Question 206973
{{{g(x) = 5x/(x^2-1)}}}
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if x = +/- 1 then x^2-1 = 0 so you will get a vertical asymptote at that point.
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the vertical asymptote is when x approaches infinity but never quite reaches it ---
the horizontal asymptote would be the value of y when x approaches infinity.
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a good treatment of this subject can be found at:
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http://www.purplemath.com/modules/asymptote.htm
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a graph of your equation to show the vertical asymptotes is as follows:
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{{{graph (400,400,-3,3,-10,10,5x/(x^2-1))}}}
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it appears that the horizontal asymptote is equal to 0 because as x approaches infinity, y approaches 0.
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a graph of your equatrion to show the horizontal asymptotes is as follows:
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{{{graph (800,400,-40,40,-4,4,5x/(x^2-1))}}}
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there is lots more information at the website i recommended so check it out if you need to know more.
they also treat slanted asymptotes.
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vertical asymptotes never cross the vertical line they approach.
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horizontal asymptote can cross the horizontal line they approach.
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enjoy