Example QF.FORM.3: Visualize a quadratic function $q$
with extreme value $q(1)= 2$ and $q(2) = 1$ with a
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 target line at the point / number, $-(x-1)^2
+ 2 = - x^2 + 2x + 1 $, which corresponds to the quadratic
function's value for $x$
When the point in the domain is $0$, the arrow points to $q(0) = 1$
visualizing the "Y-intercept" of $q$, the "initial value" of $q$.
The value at $x=1$ is the maximum value for the function.
If we consider $q(2)= 1$ then the difference in the value of $q$ for
a unit change in $x$ from the extreme point is $f(2)-f(1) = 1-2 =
Since a unit step from the extreme point is used in this mapping
diagram, we see the leading coefficient visualized in the gap
between the heads of consecutive arrows on the mapping