Notice: $A=\frac 1 2, r_1=2$ , and $r_2= -1$ and $g_P(x)=x^2 + 1$.

Draw a mapping diagram and graph of $P$ yourself or explore the GeoGebra figure.

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

When the point in the domain is $0$, the arrow points to $f(0) = \frac 1 2 *(1)*(-2)*(1) = -1$ 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 real roots, $r_1$ and $r_2$.