Question 203916
The given equation {{{f(x) = (1/5)(x+5)^2+8}}} is in the form:
{{{f(x) = a(x-h)^2+k}}}  and in this form, the vertex is located at the point (h, k), the line of symmetry is x = h.
So we have: 
{{{a = (1/5)}}}
{{{h = -5}}} 
{{{k = 8}}}
The vertex is at (-5, 8)
The equation of line of symmetry (LOS) is: {{{x = -5}}} and
{{{f(-5) = 8}}} is a minimum because the parabola opens upward. (positive coefficient of the {{{x^2}}}-term).
The graph looks like:
{{{graph(400,400,-20,5,-5,20,(1/5)(x+5)^2+8))}}}