Question 811632
<pre>
{{{1/4}}}y &#8804; |x-1|

Multiply both sides by 4

y &#8804; 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)

{{{drawing(200,400,-1,3,-2,6,graph(200,400,-1,3,-2,6),

circle(1,0,.05), circle(0,4,.05),circle(2,4,.05) )}}}

Draw the graph of the boundary  y = 4|x-1|.  We draw it solid. not
dotted, because the original inequality was &#8804;, not < , so the
points on the boundary are solutions.

{{{drawing(200,400,-1,3,-2,6,graph(200,400,-1,3,-2,6,y=4abs(x-1)),

circle(1,0,.05), circle(0,4,.05),circle(2,4,.05) )}}} 

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

{{{1/4}}}y &#8804; |x-1|
{{{1/4}}}(0) &#8804; |0-1|
0 &#8804; |-1|
0 &#8804; 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.   

{{{drawing(200,400,-1,3,-2,6,graph(200,400,-1,3,-2,6,y=4abs(x-1),y<4abs(x-1)-.3),graph(200,400,-1,3,-2,6))}}}

Edwin</pre>