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Question about vertical asymptotes

  1. Oct 19, 2006 #1
    This is a example in my book

    a rational function like f(x) = [(x+2)(x-5)]/[(x-3)(x+1)] has vertical asymptotes at x=3 and x=(-1)

    Is all you do is switch the addition and subtraction signs in [(x-3)(x+1)]?

    The same equation has a horizontal asymptotes at y=1, which i cant seem to understand where it came from

    The is the best i could explain this, thank you for reading this
    Last edited: Oct 19, 2006
  2. jcsd
  3. Oct 19, 2006 #2


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  4. Oct 19, 2006 #3
    -sorry iam unable to view the flash pictures, because i cant install macromedia plugin (wont verify)
    -could someone give me a written explanation please, iam still lost
  5. Oct 20, 2006 #4


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    No, you don't ever just arbitrarily "switch signs"! A rational function has vertical asymptotes where the denominator is 0 (and the numerator isn't). The denominator of f(x) is (x-3)(x+1) which equals 0 when x- 3= 0 or x+ 1= 0. To solve x- 3= 0, add 3 to both sides: x- 3+ 3= x= 0+ 3= 3. To solve x+ 1= 0, subtract 1 from both sides: x+ 1- 1= x= 0- 1= -1. See? I didn't "switch signs" anywhere!

    I feel I should also point out that the "+" and "-" signs in
    x= +3 and x= -1 are not "addition" and "subtraction" signs- they are "positive" and "negative" signs. The difference is subtle but important. (TI calculators have separate "subtraction" and "negative" keys for that reason.)
    Last edited: Oct 20, 2006
  6. Oct 21, 2006 #5
    thanks hallsofivy it makes sense now i see what your talking about, thanks for your help
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