Question 210400: I have a question regarding a problem. I need to graph x + abs(y) = 3. I have set up a table and have come up with xy points of = 0,3; 1,2; 2,1; 3,0; 4,-1; 5,-2 and 6,-3. Then graphing the following points I come up with a straight line. I am wondering if this is correct. Your feedback is much appreciated.
Lori
Found 3 solutions by tutor_paul, Earlsdon, stanbon:
Answer by tutor_paul(519)   (Show Source):
You can put this solution on YOUR website! You are OK with your table of values until you got to 4,-1. Since you are dealing with the absolute value of y, it must be positive. When making your table, use the given equation as you did, but take the absolute value of y after you solve for it. So for the case of x=4, y=-1, but abs(y)=1. So when you graph this, you get a "v" shape.
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Good Luck,
tutor_paul@yahoo.com
Answer by Earlsdon(6294)  (Show Source):
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! I have a question regarding a problem. I need to graph x + abs(y) = 3. I have set up a table and have come up with xy points of = 0,3; 1,2; 2,1; 3,0; 4,-1; 5,-2 and 6,-3. Then graphing the following points I come up with a straight line. I am wondering if this is correct. Your feedback is much appreciated.
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x + abs(y) = 3
|y| = 3-x
(-3,6), (-2,5), (-1,4), (0,3), (1,2), (2,1), (3,0)
(-3,-6),(-2,-5, (-1,-4),(0,-3),(1,-2),(2,-1)
x cannot be greater than 3 because |y| cannot be negative.
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Notice that when x < 3, y can have two values.
You need to include that idea in your graphing.
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Cheers,
Stan H.
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