Oblique Asymptotes

**adkinsjr** · October 28th 2009, 01:12 PM

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?

**earboth** · October 28th 2009, 01:27 PM

Originally Posted by adkinsjr

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:

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

**Danny** · October 28th 2009, 01:38 PM

Originally Posted by adkinsjr

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).

**adkinsjr** · October 28th 2009, 01:50 PM

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 $x-1$

**adkinsjr** · October 28th 2009, 01:53 PM

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

**earboth** · October 28th 2009, 11:44 PM

Originally Posted by adkinsjr

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:

...

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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