-x^2+4x-4
dy/dx=-2x+4, d2y/dx2=-2
Since acceleration is a constant negative, when dy/dx, velocity, equals zero, it is at the absolute maximum value for f(x).
dy/dx=0 when -2x+4=0, 2x=4, x=2
f(2)=-x^2+4x-4=-4+8-4=0
So there is a maximum of zero
...
.5x^2+8x+5
dy/dx=x+8, d2y/dx2=1
Since d2y/dx2, acceleration is a constant positive, when dy/dx=0, it is at the point when f(x) is at an absolute minimum.
dy/dx=0 when x+8=0, x=-8
f(-8)=.5x^2+8x+5=32-64+5=-27
So there is a absolute minimum of -27
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-2(x+1)^2+5
dy/dx=-4(x+1)=-4x-4, d2y/dx2=-4
dy/dx=9 when -4x-4=0, -4x=4, x=-1
f(-1)=5
So the vertex is (-1, 5) and the axis of symmetry is the vertical line x=-1
