Find inverse of a function

**freestar** · September 18th 2011, 01:57 PM

$y=\sqrt{x^2-2x}$

I figure the restriction on the domain would be $x\geq2$ or $x\leq0$

$x = \sqrt{y^2-2y}$

$x^2 = y(y-2)$

It's driving me insane. How do I isolate the 'y'?!?

**TheChaz** · September 18th 2011, 02:03 PM

Quadratic formula on y.
y^ - 2y - x^2 = 0
a = 1
b = -2
c = -x^2

**skeeter** · September 18th 2011, 02:11 PM

Originally Posted by freestar

$y=\sqrt{x^2-2x}$

I figure the restriction on the domain would be $x\geq2$ or $x\leq0$

$x = \sqrt{y^2-2y}$

$x^2 = y(y-2)$

It's driving me insane. How do I isolate the 'y'?!?

$x^2 = y^2 - 2y$

$x^2 + 1 = y^2 - 2y + 1$

$x^2 + 1 = (y-1)^2$

$\pm \sqrt{x^2+1} = y - 1$

$y = 1 \pm \sqrt{x^2+1}$

now all you have to do is figure out if its plus, or minus

**freestar** · September 18th 2011, 02:52 PM

ahh completing the square. Thanks a lot!

**TheChaz** · September 18th 2011, 03:04 PM

Originally Posted by freestar

ahh completing the square. And the quadratic formula works too! You guys are geniuses.
Thanks a lot!

I couldn't have said it better myself. You're welcome!

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