How do you write $y = 3 x^{2} - 6 x + 1$ $y=3x^2-6x+1$ into vertex form?

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Answer 1 · 2015-05-03T18:30:44

Vertex form for a quadratic is
$y = m (x - a) + b$ $y = m(x-a)+b$
(where $(a, b)$ $(a,b)$ is the vertex)

$y = 3 x^{2} - 6 x + 1$ $y=3x^2-6x+1$

$y = 3 (x^{2} - 2 x) + 1 extracting the m constant$ $y=3(x^2-2x)+1 " extracting the "m" constant"$

#y=3(x^2-2x+1) -3 +1 " completing the square"#

$y = 3 {(x - 1)}^{2} - 2 simplification$ $y=3(x-1)^2 -2 " simplification"$

$y = 3 {(x - 1)}^{2} + (- 2) into completely vertex form$ $y=3(x-1)^2+(-2)" into completely vertex form"$

How do you write y=3x2−6x+1y=3x^2-6x+1 into vertex form?