Question 469448
find the vertex of the parabola f(x)= 3x^2-16x+21
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f(x)= 3x^2-16x+21
completing the square
f(x)= 3(x^2-16x/3+(16/6)^2+21-3*(16/6)^2
f(x) =3(x-16/6)^2)+21-3*(16/6)^2
f(x)= 3(x-8/3)-21-21.33
Vertex at (8/3,-.33)
See  graph as a visual check on answer
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{{{ graph( 300, 300, -10,10, -10, 10, 3x^2-16x+21) }}}