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Math Help - Determine the asymptote

  1. #1
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    Determine the asymptote

    Find the asymptotes of the curve
    <br />
x^3 + 4x^2y + 4xy^2 +5x^2+15xy+10y^2-2y+1=0<br />
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  2. #2
    Super Member 11rdc11's Avatar
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    I would rearrange the equation 1st

    y^2(4x+10) +y(4x^2 +15x -2) +(x^3+5x^2+1)

    and then use the quad equation.

    When you do all the algebra work you end up with

    y = \frac{-4x^2-15x+2 ^+_- \sqrt{9x^2-76x-36}}{8x+20}

    So your vertical asymptote is

    x = \frac{-20}{8}

    and your slant asymptote is

    y= -\frac{x}{2} - \frac{5}{8}

    and

    9x^2 - 76x -36 \geq 0
    Last edited by 11rdc11; December 27th 2009 at 10:33 PM.
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  3. #3
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    How did u get the slant asymptote could u please explain
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  4. #4
    Super Member 11rdc11's Avatar
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    since the numerator is a degree bigger than denominator it has a slant asymptote.

    So use division to find out what it is

    \frac{-4x^2-15x+2}{8x-20}
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