# absolute value graph

• Feb 10th 2009, 12:14 PM
brentwoodbc
absolute value graph
|y|=|x| or |x+y|=1 how would you graph something like that.
I can do ones like y=|x+1|+1 etc. but Im stumped here.

Thanks...
• Feb 10th 2009, 01:53 PM
Jameson
One absolute value sign around one variable adds a reflection. Like y = |x| is the normal y=x line and y=-x. This is of course because |-x| = x. If both variables have an absolute value sign, that means that there are now four combinations of pluses and minuses that can satisfy the conditions (++, +-, -+,--) so for |y|=|x| you get y=x , -y=x , -y=-x , and y=-x. So I would approach these by graphing the original shape and then deciding how it is being reflected by the absolute values.
• Feb 10th 2009, 02:31 PM
brentwoodbc
would it be just a line starting at orgin with a slope of 1 contained to the upper right quadrant?

I made a x/y chart with the given equation. I'm sort of confused by the -/-, -/+ etc.

and when my teacher wrote.

y=|x|

=x if x >/= 0
= -x if x <0

what is before the ='s "y"?
• Feb 10th 2009, 02:38 PM
Jameson
Quote:

Originally Posted by brentwoodbc
would it be just a line starting at orgin with a slope of 1 contained to the upper right quadrant?

I made a x/y chart with the given equation. I'm sort of confused by the -/-, -/+ etc.

Sorry, the +- thing is confusing. If |x|=|y|, then (-1,-1),(1,1),(-1,1),(1,-1) all fit the criteria. Don't make a xy chart for every graph, that was just a conceptual thing. Absolution values are going to broaden the points in the Cartesian plane that satisfies the conditions.

edit: y=x is a line starting at the origin going into the first quadrant, BUT, what about all -x's? Whenever x<0, like you said y= -x, which makes the value positive again. This takes the third quadrant part of the line y=x and moves it to the second quadrant.
• Feb 10th 2009, 02:50 PM
Jameson
Quote:

Originally Posted by brentwoodbc
|x+y|=1 how would you graph something like that.

So now we need to find where x+y=1 and where -(x+y)=1, or x+y=-1. I think because both variables are contained in the same bracket that only two combinations exist instead of four like before. But do you see how this is graphed now?