Question 380723
  <pre><font size = 3 color = "indigo"><b>
Hi,
If I am understanding your question properly:
f(x)= 1/3(x+6)^2 + 5
the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
vertex is Pt(-6,5), line of symetry is x = -6
a = 1/3, is a positive nunber, therefore the parabola opens upward:
Minimum value of f(x) is 5
graph( 300, 300, -6, 6, -6, 6))}}}
{{{drawing(300,300, -10,10,-10,10,blue(line(-6,10,-6,-10)),
 grid(1),
circle(-6, 5,0.4),
graph( 300, 300, -10,10,-10,10,.33(x+6)^2 + 5))}}}