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Graphing absolute values

  1. Sep 28, 2005 #1
    I'm supposed to sketch this graph

    [tex] \vert x \vert + \vert y \vert = 1 + \vert xy \vert [/tex]

    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.

  2. jcsd
  3. Sep 28, 2005 #2


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    Homework Helper

    How about

    \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} [/tex]

    Then the same thing for x.
  4. Sep 28, 2005 #3


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    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.)
  5. Sep 28, 2005 #4
    I feel the pain of not having done any math for 7 years!! :cry:

    Thanks for your help guys!
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