# Thread: Graphing functions with absolute value

1. ## Graphing functions with absolute value

Graph the following:
$||x|-|y||=1$

I tried diving the sections but I just can't get those two absolute value signs off.

2. This is equivalent to $\displaystyle |y| = |x| - 1^2$

So now try to solve for $\displaystyle y$ in terms of $\displaystyle x$.

3. Originally Posted by lanierms
Graph the following:
$||x|-|y||=1$

I tried diving the sections but I just can't get those two absolute value signs off.
As usual, break into parts.

1) If |x|- |y|> 0 this is the same as |x|- |y|= 1. Now break it into parts again
a) If x, y> 0 then x- y= 1 so y= x- 1. That is a straight line through (0, -1) and (1, 0). Of course, that is only in the first quadrant- from (1, 0) up to the right.
b) If x, y< 0 then -x+ y= 1 so y= 1+ x. That is a straight line through (0, 1) and (-1, 0). Of course, that is only in the third quadrant from (-1.0) down to the left.
c) If x> 0, y< 0 then x+ y= 1 so y= 1- x. That is a straight line through (0, 1) and (1, 0). Of course, that is only in the fourth quadrant, from (1, 0) down to the right.
d) If x< 0, y> 0 then -x- y= 1 so y= -x- 1. That is a straight line thorugh (0, -1) and (-1, 0). Of course, that is only in the second quadrant, from (-1, 0) up to the left.

2) If |x|- |y|< 0 this is -(|x|-|y|)= 1 or |x|- |y|= -1.
a) If x, y> 0 then x- y= -1 so y= x+ 1. That is a straight line through (0, 1) and (1, 2). Of course, that is only in the first quadrant so it is the part of the line from (0, 1) up to the right, parallel to the line in (1a).
b) if x, y< 0 then -x+ y= -1 so y= x- 1. That is a straight line through (0, -1) and (1, 0). Of course, that is only in the third quadrant so it is the part of the line from (0, -1) down to the left, parallel to the line in (1b).
c) If x> 0, y< 0 then x+ y= -1 so y= -x- 1. That is a straight line through (0, -1) and (-1, 0). Of course, this is only in the fourth quadrant so it is the part of the line from (0, -1) down to the right parallel to the line in (1c).
d) If x< 0, y> 0 then -x- y= -1 so y= -x+ 1. That is a straight line through (0, 1) and (1, 0). Of course, this is only in the second quadrant so it is the part of the line from (0, 1) up to the left parallel to the line in (1d).

The graph is a total of eight rays.