The graph of which function will have a maximum and a y-intercept of 4? f(x) = 4x2 + 6x – 1 f(x) = –4x2 + 8x + 5 f(x) = –x2 + 2x + 4 f(x) = x2 + 4x – 4

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)

The correct answer is d, hope this