Question

How do you find the Vertical, Horizontal, and Oblique Asymptote given `f(x)=(x^3-x)/(x^3-4x)`?

Answer 1

The vertical asymptotes are `x=2` and `x=-2`
The horizontal asymptote is `y=1`

Explanation:

Let's do some simplification
`f(x)=(x^3-x)/(x^3-4x)=(x(x^2-1))/(x(x^2-4))=((x+1)(x-1))/((x+2)(x-2))`

As we cannot divide by `0`,

The degree of the numerator is identical to the degree of the denominator, so there are no oblique asymptotes.
Limit `f(x)=x^2/x^2=1`
`x->+-oo`

So a horizontal asymptote is `y=1`
graph{(x^2-1)/(x^2-4) [-10, 10, -5, 5]}