Absolute Value Graphs

Date: 11/16/2002 at 18:20:27
From: Katherine 
Subject: Finding vertices with absolute value graphs

My question is, how would you graph the inequalities:

   abs(x)+abs(y)is less than or equal to 5
   abs(x)+abs(y)is greater than or equal to 2

So far, I've tried to put it into slope intercept form and I got:

   abs(y) = -abs(x)+5
   abs(y) = -abs(x)+2

From there, I wasn't sure where I should graph it because I'm not 
sure what to do with both absolute values on each side of the 
equation. I need your help to figure out this equation, graph it, and 
find its vertices.

Date: 11/16/2002 at 21:14:38
From: Doctor Greenie
Subject: Re: Finding vertices with absolute value graphs

Hi, Katherine -

You ask how I would draw this graph. First I will give you an answer 
that will do you no good: Since I have done hundreds of problems like 
this, I simply know what the graph is going to look like, so I just 
draw it. Perhaps after doing a few problems like this, you too will be 
able to have enough of an understanding of what's happening with 
graphs like this to do the same sort of thing.

But for now, you are just learning - so let's take it slow and look at 
all the steps.

For a student just learning how to do problems like this, the safest 
way is to break each absolute value down into cases depending on 
whether the quantity inside the absolute value is positive. Then 
remember that

  if a quantity is greater than or equal to 0,
   then its absolute value is that same quantity;

  if a quantity is less than 0,
   then its absolute value is the opposite of that quantity

For a simple example:

  2 >= 0, so |2| = 2

  -2 < 0, so |-2| = -(-2) = 2

Let's look now at the first of your problems:

  graph |x| + |y| <= 5

Both x and y can be either positive or negative; so we have four 
possibilities:

(1) x>0 and y>0
(2) x<0 and y>0
(3) x<0 and y<0
(4) x>0 and y<0

I purposely chose to number the four cases in this way, because then 
case (1) is in the 1st quadrant, case (2) is in the 2nd quadrant, 
case (3) is in the 3rd quadrant, and case (4) is in the 4th quadrant.

So we have four cases to consider when we go to graph this inequality:

(1) 1st quadrant; x>0 and y>0, so |x| = x and |y| = y, so the 
    inequality is

   x+y <= 5
     y <= -x+5

(2) 2nd quadrant; x<0 and y>0, so |x| = -x and |y| = y, so the 
    inequality is

   -x+y <= 5
      y <= x+5

(3) 3rd quadrant; x<0 and y<0, so |x| = -x and |y| = -y, so the 
    inequality is

   -x-y <= 5
   -x-5 <= y
      y >= -x-5

(4) 4th quadrant; x>0 and y<0, so |x| = x and |y| = -y, so the 
    inequality is

   x-y <= 5
   x-5 <= y
     y >= x-5

If you graph these respective lines in the respective quadrants, you 
will find the boundary lines for the solution set are the sides of 
the square whose vertices are (5,0), (0,5), (-5,0), and (0,-5). 
Furthermore, in the 1st and 2nd quadrants the solution set lies on or 
below the boundary lines, while in the 3rd and 4th quadrants the 
solution set lies on or above the boundary lines. So the complete 
solution set for the inequality is the square and all points inside 
the square.

There is another way of thinking about absolute values that makes it 
easy to graph this inequality. If you are lucky, at some point in your 
math education one of your teachers will spend some time talking about 
this alternative interpretation of absolute value.

In this alternate interpretation, we think of |a-b| as the "distance 
between a and b." For example, if a is 10 and b is 3, then a-b is 7 
and b-a is -7; but the distance between a and b is 7 and the distance 
between b and a is 7. So |a-b| and |b-a| are both equal to 7.

With this interpretation of absolute values, "|x|" is the same as 
"|x-0|" so |x| is the distance from 0. And similarly, |y| is the 
distance from 0.  So when we write the inequality

  |x|+|y| <= 5

we are talking about all the points (x,y) for which

  "(the distance of x from 0) + (the distance of y from 0)"

is less than or equal to 5.

If you think of that interpretation, you can perhaps see why the graph 
turns out to be the boundary and interior of the square with vertices 
(5,0), (0,5), (-5,0), and (0,-5). This interpretation is, in fact, why 
I stated at the beginining of my response that I could sketch the 
graph of this inequality with very little effort.

But note again that I wouldn't expect a beginning student to be able 
to do that; however, with a little practice and some understanding of 
this type of problem, you may soon be able to answer a problem like 
this that easily.

Finally, note that I have only graphed one of the two inequalities in 
your problem. The other inequality will have as its graph all the 
points either on the boundary or OUTSIDE the square with vertices 
(2,0), (0,2), (-2,0), and (0,-2). The points outside the square will 
be in the solution set because this inequality is "greater than or 
equal to" 2.

So the solution set to the entire problem is all the points that are 
both inside the large square and outside the small square, or on the 
boundary of either square.

I hope all this helps. Please write back if you have any further 
questions about this.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/