# How do you find vertical, horizontal and oblique asymptotes for (-x^2 + 4x)/(x+2)?

Nov 1, 2016

vertical asymptote at x = -2
oblique asymptote at y = -x + 6

#### Explanation:

The denominator of the rational function cannot be zero as this would make the function undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve : $x + 2 = 0 \Rightarrow x = - 2 \text{ is the asymptote}$

Horizontal asymptotes occur when the degree of the numerator ≤ degree of the denominator. This is not the case here (numerator-degree 2, denominator- degree 1 ). Hence there are no horizontal asymptotes.

Oblique asymptote occur when the degree of the numerator > degree of the denominator. Hence there is an oblique asymptote.
Polynomial division gives.

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

as $x \to \pm \infty , f \left(x\right) \to - x + 6 - 0$

$\Rightarrow y = - x + 6 \text{ is the asymptote}$
graph{(-x^2+4x)/(x+2) [-10, 10, -5, 5]}

Nov 1, 2016

The vertical asymptote is $x = - 2$
the oblique asymptote is $y = - x + 6$
there is no horizontal asymptote

#### Explanation:

We cannot divide by $0$, so $x \ne 0$
Therefore $x = - 2$ is a vertical asymptote
As the degree of the numerator is greater than the degree of the denominator, we expect an oblique asymptote:
so we do a long division
$- {x}^{2} + 4 x$$\textcolor{w h i t e}{a a a a a a a}$∣$x + 2$
$- {x}^{2} - 2 x$$\textcolor{w h i t e}{a a a a a a a}$∣$- x + 6$
$\textcolor{w h i t e}{a a}$$0 + 6 x$
$\textcolor{w h i t e}{a a a a a}$$6 x + 12$
$\textcolor{w h i t e}{a a a a a a}$$0 - 12$
And finally, we have $\frac{- {x}^{2} + 4 x}{x + 2} = - x + 6 - \frac{12}{x + 2}$
So the oblique asymptote is $y = - x + 6$
graph{(y-(4x-x^2)/(x+2))(y+x-6)=0 [-52, 52.03, -26, 26.04]}