One linear equation is gotten from the case 
when what is between the absolute value bars,
3x+6, is negative, that is, less than 0.

The other linear equation is gotten from the 
case when what is between the absolute value 
bars, 3x+6, is either positive or zero, that 
is, greater than or equal to 0.

The first linear equation is gotten when 3x+6
is negative, but if 3x+6 is negative, then if 
we multiply 3x+6 by -1, then it will become 
positive, and so -1(3x+6) will be positive, 
and that will be the absolute value of 3x+6,
when 3x+6 is negative.  So the first linear 
equation is

 or 

.

So we draw the graph of that line:



However, since we are requiring that
what is between the absolute value
bars, 3x+6, is less than 0, we must
only use the part of that line where
.  So we solve that:





Therefore we must chop off the line
to the RIGHT of where x is equal to
-2.  So the graph is only this part
of the line:



and it DOES NOT include the point
(-2,0).

-----------------

Now The second linear equation is gotten 
when 3x+6 is positive or zero, and if 3x+6 
is positive or zero, then we do not need
abslute value bars at all. That is, when 
3x+6 is positive or zero, the absolute 
value of 3x+6 is simply 3x+6!  So the
second linear equation is just:

 

So we draw the graph of that line on
the same set of axes:



However, since we are requiring that
what is between the absolute value
bars, 3x+6, is greater than or equal
to 0, we must only use the part of 
that line where .  So
we solve that:





Therefore we must chop off that line
to the LEFT of where x is equal to -2.
So the FINAL graph is only this V-shaped
graph:



and and the part slanting up to the right
DOES include the point (-2,0), 

So the absolute value equation 



can be written as the piecewise function 



without using any absolute value bars!

Edwin