Question

What are the vertical and horizontal asymptotes for the function f(x)=x^2+x-6/x^3-1?

Answer 1

Answer:

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).

Answer 2

Answer:

its b

Step-by-step explanation: