Question 352607
Complete the square to put the quadratic equation into vertex form, {{{y=a(x-h)^2+k}}} where ({{{h}}},{{{k}}}) is the vertex.
{{{f(x)=-2x^2+2x+3}}}
{{{f(x)=-2(x^2-x)+3}}}
{{{f(x)=-2(x^2-x+1/4)+3+2(1/4)}}}
{{{f(x)=-2(x-1/2)^2+7/2}}}
So the vertex is ({{{1/2}}},{{{7/2}}}).
The vertex lies on the axis of symmetry, {{{x=1/2}}}
The value at the vertex is either a minimum (when {{{a>0}}}) or a maximum (when {{{a<0}}}).
In this case, {{{a=-2}}}.
{{{y[max]=7/2}}}
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{{{drawing(300,300,-5,5,-5,5,grid(1),blue(line(1/2,-10,1/2,10)),circle(1/2,7/2,0.2),graph(300,300,-5,5,-5,5,-2x^2+2x+3))}}}