Question 408223
  <pre><font size = 3 color = "indigo"><b>
Hi
Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
f(x)= -10x^2 + 5x + 6   |putting into the vertex form by completing the square
f(x)= -10[x^2 - 1/2x] + 6 
f)x)= -10[(x - 1/4)^2 - 1/16] + 6 
f)x)= -10(x - 1/4)^2 - 10/16 + 6 
f)x)= -10(x - 1/4)^2 + 10/16 + 6 
f)x)= -10(x - 1/4)^2 + 106/16   |vertex is Pt(1/4, 53/8)
{{{drawing(300,300, -10,10,-10,10,
 grid(1),
circle(.25, 6.625,0.4),
graph( 300, 300, -10,10,-10,10,-10x^2 + 5x + 6))}}}