Question 391016
  <pre><font size = 3 color = "indigo"><b>
Hi
parabola with vertex at (1/2,-2/33) and x-inercepts of (0,0) and (1,0)
Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
 y = a(x - 1/2)^2 - 2/33  using Pt(0,0) to solve for a
 0 = a(-1/2)^2 - 2/33
 2/33 = (1/4)a
  8/33 = a
 y = 8/33(x-1/2)^2 - 2/33
{{{drawing(300,300, -6, 6, -6, 6, grid(1),
circle(0, 0,0.3),
circle(0, -2/33,0.3),
circle(1, 0,0.2),
graph( 300, 300, -6, 6, -6, 6,(8/33)(x-1/2)^2 - 2/33 ))}}}