Example OAF.PFF.1 : $P(x) =2*(x-2)*(x+1)$.
Notice: $A=2, r_1=2$ , and $r_2 = -1$.
Draw a mapping diagram and graph of $P$ yourself or explore the GeoGebra figure.

Given a point / number, $x$, on the source line of the mapping diagram, there is a unique arrow  meeting the target line at the point / number, $P(x) =2*(x-2)*(x+1)$, which corresponds to the quadratic function's value for $x$

When the point in the domain is $0$, the arrow points to $f(0) = 2*(-2)*(1) = -4$ visualizing the "Y-intercept" on the graph of $P$.
The X- intercepts are the values for $a$ on the domain axis from which the arrow hits the number $0$  on the target. For this function, those values is $a=2$ and $a=-1$, the values of the roots, $r_1$ and $r_2$.
Note: From the symmetry of this quadratic polynomial, we can also see that the extreme value for P will occur midway between the X intercepts- at $x = \frac {2 + (-1)} 2 = \frac 1 2$.
You can check this as well.