407 3.6 Polynomial and Rational Inequalities Step 2 Let ƒ1x2 = x - 5 1x + 321x - 12 and graph. The vertical asymptotes have equations x = -3 and x = 1. The horizontal asymptote has equation y = 0. The y-intercept is A0, 5 3B, and the x-intercept is 15, 02. The graph intersects its horizontal asymptote at 15, 02. Additional points may be used as necessary to sketch the graph in Figure 66. Step 3 Referring to Figure 66, ƒ1x2 60 —that is, the graph lies below the x-axis for solution set 1-∞, -32 ´ 11, 52. Because the inequality is strict, the zero of ƒ1x2 is not included in the solution set. S Now Try Exercise 63. –3 1 5 0 x y x = –3 x = 1 f(x) = x – 5 (x + 3)(x – 1) (5, 0) Figure 66 −8 −4 8 (a) (b) Figure 67 Using the graph in Figure 67(a), we can quickly determine the following. ƒ1x2 = 0 for solution set 556. ƒ1x2 70 for solution set 1-3, 12 ´ 15, ∞2. ƒ1x2 60 for solution set 1-∞, -32 ´ 11, 52. See Example 5. 7 3.6 Exercises CONCEPT PREVIEW Fill in the blank to correctly complete each sentence. 1. The real solutions of ƒ1x2 60 are the x-values for which the graph lies the x-axis. 2. The real solutions of ƒ1x2 70 are the x-values for which the graph lies the x-axis. CONCEPT PREVIEW Use the graph to solve each equation or inequality. Use interval notation where appropriate. 3. 7x1x - 121x - 22 = 0 4. 7x1x - 121x - 22 60 5. 7x1x - 121x - 22 70 6. 7x1x - 121x - 22 Ú 0 x y 1 2 –2 3 0 f(x) = 7x(x – 1)(x – 2) A TI-84 Plus calculator can be used to quickly sketch the rational function from Example 5 and determine that its zero is 5. See the graph of y1 = ƒ1x2 = x - 5 1x + 321x - 12 in Figure 67(a). The table in Figure 67(b) confirms that the vertical asymptotes have equations x = -3 and x = 1.
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