# Math Help - Finding an asymptote

1. ## Finding an asymptote

how do you find the asymptotes of a curve like:

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

where the denominator is a power for eg.

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

help very appreciated thanks!!

2. Originally Posted by Yehia
how do you find the asymptotes of a curve like:

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

where the denominator is a power for eg.

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

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

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

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

however this only works if $r \neq -1$, because then denominator becomes 0. In your case, we have $y = 3{x}^{-1}$, so $r = -1$ so it doesn't work. So we use $y = ln |x|$. 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!

3. Originally Posted by Yehia
and why do you integrate $3/x$ to get 3lnx and not $-9x^{-3}$?
A better question is, why do you think that the integral of 3/x is $-9x^{-3}$?