Question

How do you find the vertical, horizontal or slant asymptotes for `(x^2-4)/(x^3+4x^2)`?

Answer 1

The vertical asymptotes are `x=0` and `x=-4`
No slant asymptotes.
The horizontal asymptote is `y=0`

Explanation:

Let `f(x)=(x^2-4)/(x^3+4x^2)`

Let's factorise the denominator
`x^3+4x^2=x^2(x+4)`

The domain of `f(x)` is `D_f(x)=RR-{0,-4}`

As we cannot divide by `0`

So `x!=0` and `x!=-4`

The vertical asymptotes are `x=0` and `x=-4`

As the degree of the numerator is `<` the degree of the denominator, there is no slant asymptotes.

For calculating the limits, we take the terms of highest degree in the numerator and the denominator

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

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

So, `y=0` is a horizontal asymptote

graph{(x^2-4)/(x^3+4x^2) [-10, 10, -5, 5]}