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Thread: Oblique Asymptotes

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    Oblique Asymptotes

    I'm having trouble figuring these out. If I have a function $\displaystyle f(x)=\frac{x^2+1}{x+1}$ and I want to find the oblique asymptote, where do I begin?
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    Quote Originally Posted by adkinsjr View Post
    I'm having trouble figuring these out. If I have a function $\displaystyle f(x)=\frac{x^2+1}{x+1}$ and I want to find the oblique asymptote, where do I begin?
    Use synthetic divison:

    $\displaystyle (x^2+1) \div (x+1)=\underbrace{x-1}_{\text{term of the asymptote}}+\dfrac2{x+1}$
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    Quote Originally Posted by adkinsjr View Post
    I'm having trouble figuring these out. If I have a function $\displaystyle f(x)=\frac{x^2+1}{x+1}$ and I want to find the oblique asymptote, where do I begin?
    Divide $\displaystyle x+1$ into $\displaystyle x^2+1$

    $\displaystyle
    \begin{array}{c|ccc}
    & x & - 1 & \\
    \hline\\
    x+1 & x^2 & 0 x & +1\\
    & x^2 & x & 0\\
    \hline\\ & & -x & +1\\
    & & -x & -1\\
    \hline\\
    & & & 2 \end{array}
    $

    so $\displaystyle \frac{x^2+1}{x+1} = x -1 + \frac{2}{x+1}$ and your slant asymptote is $\displaystyle y = x - 1$ (as was mentioned).
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    Quote Originally Posted by earboth View Post
    Use synthetic divison:

    $\displaystyle (x^2+1) \div (x+1)=\underbrace{x-1}_{\text{term of the asymptote}}+\dfrac2{x+1}$
    What I did was I used polynomial long division and I obtained $\displaystyle f(x)=x+\frac{1-x}{1+x}$. If I use synthetic division:

    [1] [1 0 1]

    .........1 1
    ----------
    .......1 1 2


    So the remainder is $\displaystyle \frac{2}{x+1}$. I don't really understand where the $\displaystyle x-1$ comes from.

    Did you do something like this:

    $\displaystyle \frac{2}{x+1}+u=\frac{x^2+1}{x+1}$

    Therefore $\displaystyle u$ is the slant asymptote, which turns out to be $\displaystyle x-1$
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    Quote Originally Posted by Danny View Post
    Divide $\displaystyle x+1$ into $\displaystyle x^2+1$

    $\displaystyle
    \begin{array}{c|ccc}
    & x & - 1 & \\
    \hline\\
    x+1 & x^2 & 0 x & +1\\
    & x^2 & x & 0\\
    \hline\\ & & -x & +1\\
    & & -x & -1\\
    \hline\\
    & & & 2 \end{array}
    $

    so $\displaystyle \frac{x^2+1}{x+1} = x -1 + \frac{2}{x+1}$ and your slant asymptote is $\displaystyle y = x - 1$ (as was mentioned).
    Ok, this helps clarify things a little. Thanks
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    Quote Originally Posted by adkinsjr View Post
    What I did was I used polynomial long division and I obtained $\displaystyle f(x)=x+\frac{1-x}{1+x}$. If I use synthetic division:

    ...
    Unfortunately you forgot to perform the last step of the division:
    Code:
                                    2
     (x^2 + 1)  (x+1) = x - 1 + --------
    -(x^2 + x)                     x+1
    -----------
           -x + 1
         -(-x - 1)
          ---------
                2
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