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

Jan 29, 2017

Vertical asymptote at $x = - 2$

#### Explanation:

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

As $x \to - 2$ from right, $\frac{2}{x + 2} \to \infty$

and as $x \to - 2$ from left, $\frac{2}{x + 2} \to - \infty$

graph{2/(x+2) [-11.71, 8.29, -5.16, 4.84]}

Hence, we have a vertical asymptote at $x = - 2$

Note that at $x = 2$, we have a hole but no asymptote as $\left(x - 2\right)$ cancels out in numerator and denominator.