# How do you find vertical, horizontal and oblique asymptotes for (x^2-4)/(x^3+4x^2)?

Nov 12, 2016

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

#### Explanation:

The denominator $= {x}^{3} + 4 {x}^{2} = \left({x}^{2}\right) \left(x + 4\right)$
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.

${\lim}_{x \to - \infty} \frac{{x}^{2} - 4}{{x}^{3} + 4 {x}^{2}} = {\lim}_{x \to - \infty} {x}^{2} / {x}^{3} = {\lim}_{x \to - \infty} \frac{1}{x} = {0}^{-}$

${\lim}_{x \to + \infty} \frac{{x}^{2} - 4}{{x}^{3} + 4 {x}^{2}} = {\lim}_{x \to + \infty} {x}^{2} / {x}^{3} = {\lim}_{x \to + \infty} \frac{1}{x} = {0}^{+}$

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