Answer
From the far left to the far right, the graph
- falls from $+\infty$, the upper far left,
- crosses the x axis at $(-2,0)$, falling,
- turns and touches the x axis at $(-1,0)$ where it
- turns down to fall through $(0,-6)$, keeps falling,
- turns and crosses the x axis at $(\displaystyle \frac{3}{2},0)$,
- and continues to rise to the far right.
Work Step by Step
$P(x)=(x+2)(x+1)^{2}(2x-3)$
End behavior:
When $ x\rightarrow-\infty$ ,
$(x+2)$ and $(2x-3)$ are negative, $(x+1)^{2}$ is positive,
$P(x)$ is positive to the far left.
When $ x\rightarrow+\infty$ , all factors are positive,
$P(x)$ is positive to the far right.
Intercepts:
$ x+2=0,\quad x+1=0\quad 2x-3=0$
x-intercepts:
$x=-2$, single (graph crosses x$)$
$x=-1$, double (graph touches x and turns)$.\\\\$
$x=3/2$, single (graph crosses x$)$
y-intercept: $P(0)=(2)(1)(-3)=-6$
$P(1)=(3)(2^{2})(-1)=-12$, so the graph falls through $(0,-6)$
From the far left to the far right, the graph
- falls from $+\infty$, the upper far left,
- crosses the x axis at $(-2,0)$, falling,
- turns and touches the x axis at $(-1,0)$ where it
- turns down to fall through $(0,-6)$, keeps faling,
- turns and crosses the x axis at $(\displaystyle \frac{3}{2},0)$,
- and continues to rise to the far right.
