How do you write $y = - 4 x^{2} + 8 x - 1$ $y= -4x^2+8x-1$ into vertex form?

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Answer 1 · 2015-05-23T13:38:38

General vertex form for a parabolic equation is
$y = m {(x - a)}^{2} + b$ $y=m(x-a)^2 +b$
where the vertex is at $(a, b)$ $(a,b)$

Given $y = - 4 x^{2} + 8 x - 1$ $y =-4x^2+8x-1$

Extract $m$ $m$
$y = (- 4) (x^{2} - 2 x) - 1$ $y = (-4)(x^2-2x) -1$

Complete the square
$y = (- 4) (x^{2} - 2 x + 1) - 1 + 4$ $y=color(blue)((-4))(x^2-2xcolor(blue)(+1)) - 1 color(blue)(+4)$

$y = (- 4) (x - 1) + 3$ $y= (-4)(x-1)+3$
is in vertex form (with the vertex at $(1, 3)$ $(1,3)$ )

How do you write y=−4x2+8x−1y= -4x^2+8x-1 into vertex form?