3/2, and – 2.


Graph 3( ) 2 5 6.x x x x f

Solution Step 2 (0) 6, so plot (0, 6). f


Graph 3( ) 2 5 6.x x x x f

Solution Step 3

The x-intercepts divide the x-axis into four intervals: (– , – 2), (– 2, – 3/2), (– 3/2, 1), and (1, ). Because the graph of a polynomial function has no breaks, gaps, or sudden jumps, the values of (x) are either always positive or always negative in any given interval.


Graph 3( ) 2 5 6.x x x x f

Solution Step 3

To find the sign of (x) in each interval, select an x-value in each interval and substitute it into the equation for (x) to determine whether the values of the function are positive or negative in that interval.


Graph 3( ) 2 5 6.x x x x f

Solution Step 3

When the values of the function are negative, the graph is below the x-axis, and when (x) has positive values, the graph is above the x-axis.


Graph 3( ) 2 5 6.x x x x f

Solution Step 3

A typical selection of test points and the results of the tests are shown in the table on the next slide. (As a bonus, this procedure also locates points that lie on the graph.)


Graph 3( ) 2 5 6.x x x x f

Solution Step 3

IntervalTest Point

Value of (x) Sign of (x)

Graph Above or Below x-

Axis

(– , – 2) – 3 – 12 Negative Below

(– 2, – 3/2) – 7/4 11/32 Positive Above

(– 3/2, 1) 0 – 6 Negative Below

(1, ) 2 28 Positive Above


Graph 3( ) 2 5 6.x x x x f

Solution Step 3

Plot the test points and join the x-intercepts, y-intercept, and test points with a smooth curve to get the graph.


Graph 3( ) 2 5 6.x x x x f

Solution Step 3

Because each zero has odd multiplicity (1), the graph crosses the x-axis each time. The graph has two turning points, the maximum number for a third-degree polynomialfunction.


Graph 3( ) 2 5 6.x x x x f

Solution Step 3

The sketch could be improved by plotting the points found in each interval in the table. Notice that the left arrow points down and the right arrow points up. This end behavior is correct since the dominating term of the polynomial is 2x3.

Graphing Polynomial Functions

Note If a polynomial function is given in factored form, such as

Step 1 of the guidelines is easier to perform, since real zeros can be determined by inspection. For this function, we see that 1 and 3 are zeros of multiplicity1, and – 2 is a zero of multiplicity 2.

2( ) ( 1)( 3)( 2) ,x x x x f

Note Since the dominating term is the end behavior of the graph is .

The y-intercept is


2 4( )( ) ,x x x x

2(0) 1( 3)(2) 12. f

Note The graph intersects the x-axis at 1 and 3 but bounces at – 2.

This information is sufficient to quickly sketch the graph of (x).


Important Relationships

We emphasize the important relationships among the following concepts.1. the x-intercepts of the graph of y = (x)2. the zeros of the function 3. the solutions of the equation (x) = 04. the factors of (x)

x-Intercepts, Zeros, Solutions, and Factors

If a is an x-intercept of the graph of then a is a zero of , a is asolution of (x) = 0, and x – a is a factor of (x).

( ),y x f

Intermediate Value Theorem for Polynomials

If (x) defines a polynomial function with only real coefficients, and if forreal numbers a and b, the values (a) and (b) are opposite in sign, then there exists at least one real zero between a and b.

Example 5 LOCATING A ZERO

Use synthetic division and a graph to show that (x) = x3 – 2x2 – x + 1 has a real zero between 2 and 3.

2 1 2 1 1 2 0 2

1 ( )1 21 0 f

3 1 2 1 1 3 3 6

31 2 71 ( ) f

Solution Since (2) is negative and (3) is positive, by the intermediate value theorem there must be a real zero between 2 and 3.

Caution Be careful how you interpret the intermediate value theorem.If (a) and (b) are not opposite in sign, it does not necessarily mean that there is no zero between a and b.

In the graph shown here, for example, (a) and (b) are both negative, but – 3 and – 1, which are between a and b, are zeros of (x).

Example 7 APPROXIMATING REAL ZEROS OF A POLYNOMIAL FUNCTION

Approximate the real zeros of (x) = x4 – 6x3 + 8x2 + 2x – 1.

Solution The greatest degree term is x4, so the graph will have end behavior similar to the graph of (x) = x4, which is positive for all values of x with large absolute values. That is, the end behavior is up at the left and the right, .There are at most four real zeros, since the polynomial is fourth-degree.



Solution Since (0) = – 1, the y-intercept is – 1. Because the end behavior is positiveon the left and the right, by the intermediate value theorem has at least one zero on either side of x = 0. To approximate the zeros, we use a graphing calculator.



Solution The graph below shows that there are four real zeros, and the table indicates that they are between – 1 and 0, 0 and 1, 2 and 3, and 3 and 4 because there is a sign change in (x) in each case.


Solution Using the capability of the calculator, we can find the zeros to a great degree of accuracy. The graph shown here shows that the negative zero is approximately – .4142136.Similarly, we find that the other three zeros are approximately

.26794919, 2.4142136, and 3.7320508.

Example 8 EXAMINING A POLYNOMIAL MODEL FOR DEBIT CARD USE

The table shows the number of transactions, in millions, by users of bank debit cards for selected years.

Year Transactions (in millions)

1995 829

1998 3765

2000 6797

2004 14,106

2009 22,120



1995 829

1998 3765

2000 6797

2004 14,106

2009 22,120

a. Using x = 0 to represent 1995, x = 3 to represent 1998, and so on, use the regression feature of a calculator to determine the quadratic function that best fits the data. Plot the data and graph.



1995 829

1998 3765

2000 6797

2004 14,106

2009 22,120

a. The best-fitting quadratic function for the data is defined by

225.53 120 453.1y x x

Solution



1995 829

1998 3765

2000 6797

2004 14,106

2009 22,120

a. The regression coordinates screen is shown below.

Solution



1995 829

1998 3765

2000 6797

2004 14,106

2009 22,120

a. The graph is shown below.

Solution



1995 829

1998 3765

2000 6797

2004 14,106

2009 22,120

b. Repeat part (a) for a cubic function.



1995 829

1998 3765

2000 6797

2004 14,106

2009 22,120

b. The best-fitting cubic function is shown below and is defined by

3 26.735 164.1 543.1 831.0y x x x

Solution



1995 829

1998 3765

2000 6797

2004 14,106

2009 22,120

b. The graph of the best-fitting cubic function is shown below.

Solution



1995 829

1998 3765

2000 6797

2004 14,106

2009 22,120

c. Repeat part (a) for a quartic function.



1995 829

1998 3765

2000 6797

2004 14,106

2009 22,120

c. The best-fitting quartic function is defined by

Solution

4 3 2.0576 5.198 151.9 571.4 829y x x x x



1995 829

1998 3765

2000 6797

2004 14,106

2009 22,120

c. The graph of the best-fitting quartic function is shown below.

Solution



1995 829

1998 3765

2000 6797

2004 14,106

2009 22,120

d. The correlation coefficient, R, is a measure of the strength of the relationship between two variables. The values of R and R2 are used to determine how well a regression model fits a set of data. The closer the value of R2 is to 1, the better the fit. Compare R2 for the three functions to decide which function fits the data.


d. Find the correlation values R2. See the graph for the quadratic function. The others are .999999265 for the cubic function and 1 for the quartic function. Therefore, the quartic function provides the best fit.

Solution

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