1. ## 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?

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:

$\displaystyle (x^2+1) \div (x+1)=\underbrace{x-1}_{\text{term of the asymptote}}+\dfrac2{x+1}$

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).

4. Originally Posted by earboth
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 $\displaystyle x-1$

5. Originally Posted by Danny
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

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