Question 1058117
I will assume f(x)=1(x+2)^2*(x-2)
(x+2) with multiplicity 2 (it will bounce on the x-axis)
The roots are -2 with multiplicity 2 and 2.
The polynomial is x^3+2x^2-4x-8 as I have written the above.  I don't know if the 1 was intended, but I can't solve for a if I don't know one point on the graph, so I will continue with this.  The domain is all x and the range is the same.  If I substitute a point other than the roots, like 0, f(0)=-8; (0,-8) is a point.
The derivative is 3x^2+4x-4
Set that equal to 0
(3x-2)(x+2)=0
x=(2/3) and x=-2.  Those are critical values.  f(-2)=0; f(2/3)=-256/27
To the left of -2 the function is negative, and to the right of -2 is negative, so -2 is a maximum. 
Pick x=1/2, and (5/2)^2*(-3/2) is -9 3/8, which is slightly larger than -256/27.  f(1)=-9, which is larger than -256/27, so x=2/3 is a local minimum. 
Second derivative is 6x+4. It is positive when x is greater than -2/3 and negative when x is less than -2/3, so it is convex up for the x=-2 (negative value) and convex down for x=2/3 (positive value) 

{{{graph(300,300,-10,10,-10,10,x^3+2x^2-4x-8)}}}