Question 380346
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Hi,
graphing 
-x^2+12x=36
 x^2 -12x +36 = 0  
 factoring
 (x-6)(x-6) = 0 Note:SUM of the inner product(-6x) and the outer product(-6x) = -12x
 (x-6)= 0
 (x-6) = 0
x = 6  Pt(6,0) the x intercept
 x^2 -12x +36 = (x-6)^2  
the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
here, the vertex is at Pt(6,0) a = 1 (a is positive, graph opens upward)
{{{drawing(300,300, -10, 10, -10, 10,  grid(1),
circle(6, 0,0.3),
graph( 300, 300,  -10, 10, -10, 10, x^2 -12x +36))}}}