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Asymptotes as the lines y=x an y=-x

  1. Nov 2, 2006 #1
    Hi there.

    I'm a maths teacher and today was having a discussion with my Head of Department about asymptotes (as you do!)

    She was me if I could think on an equation of a graph(s) which has an asymptotes at the line y=x and another at y=-x.

    Thinking about it, neither of us could come up with anything. It's really bugging me now. I ain't too sure if there is such a function...though it seems silly there not being.

    The only way I managed to get a graph like what I wanted was to rotate the graph of y=1/x by various angles.

    So can such a graph exist? Is it possible to transform equations of graphs via rotations (something I can't ever remember doing)?

    Hope I can come up with an answer by tomorrow :)
  2. jcsd
  3. Nov 2, 2006 #2
    How about

    [tex]y = x \tanh{x}[/tex] ?


    [tex] x \tanh{x} = x \frac{e^x - e^{-x}}{e^x + e^{-x}}[/tex]

    therefore, for large positive x,

    [tex]x \tanh{x} \approx x \frac{e^x}{e^x} = x[/tex]

    and for large negative x

    [tex] x \tanh{x} \approx x (\frac{-e^{-x}}{e^{-x}}) = -x[/tex]
  4. Nov 2, 2006 #3


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    Dearly Missed

    You certainly can rotate curves!

    For example, the curve described by the equation:
    has asymptotes y=x and y=-x.

    Note, however, that in this case, there exists no function of x, by which the y-coordinates of the curve could be computed, and the curve in question cannot be regarded as the graph of some function, i.e, the set of points describable as (x,f(x)), where f is some function.
  5. Nov 3, 2006 #4


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    Doesn't one branch of that hyperbola still have those lines as asymptotes?
    In a physics context, the worldline of a uniformly accelerated observer [which can be regarded as function (t,x(t)), where x(t)=sqrt(1+t^2)] is asymptotic to a light cone.
  6. Nov 3, 2006 #5
    Thanks for the help guys :D

    My head of department was so sure you could each of the 4 sections between the said asymptotes filled with a curve.

    Using what the first two posts said, I drew the graphs of

    y = sqrt(1+x^2)
    y = -sqrt(1+x^2)
    x = sqrt(1+y^2)
    x = -sqrt(1+y^2)

    to get what I needed (at least I think that is what I i if I remember correctly back to last night....)

    But I only went and forgot all about this today until just now...so I never i speak to my HofD about it...never mind :D
  7. Nov 3, 2006 #6


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    You can combine the four curves to get one that fills the whole thing. :smile:

    We already know that

    x² - y² = 1

    fills in two sections, and

    y² - x² = 1

    fills in the other two.

    We can combine them into one equation:

    (x² - y² - 1) (y² - x² - 1) = 0

    which will fill in all four quadrants.
  8. Nov 3, 2006 #7


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    ...which can be simplified to read (x² - y²)² = 1.
    Of course, the easy way to see this is to note that
    y² - x² = 1 is the same as x² - y² = -1.
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