# How do you find the vertical, horizontal and slant asymptotes of: f(x)= x^3 / (x^2-1)?

Oct 21, 2016

The vertical asymptotes are $x = 1$ and $x = - 1$
The slant asymptote is $y = x$

#### Explanation:

Firstly the domain of $f \left(x\right)$

is $\mathbb{R} - \left(1 , - 1\right)$
since we cannot divide by zero
$x - 1 \ne 0$ and $x + 1 \ne 0$
so $x \ne - 1$ and $x \ne - 1$

So $x = 1$ and $x = - 1$ are vertical asymptotes

A clant asymptote exists only if the degree of the numerator is greater than that of the numerator

We can do a long division

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

So the slant asymptote is $y = x$