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!

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