Question 359576
Don't be lost, study the "vertex form".
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The equation is already in vertex form, {{{y=a(x-h)^2+k}}}  where (h,k) is the vertex.
1) Comparing, (h,k)=(-8,2)
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2) The vertex lies on the axis of symmetry, {{{x=-8}}}
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3) Since {{{a=1/5}}} and {{{a>0}}}, then the parabola opens upwards and the value at the vertex ({{{k}}}) is a minimum. 
{{{f[min]=2}}}
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4) From 3), it's a minimum.
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{{{drawing(300,300,-12,2,-10,10,grid(1),circle(-8,2,0.2),blue(line(-8,-20,-8,20)),graph(300,300,-12,2,-10,10,(1/5)(x+8)^2+2))}}}