2-5 Rational Functions

Transcription

1 Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any. 3. f (x) = The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x 4). The real zeros of b(x) are 4 and 3. Therefore, D = {x x 4, 3, x }. Check for vertical asymptotes. Determine whether x = 4 is a point of infinite discontinuity. Find the limit as x approaches 4 from , undefined , x = 4 is a vertical asymptote of f. Determine whether x = 3 is a point of infinite discontinuity. Find the limit as x approaches 3 from undefined x = 3 is a vertical asymptote of f. Check for horizontal asymptotes. Use a table to examine the end behavior of f (x) ,417 The table suggests Therefore, there does not appear to be a horizontal asymptote. 6. f (x) = The function is undefined at the real zero of the denominator f (x) = x 4. The real zero of f (x) is 4. Therefore, D = {x x 4, x }. Check for vertical asymptotes. Determine whether x = 4 is a point of infinite discontinuity. Find the limit as x approaches 4 from , undefined , x = 4 is a vertical asymptote of f. Check for horizontal asymptotes. Use a table to examine the end behavior of f (x) The table suggests Therefore, there does not appear to be a horizontal asymptote. esolutions Manual - Powered by Cognero Page 1

2 For each function, determine any asymptotes and intercepts. Then graph the function and state its domain. 9. f (x) = 4, 5, x }. There are vertical asymptotes at x = 4 and x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal. The x-intercepts are 2 and 3, the zeros of the numerator. 12. f (x) = 0, 6, x }. There are vertical asymptotes at x = 0 and x = 6. The x-intercept is 2, the zero There is no y-intercept because f (x) is undefined for x = 0. The y-intercept is because f (0) =. Graph the asymptotes and intercepts. Then find and esolutions Manual - Powered by Cognero Page 2

3 15. h(x) = 18. h(x) can be written as h(x) =. 3, 1, x }. There are vertical asymptotes at x = 3 and x = 1. There is no horizontal asymptote since the degree of the numerator is greater than the degree of the denominator. Since the degree of the numerator is 2 greater than the denominator, there is no oblique asymptote. The function is undefined at b(x) = 0. x yields no real zeros, so D = {x x }. There are no vertical asymptotes. Since the polynomial in the numerator has no real zeros, there are no x-intercepts. The x-intercepts are 5, 0, and 2, the zeros of the numerator. The y-intercept is because Graph the asymptote and intercept. Then find and The y-intercept is 0 because h(0) = 0. Graph the asymptotes and intercepts. Then find and For each function, determine any asymptotes, holes, and intercepts. Then graph the function and state its domain. esolutions Manual - Powered by Cognero Page 3

4 and state its domain. 21. h(x) = The function is undefined at b(x) = h(x) = 3, x }. There is a vertical asymptote at x = 3. There is no horizontal asymptote since the degree of the numerator is greater than the degree of the denominator. Since the degree of the numerator is 2 greater than the denominator, there is no oblique asymptote. Thus,. The x-intercept is 0, the zero There is a vertical asymptote at. Since the polynomial in the numerator has no real zeros, there are no x-intercepts. The y-intercept is 0 because h(0) = 0. Graph the asymptote and intercepts. Then find and Thus, the y-intercept is.. esolutions Manual - Powered by Cognero Page 4

5 27. g(x) = 2, 1, or 2, x }. There is a vertical asymptote at x = 1, the real zero of the simplified denominator. Since the simplified numerator has no real zeros, there are no x-intercepts. The y-intercept is 1 because g(0) =1. There are holes at ( 2, 1) and because the original function is undefined when x = 2 and x = 2. esolutions Manual - Powered by Cognero Page 5