What is the vertex form of $y= 6x^2+x-2$ ?

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Answer 1 · 2017-03-29T05:57:34

minimum vertex at $-49/24$ and symetry at $x = - 1/12$

it can be solve by using completing a square.

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

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

$y = 6(x + 1/12)^2 - 1/24 -48/24$

$y = 6(x + 1/12)^2 - 49/24$

since coefficient of $(x + 1/12)^2$ is +ve value, it has a minimum vertex at $-49/24$ and it symetry at $x = - 1/12$

What is the vertex form of y= 6x^2+x-2 y= 6x^2+x-2 ?