Example QF.FORM.4 : Suppose $f$ is a quadratic function with leading coefficient $=A=2$ and extreme value $f(1) = 3$. Find the vertex form of the quadratic function. Visualize $f$ with a mapping diagram that illustrates the function as the composition of the four core functions  $f_{+ 3}\circ f_{*2}\circ q\circ f_{ -1}$.
Note the arrow for $f(1) =3$ and based on $A=2$, $f(2)=5$ and $f(0)=5$ are visualized by the spacing of the arrows in the mapping diagram.

Solution: We use the vertex form with $A=2$ and $f(1)=3$ to obtain $f(x) = 3 + 2*(x-1)^2.$

Draw a mapping diagram yourself or use the GeoGebra figure.
Given a point / number, $x$, on the source line, there is a unique arrow meeting the final target line at the point / number, $3 + 2*(x-1)^2 = 2x^2 -4x + 5$,  which corresponds to the function's value for $x$

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