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

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 $x \to \pm \infty$ (of course there can be two different asymptotes, one for each direction;
3. Oblique asymptotes, which are line of the form $m x + q$.

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

${\lim}_{x \setminus \to \setminus \pm \setminus \infty} \setminus \frac{3 x - 2}{x + 1} = {\lim}_{x \setminus \to \setminus \pm \setminus \infty} \setminus \frac{\cancel{x} \left(3 - \frac{2}{x}\right)}{\cancel{x} \left(1 + \frac{1}{x}\right)} = {\lim}_{x \setminus \to \setminus \pm \setminus \infty} \frac{3}{1} = 3$

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