# Vertical Asymptote

1. Mar 15, 2009

### TrueStar

1. The problem statement, all variables and given/known data

Sketch the graph of (x-1)/(1-x^2).

2. Relevant equations

Vertical Asymptotes are found in the denominator.

3. The attempt at a solution

I have all I need to sketch this graph except the vertical asymptote. The 1-x^2 is throwing me off. I thought it would come out as a difference of squares, but this can also be -x^2+1.

2. Mar 15, 2009

### CompuChip

The vertical asymptote occurs where the denominator goes to zero.
Where does the function 1 - x^2 become zero?

I don't really understand your last remark. You think that 3 - 5 is something else than -5 + 3?

3. Mar 15, 2009

### TrueStar

Yes, and trying to solve for zero is confusing me. 1-x^2 is the same as -x^2+1. If I move the 1 over, I'll get -x^2=-1 and I don't want negative square roots.

Just looking at it, I think it should be 1 and -1, but I want to show my work.

4. Mar 15, 2009

### tiny-tim

You're thinking too hard

-x^2=-1 doesn't have any negative square roots, does it?

get some sleep! :zzz:​

5. Mar 15, 2009

### TrueStar

*headdesk* Nooo... it does not have negative square roots. How embarrassing.

In my defense, I've been using my weekend to study for my college algebra exam, memorize polyatomic ion names, and prepare to give a speech about the wonderful world of autoclaves.

Is there a point where one can study too much? I think I'll finish this problem up and take a break. Thank you both. :)

6. Mar 15, 2009

### tiny-tim

mmm … too much pressure!

7. Mar 15, 2009

### Staff: Mentor

Equivalently, x^2 = 1, which has two roots, one of which is negative.

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