Absolute Value GraphsDate: 11/16/2002 at 18:20:27 From: Katherine Subject: Finding vertices with absolute value graphs My question is, how would you graph the inequalities: abs(x)+abs(y)is less than or equal to 5 abs(x)+abs(y)is greater than or equal to 2 So far, I've tried to put it into slope intercept form and I got: abs(y) = -abs(x)+5 abs(y) = -abs(x)+2 From there, I wasn't sure where I should graph it because I'm not sure what to do with both absolute values on each side of the equation. I need your help to figure out this equation, graph it, and find its vertices. Date: 11/16/2002 at 21:14:38 From: Doctor Greenie Subject: Re: Finding vertices with absolute value graphs Hi, Katherine - You ask how I would draw this graph. First I will give you an answer that will do you no good: Since I have done hundreds of problems like this, I simply know what the graph is going to look like, so I just draw it. Perhaps after doing a few problems like this, you too will be able to have enough of an understanding of what's happening with graphs like this to do the same sort of thing. But for now, you are just learning - so let's take it slow and look at all the steps. For a student just learning how to do problems like this, the safest way is to break each absolute value down into cases depending on whether the quantity inside the absolute value is positive. Then remember that if a quantity is greater than or equal to 0, then its absolute value is that same quantity; if a quantity is less than 0, then its absolute value is the opposite of that quantity For a simple example: 2 >= 0, so |2| = 2 -2 < 0, so |-2| = -(-2) = 2 Let's look now at the first of your problems: graph |x| + |y| <= 5 Both x and y can be either positive or negative; so we have four possibilities: (1) x>0 and y>0 (2) x<0 and y>0 (3) x<0 and y<0 (4) x>0 and y<0 I purposely chose to number the four cases in this way, because then case (1) is in the 1st quadrant, case (2) is in the 2nd quadrant, case (3) is in the 3rd quadrant, and case (4) is in the 4th quadrant. So we have four cases to consider when we go to graph this inequality: (1) 1st quadrant; x>0 and y>0, so |x| = x and |y| = y, so the inequality is x+y <= 5 y <= -x+5 (2) 2nd quadrant; x<0 and y>0, so |x| = -x and |y| = y, so the inequality is -x+y <= 5 y <= x+5 (3) 3rd quadrant; x<0 and y<0, so |x| = -x and |y| = -y, so the inequality is -x-y <= 5 -x-5 <= y y >= -x-5 (4) 4th quadrant; x>0 and y<0, so |x| = x and |y| = -y, so the inequality is x-y <= 5 x-5 <= y y >= x-5 If you graph these respective lines in the respective quadrants, you will find the boundary lines for the solution set are the sides of the square whose vertices are (5,0), (0,5), (-5,0), and (0,-5). Furthermore, in the 1st and 2nd quadrants the solution set lies on or below the boundary lines, while in the 3rd and 4th quadrants the solution set lies on or above the boundary lines. So the complete solution set for the inequality is the square and all points inside the square. There is another way of thinking about absolute values that makes it easy to graph this inequality. If you are lucky, at some point in your math education one of your teachers will spend some time talking about this alternative interpretation of absolute value. In this alternate interpretation, we think of |a-b| as the "distance between a and b." For example, if a is 10 and b is 3, then a-b is 7 and b-a is -7; but the distance between a and b is 7 and the distance between b and a is 7. So |a-b| and |b-a| are both equal to 7. With this interpretation of absolute values, "|x|" is the same as "|x-0|" so |x| is the distance from 0. And similarly, |y| is the distance from 0. So when we write the inequality |x|+|y| <= 5 we are talking about all the points (x,y) for which "(the distance of x from 0) + (the distance of y from 0)" is less than or equal to 5. If you think of that interpretation, you can perhaps see why the graph turns out to be the boundary and interior of the square with vertices (5,0), (0,5), (-5,0), and (0,-5). This interpretation is, in fact, why I stated at the beginining of my response that I could sketch the graph of this inequality with very little effort. But note again that I wouldn't expect a beginning student to be able to do that; however, with a little practice and some understanding of this type of problem, you may soon be able to answer a problem like this that easily. Finally, note that I have only graphed one of the two inequalities in your problem. The other inequality will have as its graph all the points either on the boundary or OUTSIDE the square with vertices (2,0), (0,2), (-2,0), and (0,-2). The points outside the square will be in the solution set because this inequality is "greater than or equal to" 2. So the solution set to the entire problem is all the points that are both inside the large square and outside the small square, or on the boundary of either square. I hope all this helps. Please write back if you have any further questions about this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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