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Horizontal Asymptote Question

  1. Oct 17, 2011 #1
    This is just a general concept question. Why is that an equation with a numerator to a greater degree than the denominator has no asymptote & the opposite does? Also why is the coefficient of the variable to the highest degree the horizontal asymptote? Maybe a proof would help here :P Anyway thanks in advance.
     
  2. jcsd
  3. Oct 17, 2011 #2
    If the numerator had a degree greater than that of the denominator, you could simply divide it down until it was no longer rational.

    The coefficient of the highest degree is used (and important) because of the way limits work I suppose. If you divide everything by the highest degree variable in the denominator, you will find that all of the parts of lesser degree will end up with a variable in their denominator. If you "take the limit" as the variable goes to +/- infinity (just imagine that x gets bigger and bigger and bigger), you will find that these parts with a number on top, and a variable on the bottom get smaller and smaller until they basically become zero. Thus, only the leading coefficients remain (as constants) and thus play a role in the H-asymptote. I hope that helps.

    This is a really good question, which you will explore more in Calculus I.
     
  4. Oct 17, 2011 #3

    Mentallic

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    Actually, for rational functions (polynomial divided by another polynomial) that has 1 degree higher in the numerator than the denominator, there exists a oblique asymptote (slanted).

    For example, the function

    [tex]f(x)=\frac{ax^2+bx+c}{dx+e}[/tex]

    can be converted into

    [tex]px+q+\frac{k}{dx+e}[/tex]

    by using polynomial division or other similar methods. This second form clearly shows that f(x) approaches px+q as x gets large positively or negatively.
     
  5. Oct 17, 2011 #4

    Mark44

    Staff: Mentor

    This is true only if each factor of the denominator also happens to be a factor of the numerator.
     
  6. Oct 19, 2011 #5
    So for example if the denominator had the factors (x+1) & (x+3) both these factors would have to be able to be a factor w/e is on top? Could you possibly show an example please :P
     
  7. Oct 19, 2011 #6

    Mentallic

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    If you have a rational function with the quadratic polynomial (x+1)(x+3) in the denominator and you can cancel out the factors so it's no longer rational, that means the factors (x+1)(x+3) must exist in the numerator too (else you can't cancel them out fully, and as I showed in post #3, the value of k would be non-zero).

    Examples:

    [tex]\frac{(x+1)(x+3)}{(x+1)(x+3)}=1[/tex]

    [tex]\frac{(x+1)(x+2)(x+3)}{(x+1)(x+3)}=x+2[/tex]

    [tex]\frac{p(x)(x+1)(x+3)}{(x^2+4x+3)}=p(x)[/tex]

    And remember, if these examples are functions, then for the second for example, you need to explicitly state that the function is equivalent to [itex]x+2, x\neq -1,-3[/itex].
     
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