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### 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

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
Middle School Graphing Equations