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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- Oct 28th 2009, 01:12 PMadkinsjrOblique 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?

- Oct 28th 2009, 01:27 PMearboth
- Oct 28th 2009, 01:38 PMJester
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). - Oct 28th 2009, 01:50 PMadkinsjr
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 $\displaystyle x-1$ - Oct 28th 2009, 01:53 PMadkinsjr
- Oct 28th 2009, 11:44 PMearboth