Example OAF.COMP.3 Suppose $f$ is a linear fractional function with $f(0) = -1$, pole at $x=1$ and $f(2)=3$.
Verify that $f =f_{+1}∘f_{∗2}∘R∘f_{−1}$.
Visualize $f$ with a mapping diagram that illustrates the function as the composition: $f = f_{+1}∘f_{∗2}∘R∘f_{−1}$.

Solution: We use the pole to see immediately that $h=1$. Then $f(0) =-1 = \frac A {-1} +k$ while $f(2)=3 = \frac A 1 +k$. Simple algebra then confirms that $2k=2, k=1,$ and $A=2$ to obtain $f(x) =\frac 2 {x-1} + 1 = f_{+1}∘f_{∗2}∘R∘f_{−1}(x).$

Draw a mapping diagram yourself or consider the GeoGebra figures below.
Given a point / number, $x$, on the source line, there is a unique arrow  meeting the target line at the point / number, $\frac 2 {x-1} + 1 = \frac {x + 1} {x-1}$,  which corresponds to the function's value for $x$

When the point in the domain is $0$, the black arrow points to $f(0) = -1$ visualizing the point $(0,-1)$ on the graph of $f$. We also have $f_{+1} \circ f_{*2} \circ R\circ f_{ -1} (2) = f_{+1} \circ f_{*2}\circ R(1)= f_{+1}(2) = 3$   which is visualized in the mapping diagram for the composition by arrows $2 \rightarrow 1 \rightarrow 1 \rightarrow 2 \rightarrow 3$ .