Question

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

Answer 1

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.