Hi,

If I am understanding your question properly:

f(x)= 1/3(x+6)^2 + 5

the vertex form of a parabola,  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))}}}



 

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