# How do I find the asymptotes of f(x)= (- 4x-x^2)/(2 +2x- x^3)?

Dec 16, 2015

asymptotes:
$x = 2$
$x = \sqrt{2}$
$x = - \sqrt{2}$

#### Explanation:

Start by simplifying the function:

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

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

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

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

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

$f \left(x\right) = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{-}}} x \left(x + 4\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{-}}} \left(x - 2\right) \left({x}^{2} - 2\right)}$

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

Take each bracketed polynomial in the denominator, set it to cannot equal to $0$, and solve for $x$.

Finding the asymptotes

$1. x - 2 \ne 0$
$\textcolor{w h i t e}{i \times \times} x \ne 2$

$2. {x}^{2} - 2 \ne 0$
$\textcolor{w h i t e}{\times \times} {x}^{2} \ne 2$
$\textcolor{w h i t e}{\times \times x} x \ne \pm \sqrt{2}$

The asymptotes are also the values which cannot be substituted into the equation such that the denominator would be $0$.

$\therefore$, the asymptotes are $x = 2$, $x = \sqrt{2}$, and $x = - \sqrt{2}$.