An expression representing the following function...

**VectorRun** · September 24th 2010, 06:51 PM

So I'm stuck on this question: find an expression for the function whose graph is the bottom half of the parabola, $x+(y-3)^2=0$ . It should be expressed in terms of x according to the text that I'm working from.

The problem I'm having is actually isolating y without encountering a negative square root...Now, I haven't learned how to deal with complex numbers much so I'm not sure if that's the only way to do this. If it's possible to do this normally, could I have some hints?

**Prove It** · September 24th 2010, 06:53 PM

If you're expressing it in terms of $x$ then

$x = -(y - 3)^2$ .

**VectorRun** · September 24th 2010, 06:56 PM

Oops, I meant expressing it as y in terms of x so that's it's $y=$

**Prove It** · September 24th 2010, 07:23 PM

$(y - 3)^2 = -x$

$y - 3 = \pm \sqrt{-x}$

$y = 3 \pm \sqrt{-x}$ .

If you want the lower half then only accept

$y = 3 - \sqrt{-x}$ .

**VectorRun** · September 24th 2010, 07:28 PM

Oh, is that really an acceptable answer? I got that but I wasn't sure if it was right due to the negative radicand...I guess I got thrown off by all the teachers always saying it's a big no-no. Thanks a lot.

**Prove It** · September 24th 2010, 07:29 PM

Is $-x$ always a negative number?

**VectorRun** · September 24th 2010, 07:47 PM

I know it isn't when it's either 0 or a negative number but I was only thinking about the notation so I didn't really concern myself with anything else...Plus, the question didn't really allow us to write a domain.

**Prove It** · September 24th 2010, 07:50 PM

Originally Posted by VectorRun

I know it isn't when it's either 0 or a negative number but I was only thinking about the notation so I didn't really concern myself with anything else...Plus, the question didn't really allow us to write a domain.

Doesn't matter, there's such a thing as an implied domain. There's nothing wrong with the notation as long as it's possible for any value of $x$ .

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