What are the vertical and horizontal asymptotes for the function f(x)=x^2+x-6/x^3-1? a. vertical asymptote: x = 1 horizontal asymptote: none b. vertical asymptote: x = 1 horizontal asymptote: y = 0 c. vertical asymptote: x = –2, x = 3 horizontal asymptote: y = 0 d. vertical asymptote: x = –2, x = –3 horizontal asymptote: none

b

Step-by-step explanation:

For clarity, f(x)=x^2+x-6/x^3-1 should be written as:

         x^2 + x - 6

f(x) =

          x^3 - 1

Note that the denominator, x^3 - 1, factors as follows:  (x - 1)(x^2 + x + 1).  If we set these results = to 0 individually, we find that there is only one real root; it is x = 1.  We cannot divide by zero, so conclude that the given function is undefined at x = 1.  Thus, the function has the vertical asymptote x = 1.

Note that x^2 in the numerator is of a higher power than is x^3 in the denominator.  As x grows large without bound, either + or - , this ratio x^2/x^3 approaches zero.  Thus, the horizontal asymptote is y = 0.

The correct answer choice is (b).

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