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Rational function hole at (0, 0)?

  1. Oct 21, 2012 #1
    The function is f(x) = 2x / x3 - 6x2 + 3x + 10

    I was taught that any rational function with a numerator of smaller degree than the denominator has a horizontal asymptote at y = 0, which would apply in this case.

    This makes sense for the end behaviors because as x approaches +/- ∞, y approaches 0. However, when x = 0, there seems to be no problem solving the equation and ending up with y = 0, which should be impossible since there's an asymptote there. Should my domain include x ≠ 0? Why is this? The equation resolves normally when x = 0, I thought you only needed restrictions to prevent dividing by 0.
     
  2. jcsd
  3. Oct 21, 2012 #2

    Dick

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    Yes. f(x)=0, there is no problem there. But your denominator is zero at some other values of x. Can you find them?
     
  4. Oct 21, 2012 #3

    Mentallic

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    Well, you're not dividing by zero at x=0, so there is no vertical asymptote at that point. As Dick mentioned, there are vertical asymptotes elsewhere however.
     
  5. Oct 21, 2012 #4
    So the horizontal asymptote at y = 0 doesn't restrict that point, where y indeed equals 0? I found the 3 vertical asymptotes, those aren't what are confusing me. I thought an asymptote at y = n meant no points can exist with a y value of n.
     
  6. Oct 21, 2012 #5

    Dick

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    No. An asymptote at y=n just means that the limit as x->infinity (or -infinity) of y(x) is n, as you said. y(x) can equal n for finite values of x.
     
  7. Oct 21, 2012 #6

    Mentallic

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    In fact, a function can cut its horizontal asymptotes infinitely many times, such as y = sin(x)/x. Vertical asymptotes can't be cut though.
     
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