Answer
From the far left to the far right, the graph
- falls from $+\infty$, the upper far left,
- touches the x axis at $(-2,0)$ where it
- turns up to rise through $(0,3)$, keeps rising,
- turns and touches the x axis at $(3,0)$, where it turns back up,
- and continues to rise to the far right.
Work Step by Step
$P(x)=\displaystyle \frac{1}{12}(x+2)^{2}(x-3)^{2}$
End behavior:
When $ x\rightarrow-\infty$ , both squared factors are positive,
$P(x)$ is positive to the far left.
When $ x\rightarrow+\infty$ , , both squared factors are positive,
$P(x)$ is positive to the far right.
Intercepts:
$ x+2=0,\quad x-3=0$
x-intercepts:
$x=-2$, double (graph touches x and turns)$.\\\\$
$x=3$, double (graph touches x and turns)$.\\\\$
y-intercept: $P(0)=\displaystyle \frac{1}{12}(2^{2})(-3^{2})=3$
$P(0.5)\approx 3.26$, so the graph rises through $(0,3)$
From the far left to the far right, the graph
- falls from $+\infty$, the upper far left,
- touches the x axis at $(-2,0)$ where it
- turns up to rise through $(0,3)$, keeps rising,
- turns and touches the x axis at $(3,0)$, where it turns back up,
- and continues to rise to the far right.
