389 3.5 Rational Functions: Graphs, Applications, and Models Step 3 ƒ102 = - 1 3 , so the y-intercept is A0, - 1 3B. Step 4 Solve ƒ1x2 = 0 to find any x-intercepts. 2x + 1 x - 3 = 0 Set ƒ1x2 = 0. 2x + 1 = 0 x = - 1 2 The x-intercept is A - 1 2 , 0B. Step 5 The graph does not intersect its horizontal asymptote because ƒ1x2 = 2 has no solution. 2x + 1 x - 3 = 2 Set ƒ1x2 = 2. 2x + 1 = 2x - 6 Multiply each side by x - 3. 1 = -6 Subtract 2x. –8–6–4–2 2 4 6 8 –8 –6 –4 4 6 8 x y x = 3 y = 2 (–4, 1) (6, ) f(x) = 2x + 1 x – 3 (1, – ) 3 2 13 3 Figure 53 If a rational expression is equal to 0, then its numerator must equal 0. A false statement results. EXAMPLE 7 Graphing a Rational Function (Intersects Its Horizontal Asymptote) Graph ƒ1x2 = 3x2 - 3x - 6 x2 + 8x + 16 . SOLUTION Step 1 To find the vertical asymptote(s), solve x2 + 8x + 16 = 0. x2 + 8x + 16 = 0 Set the denominator equal to 0. 1x + 422 = 0 Factor. x = -4 Zero-factor property The vertical asymptote has equation x = -4. Step 2 We divide all terms by x2 and consider the behavior of each term as x increases without bound to get the equation of the horizontal asymptote, y = 3 1 , Leading coefficient of numerator Leading coefficient of denominator or y = 3. Step 3 ƒ102 = - 3 8 , so the y-intercept is A0, - 3 8B. Step 4 Solve ƒ1x2 = 0 to find any x-intercepts. 3x2 - 3x - 6 x2 + 8x + 16 = 0 Set ƒ1x2 = 0. 3x2 - 3x - 6 = 0 Set the numerator equal to 0. x2 - x - 2 = 0 Divide by 3. 1x - 221x + 12 = 0 Factor. x = 2 or x = -1 Zero-factor property The x-intercepts are 1-1, 02 and 12, 02. The numerator is not 0 when x = -4. LOOKING AHEAD TO CALCULUS The rational function ƒ1x2 = 2x + 1 x - 3 , seen in Example 6, has horizontal asymptote y = 2. In calculus, the behavior of the graph of this function as x approaches -∞ and as x approaches ∞ is described using limits at infinity. As x approaches -∞, ƒ1x2 approaches 2. This is written lim xS-∞ ƒ1x2 = 2. As x approaches ∞, ƒ1x2 approaches 2. This is written lim xS∞ ƒ1x2 = 2. Steps 6 and 7 The points 1-4, 12, A1, - 3 2B, and A6, 13 3 B are on the graph and can be used to complete the sketch of this function, which decreases on every interval of its domain. See Figure 53. S Now Try Exercise 63.
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