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

1 Answer
Jun 16, 2015

Answer:

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->-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]}