SOLUTION: find the vertex of the parabola f(x)= 3x^2-16x+21

Question 469448: find the vertex of the parabola f(x)= 3x^2-16x+21
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
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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