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

1 Answer
Nov 10, 2015

There are three kinds of asymptotes:

  1. Vertical asymptotes, which are vertical lines of the form #x=k#, where #k# is a value not in the domain of the function;
  2. 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;
  3. 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 #infty# and #-infty#, there's no need to look for oblique ones.

In your case, the only point in which the function is not defined is the one which annihilates the denominator, and #x+1=0 iff x=-1#. So, #x=-1# is a vertical asymptote.

As for the horizontal ones, we have that

#lim_{x\to\pm\infty} \frac{3x-2}{x+1} =lim_{x\to\pm\infty} \frac{cancel(x)(3-2/x)}{cancel(x)(1+1/x)}=lim_{x\to\pm\infty} 3/1 = 3#

So, #y=3# is a horizontal asymptote in both directions.