Graph y=x-|x|: Is There a Solution?

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In summary, the conversation is about graphing the equation y=x-|x|. The speakers discuss how to handle absolute values and inequalities in the equation, and come to the conclusion that the graph is just the x-axis from x=0 to the right, and y=2x to the left of x=0. They also discuss how to isolate the absolute value and substitute values to graph the equation.
  • #1
mikebc
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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.
 
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  • #2
What happens when x is negative?
 
  • #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
 
  • #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!
 
  • #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
 
  • #6
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.
 
  • #7
mikebc said:
The problem is to graph this equation y=x-|x|.



From what I understand of absolute values, this x would be positive.
Surely you didn't mean to say that! x itself can be any number. |x| is always positive (or 0- don't forget that!

dranseth said:
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
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.

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

mikebc said:
The absolute value of -x would be x.
No, no, no! |-x|= |x| which may be eigther x or -x depending upon what x is.

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

Related to Graph y=x-|x|: Is There a Solution?

1. What is the graph of y=x-|x|?

The graph of y=x-|x| is a V-shaped graph that intersects the x-axis at (0,0) and has a slope of 1 on the right side and -1 on the left side.

2. Does the equation y=x-|x| have a solution?

Yes, the equation y=x-|x| has a solution. In fact, it has infinite solutions since any value of x will result in a corresponding value of y.

3. What is the domain and range of the function y=x-|x|?

The domain of the function y=x-|x| is all real numbers, and the range is all real numbers less than or equal to 0.

4. How can I graph y=x-|x|?

To graph y=x-|x|, you can plot a few points and connect them with a V-shaped line. You can also use the slope of 1 on the right side and -1 on the left side to help guide your graph.

5. What is the significance of the absolute value in the equation y=x-|x|?

The absolute value in the equation y=x-|x| is what creates the V-shape of the graph. It ensures that the output of the function is always positive, which results in the V-shape instead of a straight line.

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