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

Jun 16, 2015

First look at what makes the denominator $= 0$

#### Explanation:

When $x$ gets closer to $- 1$, either from below, or from above, the denominator will get closer to zero, and the function as a whole will get larger and larger.

Or in "the language":

${\lim}_{x \to - {1}^{+}} y = \infty \mathmr{and} {\lim}_{x \to - {1}^{-}} y = - \infty$

This is called the vertical asymptote $x = - 1$

The horizontal asymptote is when $x$ becomes incredibly large, either negative or positive. In this case $y$ will be smaller and smaller, or:

${\lim}_{x \to \infty} y = 0 \mathmr{and} {\lim}_{x \to - \infty} y = 0$
So the horizontal asymptote is $y = 0$
graph{2/(x+1) [-28.86, 28.9, -14.43, 14.43]}