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

2 Answers

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

Explanation:

#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.

May 28, 2018

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

Explanation:

#"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"#