Question 317603
The parabola is in vertex form,
{{{y=a(x-h)^2+k}}} where (h,k) is the vertex.
Comparing,
(h,k)=(-8,2)
The axis of symmetry is then {{{x=-8}}}
The coefficient of the {{{x^2}}} term is positive, so the parabola opens upward and the value at the vertex is a minimum.
{{{ymin=2}}}
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{{{drawing(300,300,-15,5,-10,10,grid(1),circle(-8,2,.3),blue(line(-8,-10,-8,10)),graph(300,300,-15,5,-10,10, (1/2)(x+8)^2+2))}}}