How do you write the quadratic in vertex form given $y=-2x^2+6x+1$ ?

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Answer 1 · 2015-05-09T02:56:56

Standard form $f(x) = - 2x^2 + 6x + 1$
Vertex form $f(x) = (x - b/(2a))^2 + f(-b/(2a))$

$-b/2a = -6/-4 = 3/2$

$f(3/2) = -2(9/4) + 6(3/2) + 1 = 11/2$

Vertex form $f(x) = (x - 3/2)^2 + 11/2$

How do you write the quadratic in vertex form given y=-2x^2+6x+1y=-2x^2+6x+1?