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

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, $R(x) = \frac {x-2}{(x+1)^3}$. 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.

When the point in the domain is $0$, the arrow points to $R(0) = \frac {(-2)^2}{1^3} = 4$ visualizing the "Y-intercept" on the graph of $R$.

The X- intercepts are the values for $a$ on the domain axis from which the arrow hits the number $0$ on the target. For this function, that value is $a=2$, the value of $h$.

Since when $x=-1$, $x+1 =0$, $R$ is not defined at $x=-1$. 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.

You can change the power of the denominator to explore the effect of that being even or odd.