391 3.5 Rational Functions: Graphs, Applications, and Models NOTE Suppose that we are asked to reverse the process of Example 7 and find the equation of a rational function having the graph shown in Figure 54 (repeated in the margin). Because the graph crosses the x-axis at its x-intercepts 1-1, 02 and 12, 02, the numerator must have factors 1x + 12 and 1x - 22, each of degree 1. The behavior of the graph at its vertical asymptote x = -4 suggests that there is a factor of 1x + 42 of even degree in the denominator. The horizontal asymptote at y = 3 indicates that the numerator and denominator have the same degree (both 2) and that the ratio of leading coefficients is 3. Verify in Example 7 that the rational function is ƒ1x2 = 31x + 121x - 22 1x + 422 ƒ1x2 = 3x2 - 3x - 6 x2 + 8x + 16 . Exercises of this type are labeled Connecting Graphs with Equations. –10 –5–1 2 5 10 15 x y y = 3 x = –4 (x + 4) is a denominator factor of degree 2. (x + 1) and (x – 2) are factors of the numerator. The quotient of the leading coefficients is 3. Figure 54 (repeated) Multiply the factors in the numerator. Square in the denominator. EXAMPLE 8 Graphing a Rational Function with an Oblique Asymptote Graph ƒ1x2 = x2 + 1 x - 2 . SOLUTION As shown in Example 4(c), the vertical asymptote has equation x = 2, and the graph has an oblique asymptote with equation y = x + 2. The y-intercept is A0, - 1 2B, and the graph has no x-intercepts because the numerator, x2 + 1, has no real zeros. The graph does not intersect its oblique asymptote because the following has no solution. x2 + 1 x - 2 = x + 2 Set the expressions defining the function and the oblique asymptote equal. x2 + 1 = x2 - 4 Multiply each side by x - 2. 1 = -4 False Using the y-intercept, asymptotes, the points A4, 17 2 B and A -1, - 2 3B, and the general behavior of the graph near its asymptotes leads to the graph in Figure 55. A rational function that is not in lowest terms often has a point of discontinuity in its graph. Such a point is sometimes called a hole. LOOKING AHEAD TO CALCULUS Different types of discontinuity are discussed in calculus. The function in Example 8, ƒ1x2 = x2 + 1 x - 2 , has an infinite discontinuity at x = 2, as indicated by the vertical asymptote there. The function in Example 9, ƒ1x2 = x2 - 4 x - 2 , is said to have a removable discontinuity at x = 2, because the discontinuity can be removed by redefining ƒ at 2. The greatest integer function has jump discontinuities because the function values “jump” from one value to another for integer domain values. 4 8 12 x y –4 –12 –8 –8 12 8 4 x = 2 y = x + 2 f(x) = x2 + 1 x – 2 (4, ) 17 2 (–1, – )2 3 Figure 55 S Now Try Exercise 87.
RkJQdWJsaXNoZXIy NjM5ODQ=