Hi,
Find the vertex of the graph
f(x)=-16x^2+8x+3
Note: the vertex form of a parabola,
where(h,k) is the vertex
f(x)=-16x^2+8x+3
f(x)=-16[x^2-x/2 - 3/16]
f(x)=-16[(x -1/4)^2 - 1/16 - 3/16]
f(x)=-16[(x -1/4)^2 - 4/16]
f(x)=-16(x -1/4)^2 + 4
vertex is (1/4,4)
