# How do you find vertical, horizontal and oblique asymptotes for (x^2-2x)/(x+1)?

Dec 8, 2016

The vertical asymptote is $x = - 1$
The oblique asymptote is $y = x$
No horizontal asymptote

#### Explanation:

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

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- 1\right\}$

As you cannot divide by $0$, $x \ne 0$

So, the vertical asymptote is $x = - 1$

The degree of the numerator is $>$ to the degree of the denominator, so we expect an oblique asymptote.

Let's do a long division

$\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 + 1$

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

$\textcolor{w h i t e}{a a a a a}$$0 - 3 x$

So,

$\frac{{x}^{2} - 2 x}{x + 1} = x - \frac{3 x}{x + 1}$

Therefore

The oblique asymptote is $y = x$

To calculate the limits as $x \to \pm \infty$, we take the terms of highest degree in the numerator and the denominator

${\lim}_{x \to \pm \infty} f \left(x\right) = {\lim}_{x \to \pm \infty} {x}^{2} / x = {\lim}_{x \to \pm \infty} x = \pm \infty$

graph{(y-(x^2-x)/(x+1))(y-x)=0 [-36.53, 36.57, -18.26, 18.27]}