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Math Help - Find the slant asymptote?

  1. #1
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    Find the slant asymptote?

    (x^3-4x+2x-5)/(x+2)
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  2. #2
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    Quote Originally Posted by puzzledwithpolynomials View Post
    Find the slant asymptote?

    (x^3-4x+2x-5)/(x+2)
    do the long division ... the quotient will be a linear expression that is the equation of the slant asymptote (ignore the remainder).

    here's a link with some examples ...

    Slant, or Oblique, Asymptotes
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  3. #3
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    Did you figure out the answer? I learned about oblique/slant asymptotes a week ago and I thought I might try this problem out for practice. Would the answer by any chance be : x-4 R 4x+3 OR x-4 + (4x+3/x^2-2) Oblique asymptote being x-4 . Not sure If I am allowed to supply answers but I would just like to see if I am right.

    Sean
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  4. #4
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    Quote Originally Posted by SeanBE View Post
    Did you figure out the answer? I learned about oblique/slant asymptotes a week ago and I thought I might try this problem out for practice. Would the answer by any chance be : x-4 R 4x+3 OR x-4 + (4x+3/x^2-2) Oblique asymptote being x-4 . Not sure If I am allowed to supply answers but I would just like to see if I am right.

    Sean
    quotient is ...

    x - 4 + \frac{3}{x^2+2}

    slant asymptote is y = x - 4
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  5. #5
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    A slant asymptote always has this form:

    y=mx+n

    Where :

    m=\lim_{x\rightarrow\inf}\frac{f(x)}{x}

    and

    n=\lim_{x\rightarrow\inf}(f(x)-mx)


    In this particular case of yours you have :

    m=\lim_{x\rightarrow\inf}\frac{\frac{x^3-4x^2+2x-5}{x^2+2}}{x}=\lim_{x\rightarrow\inf}\frac{x^3-4x^2+2x-5}{x^3+2x} = 1

    n=\lim_{x\rightarrow\inf}(\frac{x^3-4x^2+2x-5}{x^2+2}-x)=\lim_{x\rightarrow\inf}\frac{-4x^2-5}{x^2+2}=-4


    So the fianl result will be :

    y=mx+n=1*x-4 =x-4

    That's all,

    Have a nice day!
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