How do you find the asymptotes of # (1) / (x-1) ^ 2#?

1 Answer
Aug 2, 2016

The asymptotes are #x=1#.

Explanation:

Asymptotes are the places where #x# makes the denominator equal to #0#. That means that if we are given #5/(x-4)#, when #x=4#, then the denominator becomes #5/0# We cannot divide by #0# without destroying the universe or ripping a hole in the fabric of reality. All in all, we just stay away from dividing by #0#. Since our calculators, and the mathmaticians before them, can't handle the whole concept of #5/0#, we just skip over it. That is what an asyptoote it. The graph may get very, very, very, very, very close, but #x# willl never equal #4#.

In our case, we have #1/(x-1)^2#, or #1/((x-1)(x-1))#. We need to find the value that will make #x# equal to #0#. We do that by saying #0=x-1# and solving for #x#, like this: #x=1#. So, we now that when #x=1#, the graph freaks out. that is our asymptote. It makes no difference that there are two values that the same; it just changes the shape of the grapg. Let's graph it and look at what we've got:
graph{y=1/(x-1)^2}