# How do you find the asymptotes for f(x) = (x^3 + x^2 - 6x) /( 4x^2 - 8x - 12)?

Jan 26, 2016

Asymptotes are $x = 3$ , $x = - 1$ and $\frac{x}{4}$

#### Explanation:

Start by finding the factors for both numerator and denominator.
$f \left(x\right) = \frac{{x}^{3} + {x}^{2} - 6 x}{4 {x}^{2} - 8 x - 12}$

$f \left(x\right) = \frac{x \left({x}^{2} + x - 6\right)}{4 \left({x}^{2} - 2 x - 3\right)}$

$f \left(x\right) = \frac{x \left(x + 3\right) \left(x - 2\right)}{4 \left(x - 3\right) \left(x + 1\right)}$

This confirms that there are no removable asymptotes as there are no factors that cancel out.

The denominator approaches zero, and therefore the function value $\to \infty$ as $x \to 3$ or $x \to - 1$ so there are vertical asymptotes at $x = 3$ and $x = - 1$

As $x \to \infty$ the function $f \left(x\right) \to \frac{x \cancel{{x}^{2} + x - 6}}{4 \cancel{{x}^{2} - 2 x - 3}} \to \frac{x}{4}$

So there is a slant asymptote of $\frac{x}{4}$