Vertical Asymptote

  • Thread starter TrueStar
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  • #1
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Homework Statement



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


Homework Equations



Vertical Asymptotes are found in the denominator.


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.
 

Answers and Replies

  • #2
CompuChip
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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
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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
tiny-tim
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If I move the 1 over, I'll get -x^2=-1 and I don't want negative square roots.

You're thinking too hard

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

get some sleep! :zzz:​
 
  • #5
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*headdesk* Nooo... it does not have negative square roots. :frown: 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
tiny-tim
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  • #7
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You're thinking too hard

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

get some sleep! :zzz:​

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

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