Find the X-intercepts for $q$ and visualize this on the mapping diagram as well.

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, $2x^2 - 4$, which corresponds to the quadratic function's value for $x$

When the point in the domain corresponds to the number $0$, the arrow points to $q(0) = -4$ visualizes the "Y-intercept" of $q$, the "initial value" of $q$.

The value at $x=0$ is the minimum value for the function.

If we consider $q(1)= -2$ then the difference in the value of $q$ for a unit change in $x$ from the extreme point is $q(1)-q(0) = -2- (-4) = 2.$

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.

The "X- intercepts" for the graph or "zeroes' for the function $q$ are the values for $a$ on the domain axis from which the arrow hits the number $q(a) =0$ on the target. For this function, those values is $a=\pm \sqrt 2$. You can check this by entering "-sqrt(2)" $=- \sqrt 2 $ and "sqrt(2) " $= \sqrt 2$ for $a = x_0$ on the GeoGebra figure.