y=-x^2+4x-4
dy/dx=-2x+4 and d2y/dx2=-2
Since acceleration, d2y/dx2, is a constant negative, when velocity, dy/dx=0, it will be at the point when y is at an absolute maximum...
dy/dx=0 only when -2x+4=0, 2x=4, x=2, and the maximum is y(2), in this case:
y(2)=-x^2+4x-4
y(2)=-4+8-4=0
The absolute maximum is 0.
...
y=0.5x^2+8x+5
dy/dx=x+8, d2y/dx2=1
Acceleration is a constant positive, so when dy/dx=0, y(x) will be at an absolute minimum...
dy/dx=0 only when x+8=0, x=-8, then the absolute minimum will be at y(-8):
y=0.5x^2+8x+5
y(-8)=32-64+5=-27
So the minimum value for y is -27
