# Oblique Asymptotes

• October 28th 2009, 01:12 PM
Oblique Asymptotes
I'm having trouble figuring these out. If I have a function $f(x)=\frac{x^2+1}{x+1}$ and I want to find the oblique asymptote, where do I begin?
• October 28th 2009, 01:27 PM
earboth
Quote:

I'm having trouble figuring these out. If I have a function $f(x)=\frac{x^2+1}{x+1}$ and I want to find the oblique asymptote, where do I begin?

Use synthetic divison:

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

I'm having trouble figuring these out. If I have a function $f(x)=\frac{x^2+1}{x+1}$ and I want to find the oblique asymptote, where do I begin?

Divide $x+1$ into $x^2+1$

$
\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 $\frac{x^2+1}{x+1} = x -1 + \frac{2}{x+1}$ and your slant asymptote is $y = x - 1$ (as was mentioned).
• October 28th 2009, 01:50 PM
Quote:

Originally Posted by earboth
Use synthetic divison:

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

What I did was I used polynomial long division and I obtained $f(x)=x+\frac{1-x}{1+x}$. If I use synthetic division:

[1] [1 0 1]

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

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

Did you do something like this:

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

Therefore $u$ is the slant asymptote, which turns out to be $x-1$
• October 28th 2009, 01:53 PM
Quote:

Originally Posted by Danny
Divide $x+1$ into $x^2+1$

$
\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 $\frac{x^2+1}{x+1} = x -1 + \frac{2}{x+1}$ and your slant asymptote is $y = x - 1$ (as was mentioned).

Ok, this helps clarify things a little. Thanks
• October 28th 2009, 11:44 PM
earboth
Quote:

What I did was I used polynomial long division and I obtained $f(x)=x+\frac{1-x}{1+x}$. If I use synthetic division:
                                2  (x^2 + 1) ÷ (x+1) = x - 1 + -------- -(x^2 + x)                    x+1 -----------       -x + 1     -(-x - 1)       ---------             2