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

1 Answer
Jun 16, 2015

First look at what makes the denominator =0=0

Explanation:

When xx gets closer to -11, 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->-1^+) y= oo and lim_(x->-1^-) y= -oo

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->oo) y=0 and lim_(x->-oo) y=0
So the horizontal asymptote is y=0
graph{2/(x+1) [-28.86, 28.9, -14.43, 14.43]}