Question 286355
All you really need to know is where the roots of the equation are, and what the shape of the curve is.

First, we set y=0:

{{{0=(1+x)^2(2-x)}}}

So, either: {{{(1+x)^2=0}}} or {{{2-x)=0}}}
x=-1,x=-1.  There is a double root at x= -1.

if 2-x=0, then x=2. There is a single root at x=2.

Either one can be true because any number multiplied by zero is zero.

This is a negative cubic (because of the 2-x term). which means it starts in Quadrant 4 (the bottom-right) and moves up and left, to Quadrant 2 (top-left). It crosses the x-axis once at x=2, and then it comes down to touch the x-axis at x=2.

I don't know how to post an image in here, so if you want to see a sketch of it,  send me a message.