Hello, thereddevils!

Sketch the graph for each of the following functions, showing clearly the asymptotes.

$\displaystyle (1)\;\;y=\:\:\frac{x+1}{(x-1)(x-2)}$

I know that $\displaystyle x = 1, x = 2$ are asymptotes, as they make the function undefined. You're right . . . Those are the *vertical* asymptotes.

We must also look for horizontal asymptotes.

They occur when $\displaystyle y \:=\:\lim_{x\to\infty}f(x)$ has a finite value.

We have: .$\displaystyle y \:=\:\lim_{x\to\infty}\frac{x+1}{x^2-3x+2}$

Divide top and bottom by $\displaystyle x^2\!:\;\;\frac{\dfrac{x}{x^2} + \dfrac{1}{x^2}}{\dfrac{x^2}{x^2} - \dfrac{3x}{x^2} + \dfrac{2}{x^2}} $

So we have: .$\displaystyle y \:=\:\lim_{x\to\infty}\left(\frac{\frac{1}{x} + \frac{1}{x^2}}{1 - \frac{3}{x} + \frac{2}{x^2}}\right) \:=\:\frac{0+0}{1-0+0} \:=\:0$

Therefore, $\displaystyle y = 0$ (the $\displaystyle x$-axis) is a horizontal asymptote.

Code:

| : :*
| : :
| *: : *
| : : *
| : : *
| * : : *
--------------------+---:-----------:--------------------
* * : :
* *| : * :
* * | : * * :
* | : * * :
| : :
| : * * :
| : :
| : :
| :* *:
| : :