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Math Help - absolute value graph

  1. #1
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    Question 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...
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  2. #2
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    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.
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  3. #3
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    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"?
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  4. #4
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    Quote Originally Posted by brentwoodbc View Post
    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.
    Last edited by Jameson; February 10th 2009 at 02:53 PM.
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  5. #5
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    Quote Originally Posted by brentwoodbc View Post
    |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?
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