Question

How do you find the Vertical, Horizontal, and Oblique Asymptote given `(x^2-4)/(x^3+4x^2)`?

Answer 1

The vertical asymptotes are `x=0` and `x=-4`
The horizontal asymptote is `y=0`

Explanation:

Let rewrite the function
`(x^2-4)/(x^3+4x^2)=(x^2-4)/(x^2(x+4))`

As we cannot divide by `0`, the function is not defined for `x=0` and `x=-4`
So the vertical asymptotes are `x=0` and `x=-4`

The degree of the denominator is greater than the degree of the numeratos, so we don't have oblique asymptotes

Let's find the limit of the function as `x->oo`

limit `(x^2-4)/(x^3+4x^2)=x^2/x^3=1/x=0^-`
`x->-oo`

limit `(x^2-4)/(x^3+4x^2)=x^2/x^3=1/x=0^+`
`x->+oo`
So `y=0` is a horizontal asymptote
graph{(x^2-4)/(x^3+4x^2) [-10, 10, -5, 5]}