# How do you find the asymptotes for f(x)= (2x+4) / (x^2-3x-4)?

Aug 11, 2015

$x = 4 , x = - 1 \text{ and } f \left(x\right) = 0$

#### Explanation:

To find the vertical asymptotes:
${\lim}_{x \rightarrow {a}^{+}} f \left(x\right) = \pm \infty \text{ and } {\lim}_{x \rightarrow {a}^{-}} f \left(x\right) = \pm \infty$
To find the horizontal asymptotes:
${\lim}_{x \rightarrow \pm \infty} f \left(x\right) = b$

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

In this case the vertical asymptotes are when the denominator of $f \left(x\right)$ in its simplest form is equal to zero. Hence:
$\left(x - 4\right) \left(x + 1\right) = 0$
$x = 4 \text{ or } x = - 1$

Note: ${\lim}_{x \rightarrow {4}^{\pm}} f \left(x\right) = \pm \infty {\text{ and " lim_{x rarr -1^+-} f(x) = }}_{+}^{-} \infty$

To find the horizontal asymptotes:
${\lim}_{x \rightarrow \pm \infty} f \left(x\right) = {\lim}_{x \rightarrow \pm \infty} \frac{2 \left(x + 2\right)}{{x}^{2} - 3 x - 4} = {0}^{\pm}$

Hence the asymptotes for $f \left(x\right)$ are:
$x = 4 , x = - 1 \text{ and } f \left(x\right) = 0$

graph{(2x+4)/(x^2-3x-4) [-5, 8, -10, 10]}