SOLUTION: Graph ofy=|x+1|+|2x-1|

Here's what you do, step by step:

Set the expressions inside the absolute value bars equal 0,
x+1 = 0        2x-1 = 0
  x = -1         2x = 1
                  x =   
These define three intervals: x < -1, -1 < x < 1/2, and x > 1/2

For x < -1, |x+1| = -(x+1) and |2x-1| = -(2x-1)
so y = -(x+1)-(2x-1) = -x-1-2x+1 = -3x

For -1 < x < 1/2, |x+1| = x+1 and |2x-1| = -(2x-1)
so y = x+1-(2x-1) = x+1-2x+1 = -x+2

For x > 1/2, |x+1| = x+1 and |2x-1| = 2x-1
so y = x+1+2x-1 = 3x-1-2x+1 = 3x

So the graph is this piecewise graph:



So draw the three lines:





Then erase the parts of the three lines that extend 
past their part of the domain:




Edwin