Question 207614
How do I plot this equation on a graph? y=|x|+3
<pre><font size = 4 color = "indigo"><b>
Graphs of absolute value functions are V-shaped graphs.
We first find the vertex (the sharp point of the "V"}

1.  Set what's inside the absolute value bars = 0 and solve
    for x:

    x = 0

2.  Substitute that into the equation:
    
    y=|x|+3
    y=|0|+3
    y=0+3
    y=3

3.  Plot the vertex (0,3)

{{{drawing(400,400,-5,5,-2,8, graph(400,400,-5,5,-2,8),
line(0+.1,3,0-.1,3), line(0,3+.1,0,3-.1)

 )}}}

4.  Get a point on the right of 
    the vertex:

    Let x=1
    y=|x|+3
    y=|1|+3
    y=1+3
    y=4

    That's one point (1,4).  Plot it:

{{{drawing(400,400,-5,5,-2,8, graph(400,400,-5,5,-2,8),
line(0+.1,3,0-.1,3), line(0,3+.1,0,3-.1),
line(1+.1,4,1-.1,4), line(1,4+.1,1,4-.1)
 )}}}

5.  Get a point on the left of 
    the vertex:

    Let x=-1
    y=|x|+3
    y=|-1|+3
    y=1+3
    y=4

    That's another point (-1,4).  Plot it:
   
{{{drawing(400,400,-5,5,-2,8, graph(400,400,-5,5,-2,8),
line(0+.1,3,0-.1,3), line(0,3+.1,0,3-.1),
line(1+.1,4,1-.1,4), line(1,4+.1,1,4-.1),
line(-1+.1,4,-1-.1,4), line(-1,4+.1,-1,4-.1)
 )}}}

Draw the V-shaped graph:

{{{drawing(400,400,-5,5,-2,8, graph(400,400,-5,5,-2,8,abs(x)+3),
line(0+.1,3,0-.1,3), line(0,3+.1,0,3-.1),
line(1+.1,4,1-.1,4), line(1,4+.1,1,4-.1),
line(-1+.1,4,-1-.1,4), line(-1,4+.1,-1,4-.1)
 
)}}}
Edwin</pre>