Question 609010
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Hi
Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
h(x)=x^2-6x+9    |Completing the Square
h(x)=(x-3)^2 -9 +9
h(x)= (x-3)^2     |a = 1 >0, parabola open Upward 
Vertex(-3,0) also the x-intercept
y-intecept h(x) = 9, when x = 0
always recommend graphing or 'picturing' the graph in Your mind.
{{{drawing(300,300,  -10,10,-10,10,   grid(1),
circle(3, 0,0.3),
circle(0, 9,0.3),
graph( 300, 300, -10,10,-10,10,  0,(x-3)^2 ))}}}