Hello, ^_^Engineer_Adam^_^!
Sketch the graph of the function and determine its domain and range:
$\displaystyle 1)\;f(x) \:= \:\frac{x^3  2x^2}{x2}$
To repeat what Captain Black said . . .
We see that $\displaystyle x \neq 2$.
. . $\displaystyle \boxed{\text{Domain: all } x \neq 2}$ .or: $\displaystyle \boxed{(\infty,\,2) \cup (2,\,\infty)} $
Since $\displaystyle x \neq 2$, we have: .$\displaystyle f(x) \:=\:\frac{x^2(x  2)}{x2} \:=\:x^2$
We have a parabola, $\displaystyle y = x^2$, with a "hole" at (2,4). Code:

*  *

*  *
*  o(2,4)
*  *
***

Hence: .$\displaystyle \boxed{\text{Range: all }y \geq 0}$ .or: $\displaystyle [0,\,\infty)$
$\displaystyle 2)\;f(x) \:= \:3x + 2$
We know that the graph of $\displaystyle y \:=\:x$ is a $\displaystyle V$ with its vertex at the origin.
The graph of $\displaystyle y \:=\:3x$ is the same $\displaystyle V$ but it "rises faster".
The graph of $\displaystyle y \:=\:3x + 2 \:=\:\left3\left(x + \frac{2}{3}\right)\right$
. .is the same graph moved $\displaystyle \frac{2}{3}$ units to the left.
The graph looks like this: Code:
\  /
\ /
\ /
\ /
\ / 
\ / 
*+
2/3 
Therefore: .$\displaystyle \begin{array}{cc}\text{Domain} & (\infty,\,\infty) \\ \text{Range} & [0,\,\infty) \end{array}$