How do you solve for #x# in #y = 4x - x^2#?

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Answer 1 · 2017-01-05T20:59:55

#y = -(x^2 - 4x)#

#y = -(x^2 - 4x + 4 - 4)#

#y = -(x- 2)^2 + 4#

You can solve for #x# now:

#y - 4 = -(x - 2)^2#

#4 - y = (x - 2)^2#

#+-sqrt(4 - y) = x - 2#

#2+-sqrt(y - 4) = x#

Hopefully this helps!