Answer
From the far left to the far right, the graph
- rises from $-\infty$, the lower far left,
- to the point $(-1,0)$, where it touches x and turns back down,
- falling below $=-2,$ turns and
- crosses the y-axis at $(0,-2) $rising
- flattens out near $(1,0)$ where it crosses the x-axis,
- turns back down and crosses x at $(2,0)$
- and continues falling to $-\infty$ the lower far right
Work Step by Step
$P(x)=-(x+1)^{2}(x-1)^{3}(x-2)$
End behavior:
When $ x\rightarrow-\infty$ ,
$(x-1)^{3}$ and $(x-2)$ are negative, $(x+1)^{2}$ is positive,
$P(x)$ is negative to the far left.
When $ x\rightarrow+\infty$ , we have $-($all positive$)$
$P(x)$ is negative to the far right.
Intercepts:
$ x+1=0,\quad x-1=0\quad x-2=0$
x-intercepts:
$x=-1$, double (graph touches x and turns)$.\\\\$
$ x=1,$ triple (graph flattens out at the crossing)
$x=2$, single (graph crosses x$)$
y-intercept: $P(0)=-(1)(-1)(-2)=-2$
$P(0.5)\approx-0,4$,
so the graph rises through $(0,-2)$
From the far left to the far right, the graph
- rises from $-\infty$, the lower far left,
- to the point $(-1,0)$, where it touches x and turns back down,
- falling below $=-2,$ turns and
- crosses the y-axis at $(0,-2) $rising
- flattens out near $(1,0)$ where it crosses the x-axis,
- turns back down and crosses x at $(2,0)$
- and continues falling to $-\infty$, the lower far right
