Question 395567
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Hi
f(x) = 3x^2-18x+14
Finding vertex
Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} 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
{{{drawing(300,300, -15,15,-15,15 ,blue(line(3,15,3,-15))   grid(1),
circle(3, -13,0.5),
graph( 300, 300, -15,15,-15,15, 3x^2-18x+14
))}}}