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Do polynomials have asymptotes?

  1. Apr 7, 2006 #1
    First of all, do polynomials have asymptotes, including oblique ones? I know that rational functions have asymptotes, and it seems that most, if not all, of my book's exercises on this lesson contain only rational functions. So do polynomials have asymptotes? and if so, how do I determine them?

    for example, let f(x) = 2 - 15x + 9x2 - x3
    factoring, i get: f(x) = (-1)(x - 2)(x2 + 7x + 1)

    as x approaches posititve or negative infinity, f increases/decreases without bound. So I know that f does not have any horizontal asymptotes. And I'm guessing that because f is defined everywhere, f does not have any vertical asymptotes... As for oblique asymptotes, I have no idea how to find them out.

    so after all this reasoning, I'm beginning to think that polynomials do not have any asymptotes...but I wouldn't be surprised if I'm totally wrong either! :tongue2:
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  3. Apr 7, 2006 #2


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    It does depend on exactly what you mean by "asymptote". For example, in some sense, -x³ is an asymptote to 2 - 15x + 9x² - x³. And in some other sense, the vertical line x=0 is an asymptote to x².

    But let's ignore these other meanings for now!

    You have a conjecture, and the first thing you need to do is to figure out exactly what you want to prove!

    Presumably you only want to consider functions of the form g(x) = ax + b as asymptotes, right?

    So the next thing you need to do is to figure out what it means for the function g(x) = ax + b to be an asymptote to the function f(x)...
  4. Apr 7, 2006 #3
    yes, I am mainly considering horizontal and vertical asymptotes, but also oblique asymptotes like g(x) = ax + b. My book says that there are no asymptotes for f(x) = 2 - 15x + 9x2 - x3.

    When you say that -x3 is an asymptote to f, is the only reason you say because both of these function's end behavior is the same?

    By graphing the function on my calculator it looks like there are no linear asymptotes.
    Do you mean I just made a conjecture? or that this is an already made conjecture?

    So, just to be sure, is the following correct?
    A polynomial has no asymptotes, vertical, horizontal, or of the form y = mx + b.
    I'm pretty sure polynomials don't have vertical or horizontal asymptotes, but I'm not sure about y=mx +b ones.
  5. Apr 8, 2006 #4


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    Right. It's a fairly important fact that for f(x) = 2 - 15x + 9x² - x³, we have that:

    f(x) / (-x³)

    goes to one as x grows large. We would say something like "these functions are asymptotically equal". This is useful because it means that you can usually use these functions interchangably when you're interested in very large x... and -x³ is a considerably easier function to manipulate than f(x)!

    You just made one.

    I don't like that definition of conjecture you quoted: I hear and have used conjecture often simply to refer to something the speaker thinks is true, but has not yet managed to prove, and that's how I was using it here.

    Well, yes... but this is the useless wordy version that's only good for telling things to people!

    What you need to do is come up with a useful version that is a precise statement that you might be able to prove or disprove!

    I think once you can write down a definition of "The line y=mx+b is an asymptote to the graph of y=f(x)", then the proof or disproof of your conjecture would be very easy.
    Last edited: Apr 8, 2006
  6. Apr 8, 2006 #5


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    If "the line y= mx+ b is an asymptote to the graph of y= f(x)" (with f a polynomial of degree n) then f(x)- (mx+ b) must go to 0 as x goes to plus or minus infinity. Suppose n> 1. Can you show that that can't happen?
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