Question 334236
Convert to vertex form,{{{y=a(x-h)^2+k}}} where (h,k) is the vertex.
{{{y=-x^2+12x+3}}}
{{{y=-(x^2-12x+36)+3+36}}}
{{{y=-(x-6)^2+39}}}
The vertex is ({{{6}}},{{{39}}}).
The vertex lies on the axis of symmetry, {{{x=6}}}
The parabola opens downwards since the coefficient of the {{{x^2}}} term is negative and the vertex value is a maximum.
{{{ymax=39}}}
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{{{drawing(300,300,-2,12,-10,40,grid(1),circle(6,39,.3),blue(line(6,40,6,-40)),graph(300,300,-2,12,-10,40,-x^2+12x+3,-(x-6)^2+39))}}}