# What are the asymptotes of f(x)=-x/((x-1)(x^2+x)) ?

Dec 20, 2016

There is a hole at $x = 0$
The vertical asymptotes are $x = - 1$ and $x = 1$
No slant asymptote.
The horizontal asymptote is $y = 0$

#### Explanation:

Let's factorise the denominator

$\left(x - 1\right) \left({x}^{2} + x\right) = x \left(x - 1\right) \left(x + 1\right)$

Therefore,

$f \left(x\right) = \frac{\cancel{x}}{\cancel{x} \left(x + 1\right) \left(x - 1\right)}$

$= \frac{1}{\left(x + 1\right) \left(x - 1\right)}$

There is a hole at $x = 0$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- 1 , 1\right\}$

As you cannot divide by $0$, $x \ne - 1$ and $x \ne 1$

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

The degree of the numerator is $<$ than the degree of the denominator, there is no slant asymptote.

${\lim}_{x \to \pm \infty} f \left(x\right) = {\lim}_{x \to \pm \infty} \frac{1}{x} ^ 2 = {0}^{+}$

The horizontal asymptote is $y = 0$

graph{1/((x+1)(x-1)) [-14.24, 14.24, -7.11, 7.14]}