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.

For being nice i appreciate it but abc

Well what if i told you it wouldn’t matter, because in the future the pony dies, depending on what choice you make