Question

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

Answer 1

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.