A function will have an absolute maximum when the velocity, dy/dx, is equal to zero and the acceleration, d2y/dx2, is less than zero. It will have a y-intercept of 4 if f(x)=4 when x=0.
In this case only f(x)=-x^2+2x+4 is equal to 4 when x=0. So we know that this is the correct answer. However taking the two derivatives proves that the function has an absolute maximum..
df/dx=-2x+2 and d2f/dx2=-2
Since acceleration is a constant negative two, we know that when velocity is equal to zero, it will be at an absolute maximum for f(x) at that point.
-2x+2=0
2x=2
x=1
So velocity equals zero when x=1, thus the absolute maximum of the quadratic, the vertex in this context, is f(1).
f(1)=-1+2+4=5
So the absolute maximum is at the point (1,5)
