Hi

f(x) = 3x^2-18x+14

Finding vertex

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

f(x) = 3x^2-18x+14

  |completing square to put into vertex form

f(x) = 3[(x-3)^2 -9] + 14

f(x) = 3(x-3)^2 -13  Vertex is Pt(3,-13) Line of symmetry is x = 3

Vertex Pt(3,-13) is where the minimum value occurs: f(3) = -13



 

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