# How do you find the vertical, horizontal or slant asymptotes for f(x) = (-x^2 + 4x)/(x+2)?

Jan 6, 2017

The vertical asymptote is $x = - 2$
The slant asymptote is $y = - x + 6$
No horizontal asymptote

#### Explanation:

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- 2\right\}$

As we cannot divide by $0$, $x \ne - 2$

The vertical asymptote is $x = - 2$

The degree of the numerator is $>$ than the degree of the denominator, so there is a slant asymptote.

To find the slant, we start by doing a long division

$\textcolor{w h i t e}{a a a a}$$- {x}^{2} + 4 x$$\textcolor{w h i t e}{a a a a}$∣$x + 2$

$\textcolor{w h i t e}{a a a a}$$- {x}^{2} - 2 x$$\textcolor{w h i t e}{a a a a}$∣$- x + 6$

$\textcolor{w h i t e}{a a a a a a a}$$0 + 6 x$

$\textcolor{w h i t e}{a a a a a a a a a}$$+ 6 x + 12$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a}$$- 12$

So,

$f \left(x\right) = \frac{- 4 {x}^{2} + 4 x}{x + 2} = - x + 6 - \frac{12}{x + 2}$

${\lim}_{x \to - \infty} f \left(x\right) + \left(x - 6\right) = {\lim}_{x \to - \infty} - \frac{12}{x + 2} = {0}^{+}$

${\lim}_{x \to + \infty} f \left(x\right) + \left(x - 6\right) = {\lim}_{x \to + \infty} - \frac{12}{x + 2} = {0}^{-}$

The slant asymptote is $y = - x + 6$

No horizontal asymptote

graph{(y-(-x^2+4x)/(x+2))(y+x-6)(y-25x-50)=0 [-35, 38.04, -14.36, 22.2]}