386 CHAPTER 3 Polynomial and Rational Functions EXAMPLE 4 Finding Asymptotes of Rational Functions Give the equations of any vertical, horizontal, or oblique asymptotes for the graph of each rational function. (a) ƒ1x2 = x + 1 12x - 121x + 32 (b) ƒ1x2 = 2x + 1 x - 3 (c) ƒ1x2 = x2 + 1 x - 2 SOLUTION (a) To find the vertical asymptotes, set the denominator equal to 0 and solve. 12x - 121x + 32 = 0 2x - 1 = 0 or x + 3 = 0 Zero-factor property x = 1 2 or x = -3 Solve each equation. The equations of the vertical asymptotes are x = 1 2 and x = -3. To find the equation of the horizontal asymptote, begin by multiplying the factors in the denominator. ƒ1x2 = x + 1 12x - 121x + 32 = x + 1 2x2 + 5x - 3 Now divide each term in the numerator and denominator by x2. We choose the exponent 2 because it is the greatest power of x in the entire expression. ƒ1x2 = x x2 + 1 x2 2x2 x2 + 5x x2 - 3 x2 = 1 x + 1 x2 2 + 5 x - 3 x2 As x increases without bound, the quotients 1 x , 1 x2 , 5 x , and 3 x2 all approach 0, and the value of ƒ1x2 approaches 0 + 0 2 + 0 - 0 = 0. 0 2 = 0 The line y = 0 (that is, the x-axis) is the horizontal asymptote. This supports procedure 2(a) of determining asymptotes on the previous page. (b) Set the denominator x - 3 equal to 0 to find that the vertical asymptote has equation x = 3. To find the horizontal asymptote, divide each term in the rational expression by x since the greatest power of x in the expression is 1. ƒ1x2 = 2x + 1 x - 3 = 2x x + 1 x x x - 3 x = 2 + 1 x 1 - 3 x As x increases without bound, the quotients 1 x and 3 x both approach 0, and the value of ƒ1x2 approaches 2 + 0 1 - 0 = 2. The line y = 2 is the horizontal asymptote. This supports procedure 2(b) of determining asymptotes on the previous page. Stop here. Leave the expression in complex form.
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