Hello, litlmike!
I would like to know where my math went wrong with the following equation:
. . $\displaystyle y \:=\: 2x3 + 1$
So, to graph, I assumed on finding the xintercept and yintercept: set $\displaystyle y = 0$
. . $\displaystyle 2x3 + 1 \:=\:0 \quad\Rightarrow\quad 2x3 \:=\:1$
. . $\displaystyle x3 \:=\:\tfrac{1}{2}$ . Here!
An absolute value is always nonnegative (greater than or equal to zero).
So your equation has no solutions . . . There are no xintercpts.
You can sketch the graph using "transformations" . . .
We know the graph of $\displaystyle y \:=\:x$
. . It is a Vshape with its vertex at the origin.
Code:

*  *
*  *
*  *
*  *
     o      

The graph of: $\displaystyle y \:=\:x3$ is the same graph moved 3 units to the right.
Code:

*  *
* *
 * *
 * *
    +   o     
 3
The graph of: $\displaystyle y \:=\:2x3$ is the previous graph except the sides are "steeper".
Code:

 * *
 * *
 * *
 * *
    +   o     
 3
The graph of: $\displaystyle y \:=\:2x3+1$ is the previous graph moved up one unit.
Code:

 * *
 * *
 * *
 * *
 o(3,1)

    +       
 3
. . There!