Stuck on question with asymptotes Ive been stuck on this quest from textbook and I can't find an answer for it..

Question is, what may the equation of a rational function with below conditions look like? Explain how you got the equation.

$\displaystyle f(1)=0$

$\displaystyle \lim_{x \to 0}  \; f(x)= - \infty$

$\displaystyle \lim_{x\to2^+} f(x)= -\infty$  (from right side) 

$\displaystyle  \lim_{ x \to 2^-} f(x)= +\infty$  (from left side) 

$\displaystyle   \lim_{x \to \pm \infty} f(x)= 0$
For the activity around x=2 you need $(2-x)$ raised to an odd power in the denominator. To make $f(1)=0$ you need $(x-1)$ in the numerator. And for the activity at plus and minus infinity you need a power of $x$ in the denominator. Thus I would suggest
$$f(x)=\frac{(x-1)}{(2-x)^3x}.$$
Edit: Following comments below, a better suggestion is
$$f(x)=\frac{(x-1)}{(2-x)^3x^2}.$$
$(-\infty, 0)$ use $-1/x $

$(0,1]$ use $\ln(x)$

$[1,2)$ use $\ln(2-x)$  

$(2,\infty)$ use $1/(x-2)$