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Math Help - slant asymptote

  1. #1
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    slant asymptote

    show that the curve sqrt(x^2+4x) has two asymptotes: y=x+2 and y+-x-2.
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  2. #2
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    Quote Originally Posted by johntuan View Post
    show that the curve sqrt(x^2+4x) has two asymptotes: y=x+2 and y+-x-2.
    A 'nasty' way (mathematicians turn away):

    y = \sqrt{(x + 2)^2 - 4} \rightarrow \sqrt{(x+2)^2} = \pm(x + 2).

    A much better way is to expand \sqrt{x^2+4x} using the generalised binomial theorem and then take the limit.
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  3. #3
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    haha..but i dont understand how +-(x+2) becomes asymptotes?...the way i would get S.A is long division or the limit way...but this one is weird because its not a rational function
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    Quote Originally Posted by johntuan View Post
    haha..but i dont understand how +-(x+2) becomes asymptotes?...the way i would get S.A is long division or the limit way...but this one is weird because its not a rational function
    Slant asymptotes are lines that the curve approaches as x --> +oo or x --> -oo. As x --> +oo or -oo, (x+2)^2 - 4 \rightarrow (x+2)^2 and the square root of this is \pm (x+2). This is not a rigorous way of doing it. The mathematicians of MHF will pounce on this and make mince meat of it.

    But I'm not sure you're at a level where a rigorous approach would make much sense, so you'll just have to make do with the one above.
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