How do you find the asymptotes for #f(x) = x / (3x(x-1))#?

Question

Anna Allen · Accepted Answer

Vertical: #x=1#
Horizontal: #y=0# First, notice that the #x# terms will cancel, leaving the function as #f(x)=1/(3(x-1)),color(white)(xxxx)x!=0# However, canceling the #x# terms leaves a hole, or a removable discontinuity, at #x=0#. Vertical asymptotes: Vertical asymptotes will occur when the denominator equals #0#. #3(x-1)=0# Solved, this gives #x=1# Thus the vertical asymptote occurs at #x=1#. Even though #3x# is in the denominator in the original function, its cancellation makes it just a hole and not also a vertical asymptote. Horizontal asymptotes: Since the degree of the denominator is larger than the degree of the numerator, the horizontal asymptote is the line #y=0#. graph{x/(3x(x-1)) [-10, 10, -5, 5]} Don't be fooled—there is a hole at #x=0#, despite appearances.

Liam Brennan · Answer

undefined To find the asymptotes for the function $ f(x) = \frac{x}{3x(x-1)} $, first check for vertical asymptotes by setting the denominator equal to zero and solving for $ x $. Next, check for horizontal asymptotes by comparing the degrees of the numerator and denominator. If the degrees are the same, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0.