Question 394988
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Hi
Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
{{{f(x)=(1/5)(x+4)^2+8}}}  vertex is Pt(-4,8) Line of symmetry x = -4
minimum value of f(x) = 8  
Vertex (-4,8) is a minimum  |parabola opens upward as a = 1/5  > 0
{{{drawing(300,300, -10,20,-1,20,blue(line(-4,20,-4,-20)),   
 grid(1),
circle(-4, 8,0.4),
graph( 300, 300, -10,20,-1,20,(1/5)(x+4)^2+8))}}}