328 CHAPTER 3 Polynomial and Rational Functions The x-intercepts are found by setting ƒ1x2 equal to 0 and solving for x. 0 = -3x2 - 2x + 1 Set ƒ1x2 = 0. 0 = 3x2 + 2x - 1 Multiply by -1. 0 = 13x - 121x + 12 Factor. x = 1 3 or x = -1 Zero-factor property Therefore, the x-intercepts are A1 3, 0B and 1-1, 02. The graph is shown in Figure 10. f(x) = −3x2 − 2x + 1 f(x) = −3(x + ) 2 + −4.1 −6.6 4.1 6.6 1 3 4 3 This screen gives the vertex of the graph in Figure 10 as the point A - 1 3 , 4 3B. (The display shows decimal approximations.) We want the highest point on the graph, so we direct the calculator to find the maximum. NOTE It is possible to reverse the process of Example 3 and write the quadratic function from its graph in Figure 10 if the vertex and any other point on the graph are known. Because quadratic functions take the form ƒ1x2 = a1x - h22 + k, we can substitute the x- and y-values of the vertex, A - 1 3 , 4 3B, for h and k. ƒ1x2 = ac x - a- 1 3b d 2 + 4 3 Let h = - 1 3 and k = 4 3 . ƒ1x2 = aax + 1 3b 2 + 4 3 Simplify. We find the value of a by substituting the x- and y-coordinates of any other point on the graph, say 10, 12, into this function and solving for a. 1 = aa0 + 1 3b 2 + 4 3 Let x = 0 and y = 1. 1 = aa 1 9b + 4 3 Square. - 1 3 = 1 9 a Subtract 4 3. a = -3 Multiply by 9. Interchange sides. Verify in Example 3 that the vertex form of the quadratic function is ƒ1x2 = -3 ax + 1 3b 2 + 4 3 . Exercises of this type are labeled Connecting Graphs with Equations. x 1 2 –1 ( ) – , y 1 3 4 3 f(x) = –3x2 – 2x + 1 f(x) = –3(x + ) 2 + 1 3 4 3 0 2 1 x = – 1 3 1 3 (0, 1) (–1, 0) ( , 0) Figure 10 S Now Try Exercise 33.
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