Finding Oblique Asymptote

Printable View

July 13th 2007, 11:02 AM
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~
July 13th 2007, 11:18 AM
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 $y=x-3$
July 13th 2007, 11:20 AM
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:
July 13th 2007, 11:27 AM
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.

All times are GMT -8. The time now is 11:42 PM.