Presentation is loading. Please wait.

Presentation is loading. Please wait.

Section 4.2 Graphing Polynomial Functions Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc.

Similar presentations


Presentation on theme: "Section 4.2 Graphing Polynomial Functions Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc."— Presentation transcript:

1 Section 4.2 Graphing Polynomial Functions Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc.

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

3 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.)

4 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.

5 Example Graph the polynomial function f (x) = 2x 3 + x 2  8x  4. Solution: 1. The leading term is 2x 3. 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.

6 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).

7 Example continued 4. To determine the y-intercept, we find f(0): The y-intercept is (0,  4).

8 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.

9 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.

10 Example Using the intermediate value theorem, determine, if possible, whether the function has a real zero between a and b. a) f(x) = x 3 + x 2  8x; a =  4 b =  1 b) f(x) = x 3 + x 2  8x; a = 1 b = 3

11 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 = x 3 + x 2  8x

12 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 = x 3 + x 2  8x


Download ppt "Section 4.2 Graphing Polynomial Functions Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc."

Similar presentations


Ads by Google