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

1 Answer
Jan 4, 2016

Answer:

Vertical: #x=1#
Horizontal: #y=0#

Explanation:

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.