Answer
From the far left to the far right, the graph
- rises from $-\infty$, the lower far left,
- crosses the x axis at $(0,0)$, rising,
- turns and touches the x axis at $(5,0)$ where it
- turns back up to rise to $+\infty$ the upper far right
Work Step by Step
$P(x)=\displaystyle \frac{1}{5}x(x-5)^{2}$
End behavior:
When $ x\rightarrow-\infty$ ,
$\displaystyle \frac{1}{5}x$ is negative, $(x-2)^{2}$ is positive,
$P(x)$ is negative to the far left.
When $ x\rightarrow+\infty$ ,
$\displaystyle \frac{1}{5}x$ is positive, $(x-2)^{2}$ is positive,
$P(x)$ is positive to the far right.
Intercepts:
$ x=0,\quad x-5=0$
x-intercepts:
$x=0$, single (graph crosses x$)$
$x=5$, double (graph touvhes x and turns)$.\\\\$
y-intercept: $(0,0)$
From the far left to the far right, the graph
- rises from $-\infty$, the lower far left,
- crosses the x axis at $(0,0)$, rising,
- turns and touches the x axis at $(5,0)$ where it
- turns back up to rise to $+\infty$ the upper far right
