# Oblique Asymptotes

• Oct 28th 2009, 01:12 PM
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?
• Oct 28th 2009, 01:27 PM
earboth
Quote:

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:

$\displaystyle (x^2+1) \div (x+1)=\underbrace{x-1}_{\text{term of the asymptote}}+\dfrac2{x+1}$
• Oct 28th 2009, 01:38 PM
Jester
Quote:

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).
• Oct 28th 2009, 01:50 PM
Quote:

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 $\displaystyle x-1$
• Oct 28th 2009, 01:53 PM
Quote:

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
• Oct 28th 2009, 11:44 PM
earboth
Quote:

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:
                                2  (x^2 + 1) ÷ (x+1) = x - 1 + -------- -(x^2 + x)                    x+1 -----------       -x + 1     -(-x - 1)       ---------             2