# Homework Help: Graphing absolute values

1. Sep 28, 2005

### Jeff Ford

I'm supposed to sketch this graph

$$\vert x \vert + \vert y \vert = 1 + \vert xy \vert$$

I think the purpose of the exercise is to simplify this into something that resembles a typical function and be able to shift the graph over so that it looks normal. I'm having troulbe getting the y's all on one side. A push in the right direction would be appreciated.

Thanks,
Jeff

2. Sep 28, 2005

### TD

$$\begin{gathered} \left| x \right| + \left| y \right| = 1 + \left| {xy} \right| \hfill \\ \left| y \right| - \left| {xy} \right| = 1 - \left| x \right| \hfill \\ \left| y \right|\left( {1 - \left| x \right|} \right) = 1 - \left| x \right| \hfill \\ \left| y \right| = \frac{{1 - \left| x \right|}} {{1 - \left| x \right|}} = 1 \Leftrightarrow y = 1 \vee y = - 1 \hfill \\ \end{gathered}$$

Then the same thing for x.

3. Sep 28, 2005

### HallsofIvy

The usual drill for absolute value:

If x and y are both positive (Ist quadrant), |x|= x, |y|= y, |xy|= xy so
x+ y= 1+ xy. Then y-xy= y(1-x)= 1- x or y= 1. That's easy to graph!

If x is negative and y positive (IInd quadrant), |x|= -x, |y|= y, |xy|= -xy so
-x+ y= 1-xy. Then y+xy=y(1+x)= 1+ x or y= 1. Interesting!

If both x and y are negative (IIIrd quadrant), |x|= -x, |y|= -y, |xy|= xy so
-x- y= 1+xy. Then y+xy= y(1+x)= -x-1 or y= -1. I think I see a pattern!

If x is positive and y is negative (IVth quadrant), |x|= x, |y|= -y, |xy|= -xy so
x- y= 1- xy. Then xy-y= y(x-1)= 1-x or y= -1. Yes, that's very easy to graph!

(y is not, however, a function of x. This is a "relation" between x and y.)

4. Sep 28, 2005

### Jeff Ford

I feel the pain of not having done any math for 7 years!!