INTRODUCTION TO PUNCTURE POINTS (HOLES)

LESSON READ-THROUGH
by Dr. Carol JVF Burns (website creator)
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Rational functions can exhibit puncture points (also called ‘holes’).

As you'll learn in Calculus, puncture points are an example of a ‘removable discontinuity’.

Here's the idea.

The function $\displaystyle R(x) := \frac{x^3-2x^2}{x-2}\,$ certainly looks like a typical rational function.
Upon closer inspection, though, we see that there's an extra factor of $\,1\,$ in the formula: $$ \cssId{s6}{\frac{x^3 - 2x^2}{x - 2}} \quad \cssId{s7}{= \quad \frac{x^2(x-2)}{x-2}} \quad \cssId{s8}{= \quad x^2\cdot\frac{x-2}{x-2}} $$ Therefore, $\,\displaystyle R(x) = \frac{x^3-2x^2}{x-2}\,$ has exactly the same outputs as the much simpler function, $\,P(x) := x^2\,$,
except that the function $\,R\,$ isn't defined when $\,x = 2\,$.

The graphs of both $\,P\,$ and $\,R\,$ are shown below—the puncture point (hole) in $\,R\,$ is caused by that extra factor of $\,1\,$.

$P(x) = x^2$

$\displaystyle \cssId{s13}{R(x)} \cssId{s14}{= \frac{x^3 - 2x^2}{x-2}} \cssId{s15}{= x^2\cdot\frac{x-2}{x-2}}$


Puncture points are studied in more detail in a future section.

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 Vertical Asymptotes
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
PROBLEM TYPES:
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(MAX is 3; there are 3 different problem types.)