Question 330509
Complete the square to put the equation in vertex form, {{{y=a(x-h)^2+k}}}. where (h,k) is the vertex.
{{{f(x)=-2x^2+2x+8}}}
{{{f(x)=-2(x^2-x)+8}}}
{{{f(x)=-2(x^2-x+1/4)+8+2(1/4)}}}
{{{f(x)=-2(x-1/2)^2+17/2}}}
The vertex is ({{{1/2}}},{{{17/2}}}).
The vertex lies on the axis of symmetry {{{x=1/2}}}
The max or min value occurs at the vertex.
Since the coefficient for the {{{x^2}}} term is negative, the parabola opens downward and the vertex value is the maximum.
{{{ymax=17/2}}}
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.{{{drawing(300,300,-10,10,-10,10,circle(1/2,17/2,0.3),blue(line(1/2,-20,1/2,20)),grid(1),graph(300,300,-10,10,-10,10,-2x^2+2x+8))}}}