Example OAF.BRF.4 : $R(x) = \frac {(x-2)^2}{x^2(x+3)}$.
Notice: $ h=2$ , and $k = -3$.
Draw a mapping diagram and graph of $R$ yourself, or consider the GeoGebra figures and table below.
Martin Flashman, Nov. 1, 2013,
Created with GeoGebra
Notice how the points on the graph are paired with the points
on the mapping diagram.
Given a point / number, $x$, on the source line, there is a unique
arrow meeting the target line at the point / number, $R(x) =
\frac {(x-2)^2}{x^2(x+3)}$. which corresponds to the rational
function's value for $x$
Check the box in GeoGebra to see more points and arrows
corresponding to data on the table.
For this function, there is root at $a=2$, the value of $h$.
Since when $x=-3$, $x+1 =0$, since $x=-3$ is not a root of $x-2$,
$x=-3$ is a pole for $R$. Likewise $x = 0$ is a pole for $R$.
You can check this by moving
$x$ to $2$, $0$, and $-3$ on the GeoGebra mapping diagram.
Since the degrees of $x-2$ and $x$ are both even, there will be a
local extreme at $x=2$ while there will be only arbitrarily large
positive of $R(x)$ close to $x=0$.
Since the degree of $x+3$ is odd, there will be arbitrarily large
positive and negative values of $R(x)$ close to $x=-1$.
The function $R$ is not defined at $x=-3$ and $x=0$. On the GeoGebra
graph there is a dotted vertical line indicating the vertical
asymptotes at $x=-3$ and $x=0$.
On the GeoGebra mapping diagram there is an open dot on the source
line indicating that $x=-3$ and $x=0$ are in not in the domain of
$R$.
Also, when $x$ is large number in magnitude, $R(x)$ is close to 0.
So $y=0$ is a horizontal asymptote.
You can check this on the mapping diagram by moving the $x$ on the
target line to larger values and notice how the arrow point to
numbers close to 0.