388 CHAPTER 3 Polynomial and Rational Functions EXAMPLE 5 Graphing a Rational Function (x-Axis as Horizontal Asymptote) Graph ƒ1x2 = x + 1 2x2 + 5x - 3 . SOLUTION Steps 1 In Example 4(a), we found that 2x2 + 5x - 3 = 12x - 121x + 32, so the vertical asymptotes have equations x = 1 2 and x = -3, and the horizontal asymptote is the x-axis. Step 3 The y-intercept is A0, -1 3B, as justified below. ƒ102 = 0 + 1 21022 + 5102 - 3 = - 1 3 Step 4 The x-intercept is found by solving ƒ1x2 = 0. x + 1 2x2 + 5x - 3 = 0 Set ƒ1x2 = 0. x + 1 = 0 x = -1 The x-intercept is 1-1, 02. Step 5 To determine whether the graph intersects its horizontal asymptote, solve this equation. ƒ1x2 = 0 y-value of horizontal asymptote The horizontal asymptote is the x-axis, so the solution of ƒ1x2 = 0 was found in Step 4. The graph intersects its horizontal asymptote at 1-1, 02. Step 6 Plot a point in each of the intervals determined by the x-intercepts and vertical asymptotes, 1-∞, -32, 1-3, -12, A -1, 1 2B, and A 1 2, ∞B, to get an idea of how the graph behaves in each interval. The y-intercept corresponds to the ratio of the constant terms. If a rational expression is equal to 0, then its numerator must equal 0. Step 7 Complete the sketch. See Figure 52. This function is decreasing on each interval of its domain—that is, on 1-∞, -32, A -3, 1 2B, and A 1 2, ∞B. S Now Try Exercise 67. Interval Test Value x Value of ƒ1x2 Sign of ƒ1x2 Graph Above or Below x-Axis 1-∞, -32 -4 -1 3 Negative Below 1-3, -12 -2 1 5 Positive Above A -1, 1 2B 0 - 1 3 Negative Below A 1 2, ∞B 2 1 5 Positive Above y 1 –2 –5 –4 2 2 3 3 4 –2 –3 –4 x x = –3 0 x = f(x) = x + 1 2x2 + 5x – 3 ( ) ( ) –4, – 1 3 1 3 1 2 ( ) –2, 1 5 ( ) 2, 1 5 0, – Figure 52 and 2 EXAMPLE 6 Graphing a Rational Function (Does Not Intersect Its Horizontal Asymptote) Graph ƒ1x2 = 2x + 1 x - 3 . SOLUTION Steps 1 As determined in Example 4(b), the equation of the vertical asymptote is x = 3. The horizontal asymptote has equation y = 2. and 2
RkJQdWJsaXNoZXIy NjM5ODQ=