Hi

Using the vertex form of a parabola,  where(h,k) is the vertex

f(x) = 3x^2-18x+11    |completing the square to put into vertex form

f(x) = 3(x^2-6)+11

f(x) = 3[(x-3)^2 - 9]+11

f(x) = 3(x-3)^2 - 16   Vertex is Pt(3,-16), minimum at this Pt ( a>)

 Line of symmetry is x = 3



 

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