Hello, pashah!
Sketch the graph and state the domain and range.
. . $\displaystyle y \;= \;\sqrt{x+5}  2$
Since $\displaystyle x + 5$ must be nonnegative: .$\displaystyle x \geq 5$ (domain)
The minimum value of $\displaystyle y$ occurs when $\displaystyle x = 5$
Hence, the range is: .$\displaystyle y \geq 2$
We have: .$\displaystyle y + 2 \:=\:\sqrt{x+5}$
Square the equation: .$\displaystyle (y+2)^2 \:=\:x + 5$
We have a parabola with vertex $\displaystyle (5,2)$ which opens to the right: $\displaystyle \subset$
And we want only the upper half. Code:

 *
*
    *  +     
* 
* 
* 
(5,2) 
Graph and identity all intercepts.
. . $\displaystyle y \;= \;(x+2)(x2)^2$
If $\displaystyle x = 0,\;y\:=\:2(2)^2 \:=\:8$ . . . yintercept: $\displaystyle (0,8)$
If $\displaystyle y = 0,\;x\:=\:\pm2$ . . . xintercepts: $\displaystyle (\pm2,0)$
This is a cubic graph which rises to the right.
Since $\displaystyle (x2)^2$ has an even exponent,
. . the graph is tangent to the xaxis at $\displaystyle x = 2$ Code:

* * *
2 *  * *
   *   +   *  
*  2
* 