Question 371436
  <pre><font size = 3 color = "indigo"><b>
Hi,
Find the vertex of the graph
f(x)=-16x^2+8x+3

Note: the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} 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)

{{{drawing(300,300, -6, 6, -6, 6, grid(1),
graph( 300, 300,-6,6,-6,6,-16x^2+8x+3)) }}}