When graphing an absolute value linear inequality, first make sure the inequality is in terms of y. Use inverse operations to isolate y. To find points on the graph, set up a table of values. Keep in mind that absolute value linear inequalities are v-shaped. Use a wide range of values in order to get points that cover both sides of the "v". Use a dotted or solid line, depending on the inequality sign, to connect the points. After graphing using a table of values, shade in the appropriate region. Pick any point on the coordinate plane and plug it into the inequality. If the points make the inequality true, then shade in that region of the graph. If it is not true, then shade the opposite region of the graph. In other words, if y is greater than the absolute value quantity, then shade above the graph. If y is less than the absolute value quantity, then shade below the graph.
When I want to solve an inequality absolute value and graph it for when y isn't by itself it's really similar to what you guys have learned with your graphing linear equations. What that means is that before I can make my table I need to get y all by itself.
I'm going to divide both side by two so that I'll know how to work with the inequality y is greater than or equal to the absolute value of 2x. Then what I'm going to do is go through and make a table choosing x numbers and substituting them in, finding the corresponding y numbers and then we'll have the graph. You already know that an absolute value should be a v-shape as long as nothing squared or anything. So I'm hoping that these x numbers will give me points that will show me the entire v-shape. It'll show me where the v turning around.
Letâ€™s just hope that's the case, let's see so if my x numbers -2 times 2 is -4, absolute value becomes +4. -1 for x I'll have -2 inside the absolute value, when you absolute value-ize it becomes +2. If I put in zero for x I'll have zero for y. If I put in one for x, the absolute value of two is two. If I put in two for x I'll have four for y. So there are my points let's go ahead and get those on the graph.
(-2,4), (-1,2), (0,0), (1,2) and (2,4) and good I'm lucky I can see that v-shape. This one is a skinny v there we go. Hopefully you can see that a little better. I got a skinny v in this situation because it's being multiplied by two and when you guys learn about shifting rules and dilations either in this class or geometry, you'll find even more short-cuts for the graphing.
So I've got my v-shape that's good next thing I need to do is decide if this would be a solid line or a dashed line. Since itâ€™s greater than or equal to its going to be a solid v, solid line, so that's good. The last thing I need to do is shading, I want to pick a point that's not on my v and plug in the x and y values to this original problem and see if I get a true inequality or not. I'm just going to pick kind of randomly the point (1,1) I just like (1,1). So my x number is going to be one and my y number is going to be one.
Is it true that 2 times 1 is greater than or equal to the absolute value of 4 times 1? Oops letâ€™s see is 2 greater than or equal to 4? No that means don't shade here, don't shade outside the v I need to shade inside the v. Any point that I just picked that was inside this v-shape would work in this inequality absolute value statement. So the key to keep in mind is that before you make a table, you need to get the y all by itself but it's not so tricky as long as you guys remember your y=mx+b solving techniques.
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