# How do you find the asymptotes for #(3x-2) / (x+1) #?

##### 1 Answer

Nov 10, 2015

There are three kinds of asymptotes:

**Vertical asymptotes**, which are vertical lines of the form#x=k# , where#k# is a value not in the domain of the function;**Horizontal asymptotes**, which are horizontal lines of the form#y=k# , where#k# is the limit of the function as#x->pm infty# (of course there can be two different asymptotes, one for each direction;**Oblique asymptotes**, which are line of the form#mx+q# .

Horizontal asymptotes are particular oblique ones, and so if you find horizontal asymptotes at both

In your case, the only point in which the function is not defined is the one which annihilates the denominator, and

As for the horizontal ones, we have that

So,