1. ## Absolute Value Graphs

Hi

For y = abs(x+2) + abs(x) + abs(x-2), what are the cases?

I know if there were only two abs, there would be three cases to work out but I have no clue how many there will be for this question and what they might be.

Thanx

2. Dear xwrathbringerx,

the abs(x) function behaves "irregular" at x=0 that is x=0 is a singular point.
Therefore the function abs(x-2) has sigularity at x=2, hasn't it?

So in your problem you have to divide into 4 cases because there are 3 singular point.

3. If $\displaystyle x\le -2$ then $\displaystyle y=-(x+2)-x-(x-2)=-3x$.
If $\displaystyle -2\le x\le 0$ then $\displaystyle y=(x+2)-x-(x-2)=-x+4$.
If $\displaystyle 0\le x\le 2$ then $\displaystyle y=(x+2)+x-(x-2)=x+4$.
If $\displaystyle 2\le x$ then $\displaystyle y=(x+2)+x+(x-2)=3x$.

The graph is attached.

4. Hello, xwrathbringerx!

Another approach . . .

For $\displaystyle y \:=\:|x+2| + |x| + |x-2|$, what are the cases?
Make a sketch . . .

The graph of .$\displaystyle y \,=\,|x|$ .is a $\displaystyle \vee$ with its vertex at the origin.

The graph of .$\displaystyle y \,=\,|x-2|$ .is the same graph moved 2 units to the right.

The graph of .$\displaystyle y \,=\,|x+2|$ .is the same graph moved 2 units to the left.

Code:
      \    \    |    /    /
\    \   |   /    /
\    \  |  /    /
\    \ | /    /
\    \|/    /
\    \    *    /    /
\    \  /|\  /    /
\    \/ | \/    /
\   /\ | /\   /
\ /  \|/  \ /
---------*----*----*-------
-2    |    2