Example QF.FORM.2: Visualize a quadratic function $q$
with leading coefficient $A=2$ and extreme value $f(-1)= 3$ 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, $2(x+1)^2
+ 3 = 2x^2 + 4x + 5 $, which corresponds to the quadratic function's
value for $x$
When the point in the domain is $0$, the arrow points to $q(0) = 5$
visualizing the "Y-intercept" of $q$, the "initial value" of $q$.
The value at $x=-1$ is the minimum value for the function.
If we consider $q(0)= 5$ then the difference in the value of $q$ for
a unit change in $x$ from the extreme point is $f(0)-f(-1) = 5-3 =
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
Since the minimum value of $q$ is 3, there are no values for $a$ on
the domain axis from which the arrow hits the number $0$ on