366 CHAPTER 3 Polynomial and Rational Functions x 1 3 –2 0 –6 –12 12 6 y f(x) = –(x – 1)(x – 3)(x + 2)2 Figure 32 EXAMPLE 4 Graphing a Polynomial Function Graph ƒ1x2 = -1x - 121x - 321x + 222. SOLUTION Step 1 Because the polynomial is given in factored form, the zeros can be determined by inspection. They are 1, 3, and -2. Plot the corresponding x-intercepts of the graph of ƒ1x2. See Figure 32. Step 2 ƒ102 = -10 - 1210 - 3210 + 222 Find ƒ102. ƒ102 = -1-121-321222 Simplify in parentheses. ƒ102 = -12 The y-intercept is 10, -122. Plot the y-intercept 10, -122. See Figure 32. Step 3 The dominating term of ƒ1x2 can be found by multiplying the factors and identifying the term of greatest degree. Here it is -1x21x21x22 = -x4, indicating that the end behavior of the graph is . Because 1 and 3 are zeros of multiplicity 1, the graph will cross the x-axis at these zeros. The graph of ƒ1x2 will touch the x-axis at -2 and then turn and change direction because it is a zero of even multiplicity. Begin at either end of the graph with the appropriate end behavior and draw a smooth curve that crosses the x-axis at 1 and 3 and that touches the x-axis at -2, then turns and changes direction. The graph will also pass through the y-intercept 10, -122. See Figure 32. Using test values within intervals formed by the x-intercepts is a good way to add detail to the graph and verify the accuracy of the sketch. A typical selection of test points is 1-3, -242, 1-1, -82, 12, 162, and 14, -1082. S Now Try Exercise 33. NOTE It is possible to reverse the process of Example 4 and write the polynomial function from its graph if the zeros and any other point on the graph are known. Suppose that we are asked to find a polynomial function of least degree having the graph shown in Figure 32. Because the graph crosses the x-axis at 1 and 3 and bounces at -2, we know that the factored form of the function is as follows. ƒ1x2 = a1x - 1211x - 3211x + 222 Now find the value of a by substituting the x- and y-values of any other point on the graph, say 10, -122, into this function and solving for a. ƒ1x2 = a1x - 121x - 321x + 222 -12 = a10 - 1210 - 3210 + 222 Let x = 0 and y = -12. -12 = a1122 Simplify. a = -1 Divide by 12. Interchange sides. Verify in Example 4 that the polynomial function is ƒ1x2 = -1x - 121x - 321x + 222. Exercises of this type are labeled Connecting Graphs with Equations. Multiplicity one Multiplicity two
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