Notice: $ h=2$ , and $k = -1$.

Draw a mapping diagram and graph of $R$ yourself, 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, $R(x) = \frac {(x-2)^2}{(x+1)^2}$. 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 $x=-1$, $x+1 =0$, since $x=-1$ is not a root of $x-2$, $x=-1$ is a pole for $R$.

You can check this by moving $x$ to $2$ and $-1$ on the GeoGebra mapping diagram.

The function $R$ is not defined at $x=-1$. On the GeoGebra graph there is a dotted vertical line indicating the vertical asymptote at $x=-1$.

On the GeoGebra mapping diagram there is an open dot on the source line indicating that $x=-1$ in not in the domain of $R$.

Since the degrees of $x-2$ and $x+1$ are both $2$, there will be only positive values close to $0$ for $R(x)$ when $x$ is close to $2$.

There will be arbitrarily large positive values of $R(x)$ when $x$ is close to $-1$.

Also, when $x$ is large number in magnitude, $R(x)$ is close to 1. So $y=1$ 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 1.

Since the degrees of $x-2$ and $x+1$ 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 $x=-1$.