250 CHAPTER 4 Polynomial and Rational Functions Step 5 Since the degree of the numerator, 2, is one more than the degree of the denominator, 1, the rational function R is improper and has an oblique asymptote. To find the oblique asymptote, use long division. ) − − x x x x 1 1 2 2 The quotient is x, so the line = y x is an oblique asymptote of the graph. To determine whether the graph of R intersects the asymptote = y x, solve the equation ( ) = R x x. ( ) = − = − = − = R x x x x x x 1 1 1 0 2 2 2 Impossible The equation − = x x x 1 2 has no solution, so the graph of ( ) R x does not intersect the line = y x. Step 6 See Figure 45 for the graph using Geogebra. We see from the graph that there is no y-intercept and there are two x-intercepts, −1 and 1. We can also see that there is a vertical asymptote at = x 0. Step 7 Using the information gathered in Steps 1 through 6, we obtain the graph of R shown in Figure 46. Notice how the oblique asymptote is used as a guide in graphing the rational function by hand. Figure 46 ( ) = − R x x x 1 2 x 3 –3 y 3 –3 (1, 0) (–1, 0) y = x NOTE Because the denominator of the rational function is a monomial, we can also find the oblique asymptote as follows: − = − = − x x x x x x x 1 1 1 2 2 Since → x 1 0 as →∞ = x y x , is an oblique asymptote. j Analyzing the Graph of a Rational Function Analyze the graph of the rational function: ( ) = + R x x x 1 4 2 EXAMPLE 3 Figure 45 ( ) = − R x x x 1 2 Now Work PROBLEM 15 Notice that R in Example 2 is an odd function. Do you see the symmetry about the origin in the graph of R? Solution Step 1 R is completely factored. The domain of R is { } ≠ x x 0 . Step 2 R is in lowest terms. Step 3 There is no y-intercept. Since + = x 1 0 4 has no real solutions, there are no x-intercepts. Step 4 R is in lowest terms, so = x 0 (the y-axis) is a vertical asymptote of R. The multiplicity of 0 is even, so the graph approaches either ∞ or −∞ on both sides of the asymptote. Step 5 Since the degree of the numerator, 4, is two more than the degree of the denominator, 2, the rational function does not have a horizontal or oblique asymptote. Find the end behavior of R. As →∞ x , ( ) = + ≈ = R x x x x x x 1 4 2 4 2 2 The graph of R approaches the graph of = y x2 as →−∞ x and as →∞ x . The graph of R does not intersect = y x .2 Do you know why?
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