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

Nov 23, 2016

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

#### Explanation:

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

Let's factorise the denominator
${x}^{3} + 4 {x}^{2} = {x}^{2} \left(x + 4\right)$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{0 , - 4\right\}$

As we cannot divide by $0$

So $x \ne 0$ and $x \ne - 4$

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

${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} {x}^{2} / {x}^{3} = {\lim}_{x \to - \infty} \frac{1}{x} = {0}^{-}$

${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} {x}^{2} / {x}^{3} = {\lim}_{x \to + \infty} \frac{1}{x} = {0}^{+}$

So, $y = 0$ is a horizontal asymptote

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