Question 919341
the vertex form of a Parabola opening up(a>0) or down(a<0), 
{{{y=a(x-h)^2 +k}}}. V(h, k)
 Completing the Square to Obtain the Vertex Form:
y = ax^2 + bx + c = 0
y = x^2-5x+6 = 0  {{{b/(-2a) = (-5)/(-2) = 5/2}}} 5/2 the x-value for the Vertex
y = (x - 5/2)^2 - (5/2)^2 + 6 = 0
y = (x - 5/2)^2 - 25/4 + 24/4= 0   {{{-a(b/(-2a))^2 }}}= -25/4
y = (x - 5/2)^2 - 1/4 = 0
V(5/2, -1/4)

{{{drawing(300,300,   -6, 6, -6, 6, grid(1), blue(line(2.5,6,2.5,-6)) ,
circle(2.5, -.25,0.2),
graph( 300, 300, -6, 6, -6, 6,0, x^2-5x+6) )}}}