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Confused . y=x-|x|

  1. Feb 19, 2008 #1
    confused ..... y=x-|x|

    The problem is to graph this equation y=x-|x|.



    From what I understand of absolute values, this x would be positive. If it is positive then y=0 and there would be no points to graph. Is there something that I am missing? The question is worth 4 points, so I can't see the answer just being 0. Thanks for any suggestions.
     
  2. jcsd
  3. Feb 19, 2008 #2

    D H

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    What happens when x is negative?
     
  4. Feb 19, 2008 #3
    What I was taught to do when dealing with absolute value, is to rewrite the equation so that the absolute value is isolated, then find the 2 equations.

    so you'll have:
    x=x-y
    y=o

    and also:
    x=-x+y
    2x=y
     
  5. Feb 19, 2008 #4
    The absolute value of -x would be x. But I remember with inequalities there are 2 possible answers with absolute value (-,+). From what you are explaining it sounds like that is what you are saying to do, use both possible values. Then I would graph by beginning with y at 0 and continue by substituting values into 2x=y? That seems to make sense to me. Thank you both for your help!
     
  6. Feb 19, 2008 #5
    Also just try plugging in some numbers:
    For 2 -> Y = 2 - |2| = 0
    For -2 - > Y = -2 -|-2| = -2 -2 = -4
    see?
    So for negatives you have Y = 2X, X<0
     
  7. Feb 19, 2008 #6

    symbolipoint

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    Solve or graph y = x - |x|.

    If x>0, then y = x - x, meaning y=0.

    If x<0, then y = x - (-x) [ notice those are parentheses, not absolute value notation symbols ], meaning y = x + x = 2x.
     
  8. Feb 19, 2008 #7

    HallsofIvy

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    Surely you didn't mean to say that! x itself can be any number. |x| is always positive (or 0- don't forget that!

    Why did you switch to x=? If x[itex]\ge 0[/itex] y= x- x= 0. The graph is just the x axis from x= 0 to the right.

    If x< 0, don't forget that. Then y= x- (-x)= 2x.

    No, no, no! |-x|= |x| which may be eigther x or -x depending upon what x is.

    Draw the graph of y= 2x, to the left of x= 0. To the right, the graph is just y= 0, the x-axis.
     
  9. Feb 19, 2008 #8
    I rearranged the formula to isolate the absolute value.
     
  10. Feb 19, 2008 #9
    Wow, you guys couldn't have made it any more clear for me. Thanks alot!
     
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