# How to find the asymptotes of F(x)=(x^2+x-12) /( x^2-4)?

Jan 30, 2016

Vertical asymptotes: $x = - 2 \textcolor{w h i t e}{\text{XXX")andcolor(white)("XXX}} x = + 2$
Horizontal asymptote: $F \left(x\right) = 1$

#### Explanation:

For an expression composed of a fraction with a polynomial numerator $N \left(x\right)$ and a polynomial denominator $D \left(x\right)$

Vertical asymptotes exist as $x = a$
for any and all point for which $D \left(a\right) = 0$ and $N \left(a\right) \ne 0$

With $F \left(x\right) = \frac{N \left(x\right)}{D \left(x\right)} = \frac{{x}^{2} + x - 12}{{x}^{2} - 4}$

$\textcolor{w h i t e}{\text{XXX}} D \left(x\right) = 0$ for $x = \pm 2$
$\textcolor{w h i t e}{\text{XXXXXXX}}$hopefully this is obvious
and
$\textcolor{w h i t e}{\text{XXX}} N \left(x\right) \ne 0$ for either $x = - 2$ or $x = 2$

Therefore both $x = - 2$ and $x = 2$ are vertical asymptotes.

Horizontal asymptotes exist as $F \left(x\right) = b$
if ${\lim}_{x \rightarrow \infty} F \left(x\right) = b$

$\textcolor{w h i t e}{\text{XXX}} = {\lim}_{x \rightarrow \infty} F \left(x\right)$

$\textcolor{w h i t e}{\text{XXX}} {\lim}_{x \rightarrow \infty} \frac{{x}^{2} + x - 12}{{x}^{2} - 4}$

$\textcolor{w h i t e}{\text{XXX}} = {\lim}_{x \rightarrow \infty} 1 + \frac{x - 8}{{x}^{2} - 4}$

$\textcolor{w h i t e}{\text{XXX}} = 1 + {\lim}_{x \rightarrow \infty} \frac{x - 8}{{x}^{2} - 4}$

$\textcolor{w h i t e}{\text{XXX}} = 1 + 0 = 1$
graph{(x^2+x-12)/(x^2-4) [-5.45, 7.033, -1.27, 4.98]}