section 4.2 graphing polynomial functions copyright 2013, 2009, 2006, 2001 pearson education, inc

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Graphing Polynomial Functions If P(x) is a polynomial function of degree n, the graph of the function has:  at most n real zeros, and thus at most n x- intercepts;  at most n  1 turning points. (Turning points on a graph, also called relative maxima and minima, occur when the function changes from decreasing to increasing or from increasing to decreasing.)

TRANSCRIPT

Section 4.2

Graphing Polynomial Functions

Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc.

Objectives

Graph polynomial functions. Use the intermediate value theorem to determine whether a function has a real zero between two given real numbers.

Graphing Polynomial Functions

If P(x) is a polynomial function of degree n, the graph of the function has:

at most n real zeros, and thus at most n x- intercepts; at most n 1 turning points.

(Turning points on a graph, also called relative maxima and minima, occur when the function changes from decreasing to increasing or from increasing to decreasing.)

To Graph a Polynomial Function 1. Use the leading-term test to determine the end behavior.2. Find the zeros of the function by solving f (x) = 0. Any real

zeros are the first coordinates of the x-intercepts.3. Use the x-intercepts (zeros) to divide the x-axis into intervals

and choose a test point in each interval to determine the sign of all function values in that interval.

4. Find f (0). This gives the y-intercept of the function.5. If necessary, find additional function values to determine the

general shape of the graph and then draw the graph.6. As a partial check, use the facts that the graph has at most n

x-intercepts and at most n 1 turning points. Multiplicity of zeros can also be considered in order to check where the graph crosses or is tangent to the x-axis. We can also check the graph with a graphing calculator.

Example

Graph the polynomial function f (x) = 2x3 + x2 8x 4.

Solution:1. The leading term is 2x3. The degree, 3, is odd, the coefficient, 2, is positive. Thus the end behavior of the graph will appear as:

2. To find the zero, we solve f (x) = 0. Here we can use factoring by grouping.

Example continued

Factor:

The zeros are 1/2, 2, and 2. The x-intercepts are (2, 0), (1/2, 0), and (2, 0).

3. The zeros divide the x-axis into four intervals:(, 2), (2, 1/2), (1/2, 2), and (2, ).We choose a test value for x from each interval and find f(x).

3 2

2

2

2 8 4 0

(2 1) 4(2 1) 0

(2 1)( 4) 0(2 1)( 2)( 2) 0

x x x

x x x

x xx x x

Example continued

4. To determine the y-intercept, we find f(0):

The y-intercept is (0, 4).

3 2

3 2

( ) 2 8 4

( ) 2( ) 8( )0 0 0 0 4 4

f x x x x

f

Example continued

5. We find a few additional points and complete the graph.

6. The degree of f is 3. The graph of f can have at most 3 x-intercepts and at most 2 turning points. It has 3 x-intercepts and 2 turning points. Each zero has a multiplicity of 1; thus the graph crosses the x-axis at 2, 1/2, and 2. The graph has the end behavior described in step (1). The graph appears to be correct.

Intermediate Value Theorem

For any polynomial function P(x) with real coefficients, suppose that for a b, P(a) and P(b) are of opposite signs. Then the function has a real zero between a and b.

Example

Using the intermediate value theorem, determine, if possible, whether the function has a real zero between a and b. a) f(x) = x3 + x2 8x; a = 4 b = 1 b) f(x) = x3 + x2 8x; a = 1 b = 3

Solution

We find f(a) and f(b) and determine where they differ in sign. The graph of f(x) provides a visual check.

f(4) = (4)3 + (4)2 8(4) = 16

f(1) = (1)3 + (1)2 8(1) = 8

By the intermediate value theorem, since f(4) and f(1) have opposite signs, then f(x) has a zero between 4 and 1.

zero

y = x3 + x2 8x

Solution

f(1) = (1)3 + (1)2 8(1) = 6

f(3) = (3)3 + (3)2 8(3) = 12

By the intermediate value theorem, since f(1) and f(3) have opposite signs, then f(x) has a zero between 1 and 3.

zero

y = x3 + x2 8x

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