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.
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.
Another approach . . .
Make a sketch . . .For , what are the cases?
The graph of . .is a with its vertex at the origin.
The graph of . .is the same graph moved 2 units to the right.
The graph of . .is the same graph moved 2 units to the left.
Code:\ \ | / / \ \ | / / \ \ | / / \ \ | / / \ \|/ / \ \ * / / \ \ /|\ / / \ \/ | \/ / \ /\ | /\ / \ / \|/ \ / ---------*----*----*------- -2 | 2
Then add the ordinates.
We get the graph that james_bond so thoughtfully provided.