Make $x$ $x$ the subject of $y = {(x - 4)}^{2} + 6$ $y= ( x - 4) ^ { 2} + 6$ ?

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Answer 1 · 2016-12-03T13:38:47

$x = \pm \sqrt{y - 6} + 4$ $x = +-sqrt(y-6) +4$

$y = {(x - 4)}^{2} + 6$ $y=(x-4)^2 +6$

${(x - 4)}^{2} = y - 6$ $(x-4)^2=y-6$

Take the square root of both sides:

$\sqrt{{(x - 4)}^{2}} = \sqrt{y - 6}$ $sqrt((x-4)^2) = sqrt(y-6)$

$x - 4 = \pm \sqrt{y - 6}$ $x-4= +- sqrt(y-6)$

Add $4$ $4$ on both sides of the equation

$x = \pm \sqrt{y - 6} + 4$ $x= +-sqrt(y-6) +4$

Make xx the subject of y=(x−4)2+6y= ( x - 4) ^ { 2} + 6?