Question 362551
I'm going to assume that this is your function.
{{{f(x)=(x+2)^2+8}}}
The quadratic is already in vertex form, {{{y=a(x-h)^2+k}}}
where ({{{h}}},{{{k}}}) is the vertex.
Comparing,
({{{h}}},{{{k}}})=({{{-2}}},{{{8}}})
The vertex lies on the axis of symmetry, {{{x=-2}}}.
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{{{drawing(300,300,-10,10,-2,18,grid(1),circle(-2,8,0.3),blue(line(-2,20,-2,-20)),graph(300,300,-10,10,-2,18,(x+2)^2+8))}}}