Example QF.FORM.2:  Visualize  a quadratic function \$q\$ with leading coefficient \$A=2\$ and extreme value \$f(-1)= 3\$ with a mapping diagram.

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 = 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 diagram.

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 the target.