# How do you identify all asymptotes or holes and intercepts for f(x)=(x^2-2x)/(x^3+1)?

Oct 24, 2016

The vertical asymptote is $x = - 1$
The intercepts with the x axis are $\left(0 , 0\right)$ and $\left(2 , 0\right)$

#### Explanation:

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

As we cannot divide by $0$, the denominator $\ne 0$
${x}^{3} + 1 \ne 0$ $\implies$ ${x}^{3} \ne - 1$ $\implies$ $x \ne - 1$
So x=-1 is a vertical asymptote

To see the intercept with the x axis,
$f \left(x\right) = 0$
$\frac{{x}^{2} - 2 x}{{x}^{3} + 1} = 0$
${x}^{2} - 2 x = 0$
$x \left(x - 2\right) = 0$
so $x = 0$ and $x = 2$
The points are $\left(0 , 0\right)$ and $\left(2 , 0\right)$

We can find the limits at $\pm \infty$
limit f(x) = ${x}^{2} / {x}^{3} = \frac{1}{x} = 0$
$x \rightarrow - \infty$

limit f(x) = ${x}^{2} / {x}^{3} = \frac{1}{x} = 0$
$x \rightarrow + \infty$