Sometimes, students are fooled into thinking that a rational function has a vertical asymptote
when—in reality—it has a puncture point (also called a ‘hole’).

This section provides detail and practice with punctured graphs.

a graph with a ‘puncture point’
(also called a ‘hole’)

First, an example:

Does the function $R(x) = \frac{x^2 - x - 2}{x - 2} $ have a vertical asymptote at $\,x = 2\,$?   No!

Some well-intentioned web sites conclude that this function has a vertical asymptote at $\,x = 2\,$,
simply because the denominator is zero there.
This is incorrect.

In order to find a vertical asymptote,
it is NOT ENOUGH to just set the denominator equal to zero.

You must ALSO check that the numerator is nonzero!

For this particular function, substitution of $\,x = 2\,$ into the numerator gives: $$2^2 - 2 - 2\ \ = \ \ 4 - 2 - 2 \ \ = \ \ 0$$ Therefore, both the numerator and denominator are zero when $\,x = 2\,$.

When BOTH the numerator and denominator of a rational function are zero,
then you must work harder to figure out what's going on.

You must determine what's happening CLOSE TO the $x$-value of interest.
Further analysis shows: $$R(x) = \frac{x^2-x-2}{x-2} = \frac{(x+1)(x-2)}{x-2} = (x+1)\cdot\frac{x-2}{x-2} $$ Cancelling off the extra factor of one: $$R(x) = x + 1\,, \ \text{providing } x\ne 2$$ The function $\,R\,$ isn't defined when $\,x = 2\,$, since division by zero isn't allowed.
For all values of $\,x\ne 2\,$, ‘$\,\frac{x-2}{x-2}\,$’ is just the number $\,1\,$!
The graph of $\,R\,$ is therefore the same as the line $\,y = x + 1\,$, except that it has been ‘punctured’ at $\,x = 2\,$.
From this re-naming, it is now clear that when $\,x\,$ is close to $\,2\,$, the outputs are close to $\,2 + 1 = 3\,$.
The graph of the function $\,R\,$ is shown below:

DEFINITION puncture points (holes)
Let $\displaystyle\,R(x) = \frac{N(x)}{D(x)}\,$ be a function.

The function $\,R\,$ has a puncture point (hole) at $\,x = c\,$ if and only if the following three conditions hold:
  • $D(c) = 0\,$   (the denominator is zero at $\,c\,$)
  • $N(c) = 0\,$   (the numerator is zero at $\,c\,$)
  • as $\,x\,$ approaches $\,c\,$, the values $\,R(x)\,$ approach a finite number


Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Finding Horizontal Asymptotes
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
1 2 3 4 5 6 7 8 9 10 11

(MAX is 11; there are 11 different problem types.)