# Finding an asymptote

#### Yehia

how do you find the asymptotes of a curve like:

$$\displaystyle (x+3)/(x^2+4)$$

where the denominator is a power for eg.

and why do you integrate $$\displaystyle 3/x$$ to get 3lnx and not $$\displaystyle -9x^{-3}$$??

help very appreciated thanks!!

Last edited by a moderator:

#### eddie2042

how do you find the asymptotes of a curve like:

$$\displaystyle (x+3)/(x^2+4)$$

where the denominator is a power for eg.

and why do you integrate $$\displaystyle 3/x$$ to get 3lnx and not $$\displaystyle -9x^-3$$??

help very appreciated thanks!!
To get the asymptotes, test its end behaviors by taking both limits as $$\displaystyle x\rightarrow \infty$$ and as $$\displaystyle x\rightarrow -\infty$$. Both of which are 0, therefore there is only one asymptote which is $$\displaystyle y = 0$$.

To answer your second question, the reason it turns into $$\displaystyle y = ln |x|$$ is because when you are integrating using the reverse power rule which says

$$\displaystyle \int {x}^{r}dx = \frac{{x}^{r+1}}{r+1}$$

however this only works if $$\displaystyle r \neq -1$$, because then denominator becomes 0. In your case, we have $$\displaystyle y = 3{x}^{-1}$$, so $$\displaystyle r = -1$$ so it doesn't work. So we use $$\displaystyle y = ln |x|$$. Make sure that it's absolute value of x and not just regular x. If you don't put the absolute value signs, you'll get it wrong.

Cheers!

#### matheagle

MHF Hall of Honor
and why do you integrate $$\displaystyle 3/x$$ to get 3lnx and not $$\displaystyle -9x^{-3}$$?
A better question is, why do you think that the integral of 3/x is $$\displaystyle -9x^{-3}$$?