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

Oct 28, 2016

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

#### Explanation:

Let rewrite the function
$\frac{{x}^{2} - 4}{{x}^{3} + 4 {x}^{2}} = \frac{{x}^{2} - 4}{{x}^{2} \left(x + 4\right)}$

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 $x \to \infty$

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

limit $\frac{{x}^{2} - 4}{{x}^{3} + 4 {x}^{2}} = {x}^{2} / {x}^{3} = \frac{1}{x} = {0}^{+}$
$x \to + \infty$
So $y = 0$ is a horizontal asymptote
graph{(x^2-4)/(x^3+4x^2) [-10, 10, -5, 5]}