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

##### 1 Answer

Aug 20, 2016

#### Explanation:

#f(x) = (x^2+1)/(x+1)#

#=(x^2-1+2)/(x+1)#

#=((x-1)(x+1)+2)/(x+1)#

#=x-1+2/(x+1)#

As

So

When

Hence

graph{(y-(x^2+1)/(x+1))(y-x+1) = 0 [-20, 20, -10, 10]}