# Finding Oblique Asymptote

• Jul 13th 2007, 12:02 PM
JonathanEyoon
Finding Oblique Asymptote
Hi all,

I'm currently having the hardest time understanding how to find the Oblique Asymptote.

An example is x^2 -1 / x + 3. I know I have to use long division method to figure this out but I don't even know how to start on such a simple one as this. Can someone give me a detailed method of doing this? Thanks alot~
• Jul 13th 2007, 12:18 PM
galactus
If you long divide you get:

$\frac{x^{2}-1}{x+3}=\underbrace{\frac{8}{x+3}}_{\text{r(x)/h(x)}}+\underbrace{x-3}_{\text{ax+b}}$

$\lim_{x\rightarrow{\infty}}\frac{r(x)}{h(x)}=0$

Therefore, f(x) approaches the line y=ax+b as x increases or decreases without bound.

Hence, your asymptote is $y=x-3$
• Jul 13th 2007, 12:20 PM
JonathanEyoon
Quote:

Originally Posted by galactus
If you long divide you get:

$\frac{x^{2}-1}{x+3}=\underbrace{\frac{8}{x+3}}_{\text{r(x)/h(x)}}+\underbrace{x-3}_{\text{ax+b}}$

$\lim_{x\rightarrow{\infty}}\frac{r(x)}{h(x)}=0$

Therefore, f(x) approaches the line y=ax+b as x increases or decreases without bound.

Hence, your asymptote is $y=x-3$

Thanks but I was wondering how to do the division part step by step :confused:
• Jul 13th 2007, 12:27 PM
galactus
Oh. I would suggest getting an algebra book and practicing. You really should be able to do that at this level.

But here goes this one:

Code:

          x    -3       ____________________ x+3|x^2      -1       x^2+3x     ---------------------------             -3x-1             -3x-9     -------------------------------                   8
So, you have: $\frac{8}{x+3}+x-3$

It works just like other division.