How do you find vertical, horizontal and oblique asymptotes for # x / (3x(x-1))#?

1 Answer
May 6, 2016

Answer:

There is a vertical asymptote at #x=1# and a horizontal asymptote at #y=0#

Explanation:

To find all the asymptotes for function #y=x/(3x(x-1))#, we first observe that #x# cancels out from numerator and denominator and the function is primarily #x/(3(x-1))#, but there is a hole at #x=0#

Let us first start with vertical asymptotes, which are given by putting denominator equal to zero or #x-1=0# i.e. #x=1#.

Further as in #y=3/(x-1)#, there is no variable in numerator, we have a horizontal asymptote at #y=0#

graph{x/(3x(x-1)) [-10, 10, -5, 5]}