Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

    Do you mean, a horizontal asymptote? Here, this rough sketch might help: http://archives.math.utk.edu/visual.calculus/1/horizontal.5/index.html" [Broken].
    Last edited by a moderator: May 2, 2017
  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


    User Avatar
    Science Advisor

    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 by a moderator: Oct 20, 2006
  6. Oct 21, 2006 #5
    thanks hallsofivy it makes sense now i see what your talking about, thanks for your help
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook