f(x) = -2(x-2)²(x+1)

To find the y-intercept substitute 0 for x

f(0) = -2(0-2)²(0+1)
f(0) = -2(-2)²(1)
f(0) = -2(4)
f(0) = -8

y-intercept (0,-8)

To find the x-intercept substitute 0 for f(x),
and remember that y and f(x) are the same thing.

f(x) = y = -2(x-2)²(x+1)
       0 = -2(x-2)²(x+1)
    
Use the zero-factor property:

   -2(x-2)² = 0           x+1 = 0
     (x-2)² =      x = -1
 x-2=0  x-2=0  
   x=2    x=2

So the x-intercepts are (2,0) and (-1,0)

The x-value of the x-intercept 2 is a "double-zero" 
since we got it twice. So the correct terminology is 
"2 is a zero of multiplicity 2", which means
that the graph just touches the x-axis ("bounces off")
the x-axis, and does not cut through the x-axis.
 
The x-value of the x-intercept -1 is a "single-zero" 
since we got it only once. So the correct terminology is 
"2 is a zero of multiplicity 1", which means
that the graph cuts through the x-axis at that point.

We plot those intercepts, and use the fact that it cuts through the x-axis
at -1.  It has to go down through (-1,0) to get to the y-intercept, (0,-8),
then it has to come back up to the x-axis and bounce off the x-axis at (2,0),
and then go back down.



Edwin