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)





 

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