1. Chapter 4 Polynomial (Poly) & Rational Functions Copyright ©2013, 2009, 2006, 2005 Pearson Education, Inc.

3. Sec 4.2 Graphing Polynomial Functions

4. If P(x) is a polynomial function of degree “n”, the graph of the function has:  at most n real zeros, (x-intercepts);  at most n  1 turning points (extrema). (Turning points on a graph are also called maxima/minima, and occur when the function changes from decreasing to increasing slope or vice versa)

5. 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.

6. 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.

7. 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).

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

9. 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.

10. Intermediate Value Theorem (IVT) 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 at least one real zero between x=a and x=b.

11. Example Using the Intermediate Value Theorem, determine whether the function shown has a real zero between x=a and x=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 After you have tried this manually, try it again using the Calc: Value fn on the calculator.

12. Solution: Part a) We find f(a) and f(b) and determine where they differ in sign at x=a and x=b. 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 at least one zero between x=  4 and x=  1. zero y = x 3 + x 2  8x

13. Solution: Part b) 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