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

Dec 9, 2016

The vertical asymptotes are $x = - 1$ and $x = 4$
No slant asymptote
The horizontal asymptote is $y = 0$

#### Explanation:

Let's factorise the denominator

${x}^{2} - 3 x - 4 = \left(x + 1\right) \left(x - 4\right)$

Let $f \left(x\right) = \frac{2 x + 4}{\left(x + 1\right) \left(x - 4\right)}$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- 1 , 4\right\}$

As we cannot divide by $0$, $x \ne - 1$ and $x \ne 4$

The vertical asymptotes are $x = - 1$ and $x = 4$

The degree of the numerator is $<$ than the degree of the denominator, so there is no slant asymptote

To calculate the limits as $x \to \pm \infty$, we take the terms of highest degree in the numerator and the denominator.

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

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

The horizontal asymptote is $y = 0$

graph{(2x+4)/(x^2-3x-4) [-11.25, 11.25, -5.625, 5.625]}