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 $\displaystyle R(x) := \frac{x^32x^2}{x2}\,$ certainly
looks like a typical rational function.
Upon closer inspection, though, we see that there's an extra factor of
$\,1\,$ in the formula:
$$
\cssId{s6}{\frac{x^3  2x^2}{x  2}} \quad
\cssId{s7}{= \quad \frac{x^2(x2)}{x2}} \quad
\cssId{s8}{= \quad x^2\cdot\frac{x2}{x2}}
$$
Therefore, $\,\displaystyle R(x) = \frac{x^32x^2}{x2}\,$ 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 $\,1\,$.
$P(x) = x^2$ 
$\displaystyle
\cssId{s13}{R(x)}
\cssId{s14}{= \frac{x^3  2x^2}{x2}}
\cssId{s15}{= x^2\cdot\frac{x2}{x2}}$ 
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
