# Thread: Finding Oblique Asymptote

1. ## 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~

2. 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 $y=x-3$

3. 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

4. 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.