Answer
Decreasing: $(-\infty,0)$
Increasing: nowhere.
Work Step by Step
$\mathrm{First\:Part:}\:\:$ According to the definitions:
A function $\ f\ $ defined on an interval is increasing on $\ (a, b)\ $ if for every $\ x_1, x_2\ $ $\in$ $(a, b)$ $\ x_1\le x_2\ $ implies that $\ f(x_1)\le f(x_2).\ $
A function $\ f\ $ defined on an interval is decreasing on $\ (a, b)\ $ if for every $\ x_1, x_2\ $ $\in$ $(a, b)$ $\ x_1\le x_2\ $ implies that $\ f(x_1)\ge f(x_2).\ $
We know that we can take the square root only of positive numbers, i.e $\ \ \sqrt{f(x)}\:\Rightarrow\:f(x)\ge0.$
For the given function $\ y=\sqrt{-x}\ $, we have:
$-x\ge0$
$\Rightarrow\:x\le0$
$\mathrm{Domain:}\:\:(-\infty,0]$
First of all create a table with a few points to sketch the graph.
$\quad \mathrm{See\:the\:table\:and\:graph\:above.}$
$\mathrm{Second\:Part:}\:\:$ The graph does not have any symmetry.
$\mathrm{Third\:Part:}\:\:$ The graph of the given function $\ y=\sqrt{-x}\ $ is decreasing on $\ (-\infty,0).$
