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
