Question 323970
Convert to vertex form, {{{y=a(x-h)^2+k}}} by completing the square.
{{{f(x)=-5x^2-10x + 6}}}
{{{f(x)=-5(x^2+2x)+6}}}
{{{f(x)=-5(x^2+2x+1)+6+5}}}
{{{f(x)=-5(x+1)^2+11}}}
The vertex occurs at (-1,11).
The coefficient for the {{{x^2}}} term is negative and the parabola opens downward, the value at the vertex is the maximum.
{{{ymax=11}}}
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{{{drawing(300,300,-5,5,-2,12,grid(1),circle(-1,11,0.2),graph(300,300,-5,5,-2,12, -5x^2-10x+6))}}}