Question 334220
{{{f(x)=-2x^2+2x+3}}}
Complete the square to convert to 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+1/2}}}
{{{f(x)=-2(x-1/2)^2+7/2}}}
Vertex:({{{1/2}}},{{{7/2}}})
The vertex lies on the axis of symmetry,{{{x=1/2}}}.
The vertex value is the function max or min depending on whether the parabola opens upwards or downwards.
The sign of the {{{x^2}}} term determines direction of opening.
Positive upwards, negative downwards.
Since the coefficient is {{{-2}}}, the parabola opens downwards and the vertex value is a maximum.
{{{ymax=7/2}}}
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{{{drawing(300,300,-5,5,-5,5,grid(1),circle(1/2,7/2,0.2),blue(line(1/2,-10,1/2,10)),graph(300,300,-5,5,-5,5,-2(x-1/2)^2+7/2,-2x^2+2x+3))}}}