How do you write the quadratic $y=2x^2 +11x-6$ in vertex form?

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Answer 1 · 2015-04-29T14:59:50

The vertex form of a parabola is
$y = k(x-a)^2 +b$
where the vertex is $(a,b)$

$y=2x^2+11x-6$

$y= 2(x^2+11/2x+(11/4)^2) -2(11/4)^2 -6$

$y = 2(x-(-11/4))^2 +(-169/8)$

How do you write the quadratic y=2x^2 +11x-6y=2x^2 +11x-6 in vertex form?