How do you find vertical, horizontal and oblique asymptotes for #x/(1-x)^2#?

1 Answer
Jun 13, 2017

Answer:

#y=0#

#x=1#

Explanation:

The degree of the bottom is 2, and the degree of the top is 1. Therefore, there will be a horizontal asymptote at #y=0# since the function will tend towards 0 as x goes to #oo#. This also means there will be no oblique asymptote.

As far as vertical asymptotes, they will occur when the denominator is 0 and the numerator is non-zero. For this function, this will happen when #x=1# since this makes the bottom equal to 0 and the top equal to 1.

So our asymptotes are #y=0# and #x=1#