# What are the asymptotes of f(x)=-x/((x^2-8)(x-2)) ?

##### 1 Answer
Oct 25, 2016

the asymptotes are;
Vertical at $x = - \sqrt{8} , 2 , \sqrt{8}$
Horizontal at $0$ as $x \to \pm \infty$

#### Explanation:

$f \left(x\right) = \frac{x}{\left({x}^{2} - 8\right) \left(x - 2\right)}$
$\therefore f \left(x\right) = \frac{x}{\left({x}^{2} - {\sqrt{8}}^{2}\right) \left(x - 2\right)}$
$\therefore f \left(x\right) = \frac{x}{\left(x + \sqrt{8}\right) \left(x - \sqrt{8}\right) \left(x - 2\right)}$
So these will be vertical asymptotes.

So we can see that the denominator is zero when $x = - \sqrt{8} , 2 , \sqrt{8}$

Also We have $f \left(x\right) = \frac{x}{{x}^{3} + \text{lower terms}}$
So as $x \to \pm \infty \implies f \left(x\right) \to 0$

Hence, the asymptotes are;
Vertical at $x = - \sqrt{8} , 2 , \sqrt{8}$
Horizontal at $0$ as $x \to \pm \infty$

The graph will help to visualise the results;
graph{x/((x^2-8)(x-2)) [-10, 10, -5, 5]}