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Math Help - crossing the aymptote

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    crossing the aymptote

    When will a function cross the slant or horizontal asymptote? Is there a way to tell from a given function that it will cross the asymptote.
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    Quote Originally Posted by shane99 View Post
    When will a function cross the slant or horizontal asymptote? Is there a way to tell from a given function that it will cross the asymptote.
    By definition, an asymptote does not get crossed...
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  3. #3
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    Hello everyone
    Quote Originally Posted by shane99 View Post
    When will a function cross the slant or horizontal asymptote? Is there a way to tell from a given function that it will cross the asymptote.
    Quote Originally Posted by Prove It View Post
    By definition, an asymptote does not get crossed...
    Sorry, Prove It, but this is not true. Of course an asymptote can be crossed. The definition of an asymptote is (loosely) at tangent to the curve 'at infinity'; but that doesn't preclude the curve from crossing the asymptote at other points.

    Consider, for example, the graph of
    y =(x-1)e^{-x}
    As x \to \infty, y\to 0. So the x-axis is an asymptote.

    But when x = 1, y = 0. So the curve crosses this asymptote at (1,0).

    As far as your original question is concerned, you would just need to look for solutions to an equation in the usual way. If, for example, the graph of y = f(x) has y = mx+c as an asymptote, then the graph crosses the asymptote at any finite values of x that satisfy the equation f(x) = mx+c. Of course, there may not be any such values - but that's another issue!

    Grandad
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    Quote Originally Posted by Prove It View Post
    By definition, an asymptote does not get crossed...
    By definition, a vertical asymptote does not get touched or crossed. However, there is nothing preventing a function crossing or tiuching a horizontal or oblique asymptote. eg. The function f(x) = \frac{x}{x^2 + 1} has a horizontal asymptote y = 0. It is also clear that the function crosses this asymptote (at the origin).
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  5. #5
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    I stand corrected.
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