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Vertical asymptote

  1. Sep 25, 2007 #1
    its asking me to find the value of p so that the function won't have a vertical asymptote:

    g(x) = (3x + p) / (x^3 + 8)

    however, i'm not sure how to figure that out...any help?
     
  2. jcsd
  3. Sep 25, 2007 #2
    For what value of x is the vertical asymptote? What value should the numerator take there to get rid of it?
     
  4. Sep 25, 2007 #3

    HallsofIvy

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    Do you understand what a " vertical asymptote" IS and why a graph of this function might have one? Answering this will involve factoring the denominator.
     
  5. Sep 25, 2007 #4
    after i factor the denominator though...where do i go from there?
     
  6. Sep 25, 2007 #5

    rock.freak667

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    What values of x would make the entire denominator be equal to zero??

    ..........................................

    and hence make the function tend to infinity?
     
  7. Feb 29, 2008 #6
    i know if i set x equal to 2 that the denominator will equal 0 but then the point is not to have a vertical asymptote so how do i find the value of p?
     
  8. Feb 29, 2008 #7

    Dick

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    If the numerator is zero at x=2, then you might not have a vertical asymptote.
     
  9. Feb 29, 2008 #8

    symbolipoint

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    WHAT?
    Look at the denominator. If x=-2, then x^3 + 8 = 0. The denominator makes the function undefined at x=-2, so this is the vertical asymtote.
     
  10. Feb 29, 2008 #9

    Dick

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    Sure. Sorry. x=-2. I was only looking at the OP's last post.
     
  11. Feb 29, 2008 #10

    HallsofIvy

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    You still might not have a vertical asymptote at x= -2!

    Yes, at x= -2, the denominator goes to 0. As long as the numerator has a non-zero value at x= -2, there is a vertical asymptote there. Now, what value of p will guarentee that x= -2 is NOT a vertical asymptote?
     
  12. Mar 1, 2008 #11
    Say you had a number set 1-10. X can equal any of these numbers other than 5, how would you show that?

    Hint: <,>
     
  13. Mar 1, 2008 #12

    HallsofIvy

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    Since the only possible vertical asymptote would be at x= -2, I have no idea what that has to do with this problem!
     
  14. Mar 1, 2008 #13

    symbolipoint

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    You want p to have a value? If p were to contain a factor of x^2 or higher power, then g(x) would have a slant asymtote; but that seems not exactly to be what you ask.

    HallsofIvy - help us here maybe, since we seem to not be so advanced. What are we missing?
     
  15. Mar 1, 2008 #14

    Dick

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    p=(-6)?
     
  16. Mar 1, 2008 #15

    HallsofIvy

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    Not quite, Dick, you've lost track of sign somewhere.

    At x= -2, the denominator, [itex]x^3+ 2[/itex] is equal to 0. Obviously, as I said above, as long as the numerator is not also 0, the value of the function goes to infinity as we approach x= -2: a vertical asymptote. The only way not to have a vertical asymptote is to make sure the numerator also goes to 0 at x= -2: that 3x+ p= 3(-2)+ p= -6+ p= 0 or p= 6. If p= 6, then
    [tex]\frac{3x+ 6}{x^3+ 8}= \frac{3(x+ 2)}{(x+ 2)(x^2- 2x+ 4)}= \frac{3}{x^2- 2x+ 4}[/tex]
    Rather than a vertical asymptote, we have a curve with a "hole" at (-3, 3/7).
     
    Last edited: Mar 2, 2008
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