# How do you find the vertical, horizontal and slant asymptotes of: f(x)=(x^4-x^2)/(x(x-1)(x+2))?

Feb 18, 2018

See below.

#### Explanation:

Start by simplifying the expression:

$\frac{{x}^{4} - {x}^{2}}{x \left(x - 1\right) \left(x + 2\right)}$

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

$\left({x}^{2} - 1\right) = \left(x + 1\right) \left(x - 1\right)$ ( Difference of two squares )

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

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

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

Vertical asymptotes occur where the function is undefined. This can be seen to be when $x = - 2$ (division by zero)

So the line $x = - 2$ is a vertical asymptote.

We now divide:

$\frac{{x}^{2} + x}{x + 2}$

The quotient of this will be the oblique asymptote:

$\left(x + 2\right) | \overline{{x}^{2} + x}$

$\textcolor{w h i t e}{888888888} x$
$\left(x + 2\right) | \overline{{x}^{2} + x}$

$\textcolor{w h i t e}{888888888} x - 1$
$\left(x + 2\right) | \overline{{x}^{2} + x}$
$\textcolor{w h i t e}{88888888} {x}^{2} + 2 x$
$\textcolor{w h i t e}{88888888888} - x$
$\textcolor{w h i t e}{88888888888} - x - 2$
$\textcolor{w h i t e}{88888888888.888888} 2$

We do not need to be concerned with the remainder. So the oblique asymptote is the line:

$y = x - 1$

This is the line $\frac{{x}^{2} + x}{x + 2}$ approaches as $x \to \pm \infty$

Graph: