y ≤ |x-1|

Multiply both sides by 4

y ≤ 4|x-1|

Draw the boundary graph of y = 4|x-1|

It's vertex is when what's between the | |'s equals 0
 
                                x-1 = 0
                                  x = 1

Substitute in y = 4|x-1|
              y = 4|1-1|
              y = 4|0|
              y = 4(0)
              y = 0

So the vertex is (1,0)

We get a point on each side, let x=0, y = 4|0-1| = 4|-1| = 4(1) = 4
                             let x=2, y = 4|2-1| = 4|1| = 4(1) = 4

Plot points vertex (1,0) and points on each side (0,4) and (2,4)



Draw the graph of the boundary  y = 4|x-1|.  We draw it solid. not
dotted, because the original inequality was ≤, not < , so the
points on the boundary are solutions.

 

Test a point, say the origin (0,0) in the original inequality,
to see if it's a solution.

y ≤ |x-1|
(0) ≤ |0-1|
0 ≤ |-1|
0 ≤ 1

That's true.  The origin is a solution and therefore all points
on the same side of the graph that the origin is on are also 
solutions. So we shade below the graph.   



Edwin