Question 363181
y = 2|3x + 6| - 4

Finding the vertex:

1.  Set what's between the absolute value bars = 0

3x + 6 =  0
    3x = -6
     x = -2

That's the x-coordinate of the vertex.

2.  Substitute that in the equation to find the y-coordinate 
    of the vertex:

    y = 2|3(-2)+6| - 4
    y = 2|-6+6| - 4
    y = 2(0) - 4
    y = -4

(Shortcut for this step: the y-coordinate of the vertex is
just the number that's added to the absolute value term, in this
case -4, because when you substitute the x-coordinate the 
absolute value term will always be 0)

3. The vertex is (-2,-4)

Plot that:

{{{drawing(800/3,400,-5,5,-5,10, graph(800/3,400,-5,5,-5,10),
 locate(-2,-4,"(-2,-4)"),

circle(-2,-4,.1) )}}}

4. Find one other point, say the y-intercept, so let x=0,

y = 2|3(0) + 6| - 4
y = 2|0 + 6| - 4
y = 2|6| - 4
y = 2(6) - 4
y = 12 - 4
y = 8
    
5. Plot the y-intercept (0,8)     


{{{drawing(800/3,400,-5,5,-5,10, graph(800/3,400,-5,5,-5,10),
 locate(-2,-4,"(-2,-4)"),

circle(-2,-4,.1),locate(0,8,"(0,8)"),

circle(0,8,.1) )}}}

6.  Draw a line through that point stopping at the vertex (-2,-4)

{{{drawing(800/3,400,-5,5,-5,10, graph(800/3,400,-5,5,-5,10),
 locate(-2,-4,"(-2,-4)"),

circle(-2,-4,.1),locate(0,8,"(0,8)"), green(line(10,68,-2,-4)),

circle(0,8,.1) )}}}

7.  draw another line to make a symmetrical V-shaped graph, with vertex (-2,-4):

{{{drawing(800/3,400,-5,5,-5,10, graph(800/3,400,-5,5,-5,10),
 locate(-2,-4,"(-2,-4)"),green(line(10,68,-2,-4)),


circle(-2,-4,.1),locate(0,8,"(0,8)"), green(line(-8,32,-2,-4)),

circle(0,8,.1) )}}}

That's it!

Edwin</pre>