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

Dec 1, 2016

We have a vertical asymptote at $x = - 1$ and

an oblique asymptote $y = x + 1$

#### Explanation:

As in $f \left(x\right) = \frac{{x}^{2} + 2 x}{x + 1}$ we have the denominator $x + 1$,

when $x \to - 1$, $\left(x + 1\right) \to 0$ and $f \left(x\right) \to \infty$

As such we have a vertical asymptote at $x = - 1$

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

= $x + 1 - \frac{1}{x + 1}$

Hence when $x \to \infty$, $f \left(x\right) \to x + 1$

Hence, we have an oblique asymptote $y = x + 1$

There is no other asymptote.
graph{(x^2+2x)/(x+1) [-10, 10, -5, 5]}