Question 448619
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Hi
Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
maximum is (2,8) and the start of the parabola is (0,6) ?
Standard form is {{{(x -h)^2 = 4p(y -k)}}}, where  the focus is (h,k + p) 
  y = a(x-2)^2 +8  |Using point (0,6) to solve for a
  6 = a*4 + 8
 -2 = 4a
  -1/2 = a
y = -.5(x-2)^2 +8   OR -2(y-8) = (x-2)^2 
{{{drawing(300,300,  -10,10,-10,10, grid(1),
circle(2, 8,0.3),
circle(0, 6,0.3),
graph( 300, 300,-10,10,-10,10,0,-.5(x-2)^2 +8  ))}}}