How do you write $y=-3(x-2)^2+8$ into standard form?

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Answer 1 · 2018-05-28T14:37:36

$color(indigo)(y = -3x^2 + 12x - 4$

$y = -3 * (x - 2)^2+8$

$y = -3 * (x^2 - 4x + 4)+8$ , expanding.

$y = -3x^2 + 12x - 12+8$ , removing bracket and multiplying.

$y = -3x^2 + 12x - 4$ , final result.

Answer 2 · 2018-05-28T14:55:20

$y=-3x^2+12x-4$

$"expand "(x-2)^2=x^2-4x+4$

$y=-3(x^2-4x+4)+8$

$color(white)(y)=-3x^2+12x-12+8$

$color(white)(y)=-3x^2+12x-4larrcolor(blue)"in standard form"$

How do you write y=-3(x-2)^2+8y=-3(x-2)^2+8 into standard form?