Question

How do you find vertical, horizontal and oblique asymptotes for `(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:

The denominator `=x^3+4x^2=(x^2)(x+4)`
as you cannot divide by `0`,
So, `x=0` and `x=-4` are vertical asymptotes.

The degree of the numerator is `<` than the degree of the denominator, there is no oblique asymptote.

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

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

Therefore `y=0` is a horizontal asymptote
graph{(x^2-4)/(x^3+4x^2) [-7.436, 8.364, -3.85, 4.05]}