How do you find the vertex of a parabola $f(x) = 3x^2 + 6x + 1$ ?

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Answer 1 · 2015-07-04T23:30:46

$f(x) = 3x^2 + 6x + 1$

x-coordinate of vertex: $x = (-b/(2a)) = -6/6 = - 1$

y-coordinate of vertex: y = f(-1) = 3 - 6 + 1 = - 2

How do you find the vertex of a parabola f(x) = 3x^2 + 6x + 1f(x) = 3x^2 + 6x + 1?