Hey guys, please tell me if I'm correct in my calculations

Given the function $\displaystyle \frac{x^2}{x-1}$

a) Determine the domain and all the intercepts of the function. Does the function have any symmetry?

This is what I did: Domain is all real #s except 1, because x-1=0, x=1.

intercepts are/is: (0,0) because $\displaystyle \frac{(0)^2}{(0)-1}$=0

it's symmetric because $\displaystyle \frac{(-x)^2}{(-x)-1} $=$\displaystyle \frac{x^2}{-x-1} $

Is this right?

I got that lim x-->inf it goes to inf and lim x-->-inf goes to -inf.

b) Determine $\displaystyle \lim_{x \to \infty} $f(x)and $\displaystyle \lim_{x \to -\infty} $f(x).Determine all horizontal and vertical asymptotes of the function.

Vertical asymptote is 1 as shown above.

Can someone show me how you get the horizontal asymptote for this please?

c) find all critical numbers.Quotient rule: $\displaystyle \frac{(x-1)(2x)-(x^2)(1)}{(x-1)^2} = \frac{x^2-2x}{(x-1)^2 }$

$\displaystyle x^2-2x=0, x=0,2$ critical numbers are 0 and 2

d) Determine the intervals on which the function is increasing and where it is decreasing. Find all maximum and minimum values of f.I use a chart.

______0__2

x __|- | + | +

x-2 |- | - | +

f '_ |+ | - | +

___up, down, up. so there must be a maximum at 0 and a minimum and 2. right?