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

The horizontal asymptotes will occur when x approaches ±oo

The easy way to see this is to divide all terms by the highest order variable ...

(x^2/x^3 + x/x^3-6/x^3) / (x^3/x^3-1/x^3) and all of that mess boils down to

0/1 or just 0 as x approaches ±oo

So the horizontal asymptote is the horizontal line y=0

The vertical asymptote will occur when there is division by zero, which is undefined because it is not a real value.

x^3-1=0

(x-1) (x^2+x+1) = 0

So this only occurs as x approaches - 1, so the vertical asymptote is the vertical line x=-1.