Question 379471
The equation is already in vertex form, {{{y=a(x-h)^2+k}}}.
Vertex : ({{{-2}}},{{{-4}}})
The vertex lies on the axis of symmetry {{{x=-2}}}.
Since {{{a=-1<0}}}, the value at the vertex is a maximum.
{{{y[max]=-4}}}
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{{{drawing(300,300,-8,6,-12,2,grid(1),circle(-2,-4,0.2),blue(line(-2,-100,-2,100)),graph(300,300,-8,6,-12,2,0,-(x+2)^2-4))}}}