How do you convert $y = - 2 x^{2} + 4 x - 5$ $y = -2x^2 + 4x - 5$ into vertex form?

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Answer 1 · 2016-11-03T13:03:25

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

Simply complete the square.

$y = - 2 (x^{2} - 2 x) - 5$ $y= -2(x^2 - 2x) - 5$

$y = - 2 (x^{2} - 2 x + 1 - 1) - 5$ $y = -2(x^2 - 2x + 1 - 1) - 5$

$y = - 2 (x^{2} - 2 x + 1) + 2 - 5$ $y = -2(x^2 - 2x + 1) + 2 - 5$

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

Hopefully this helps!

How do you convert y=−2x2+4x−5y = -2x^2 + 4x - 5 into vertex form?