What is the vertex form of $y= 3x^2+x-55$ ?

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Answer 1 · 2017-03-02T09:14:37

$y = 3 x^2 + x - 55$ has a minimum $-661/12$ at $(-1/6, -661/12)$

$y = 3 x^2 + x - 55$

$y = [3(x^2 + x/3)] - 55$

solve using completing a square,
$y = [3(x + 1/6)^2 - 3*(1/6)^2] - 55$

$y = 3(x + 1/6)^2 - 3*(1/36) - 55$

$y = 3(x + 1/6)^2 - 1/12 - 55$

$y = 3(x + 1/6)^2 - 661/12$

Therefore,
$y = 3 x^2 + x - 55$ has a minimum $-661/12$ at $(-1/6, -661/12)$

What is the vertex form of y= 3x^2+x-55y= 3x^2+x-55?