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

##### 1 Answer

Dec 1, 2016

We have a vertical asymptote at

an oblique asymptote

#### Explanation:

As in

when

As such we have a vertical asymptote at

Further

=

Hence when

Hence, we have an oblique asymptote

There is no other asymptote.

graph{(x^2+2x)/(x+1) [-10, 10, -5, 5]}