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Math Help - How do you find the oblique asymptote for this specific rational function

  1. #1
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    Question How do you find the oblique asymptote for this specific rational function

    The problem I'm trying to find the oblique asymptote for is: 8x^3+1/x^2-5x+6
    I know the answers to this specific problem(vertical asymptotes at: x=2,x=3; oblique asymptote at:y=x+5) but I don't know how to arrive at the answer for the oblique asymptote for this specific problem. Can anyone help? I know about how to use long division to solve most of these types of problems, but this one I don't know how to achieve the correct answer for the oblique asymptote even with long division.
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     \frac{8x^3+1}{x^2-5x+6}

    Use polynomial long division to change this into a form where the order of the numerator is less than the order of the denominator. It should be clearer then.
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    I understand that I should use polynomial long division, but even so I still don't arrive at the correct answer. Perhaps my calculations are wrong. If you've tried it already using long division and arrived at the oblique asymptote I have listed above, would you mind showing me your steps?
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    Check this for the method Polynomial Long Division

    I get (did it rather quick)

     \frac{8x^3+1}{x^2-5x+6} = 8x+40+\frac{241}{x^2-5x+6}<br />
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    Yea that's what I get when I work it out too. But my book is listing the answer for the oblique asymptote as y=x+5 and shows that I need to work it out as x^3-8/x^2-5x+6
    Which I'm not understanding where x^3-8 comes from. Any ideas?
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    Quote Originally Posted by micbal9388 View Post
    I'm not understanding where x^3-8 comes from. Any ideas?
    No I can't see how, maybe 8 has been taken out as a common factor. But not even.

    \frac{8x^3+1}{x^2-5x+6} = 8x+40+\frac{241}{x^2-5x+6} = 8(x+5)+\frac{241}{x^2-5x+6}
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    Yea, thanks anyway. Ha, I've asked a bunch of people and no one knows. It's quite baffleing to me.
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  8. #8
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    Yea I see what the problem is. My teacher created an incorrect answer sheet and confused the hell out of me. Thanks, you were right.
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