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Vertical Asymptote

  1. Mar 15, 2009 #1
    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. jcsd
  3. Mar 15, 2009 #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?
     
  4. Mar 15, 2009 #3
    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.
     
  5. Mar 15, 2009 #4

    tiny-tim

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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:​
     
  6. Mar 15, 2009 #5
    *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. :)
     
  7. Mar 15, 2009 #6

    tiny-tim

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    mmm … too much pressure! :rolleyes:
     
  8. Mar 15, 2009 #7

    Mark44

    Staff: Mentor

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