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INSTRUCTORSCarleen EatonGrant Fraser
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Lecture Comments (3)

1 answer

Last reply by: Magdy Mettias
Sun Sep 23, 2012 3:35 PM

Post by Nancy Dempsey on March 18, 2011

Thanks

0 answers

Post by Jeff Mitchell on March 5, 2011

At about 3:19 into the lecture one of the coordinate points was incorrectly graphed at (1,0) rather than (1,1). I am really enjoying your course program. Thanks,

Graphing Inequalities

  • The graph of a linear inequality is a half plane bounded by a straight line.
  • If the inequality is strict (< or >), then draw the boundary as a dashed line, otherwise draw it as a solid line.
  • Use a test point, normally the origin, to determine which half plane is the solution of the inequality. Shade that region.

Graphing Inequalities

Graph y ≤ [1/4]x + 5
  • Step 1:Graph the boundary line using the slope, m = [1/4] and the y - intercept (0,5). Starting at (0,5),
  • go Up 1 and Right 4. If you don't prefer this method, you may create a table of values.
  • The boundary line must be a solid line becasue you have the inequality ≤
  • Step 2: Test the boundary using a point not on the line, the easiest point to test is (0,0). Always check that is not on the line, or else the test will be useless.
  •  Test (0,0)
     y ≤ [1/4]x+5 0 ≤ [1/4]+5
      0 ≤ 0+5
      0 ≤ 5
      True, Shade towards (0,0) since (0,0) is part of the solution set.
Graph y ≥ 3x + 2
  • Step 1:Graph the boundary line using the slope, m = 3 = [3/1] and the y - intercept (0,2). Starting at (0,2),
  • go Up 3 and Right 1. If you don't prefer this method, you may create a table of values.
  • The boundary line must be a solid line becasue you have the inequality ≤
  • Step 2: Test the boundary using a point not on the line, the easiest point to test is (0,0). Always check that is not on the line, or else the test will be useless.
  •  Test (0,0)
     0 ≥ 3(0)+2
    y ≥ 3x+2 0 ≥ 0+2
      0 ≥ 2
      Not True, shade away from (0,0) since (0,0) is not part of the solution set
Graph y < − [1/3]x − 2
  • Step 1:Graph the boundary line using the slope, m = − [1/3] and the y - intercept (0, − 2).
  • Starting at (0, − 2), go Down 1 and Right 3. If you don't prefer this method, you may create a table of values.
  • The boundary line must be a dashed line becasue you have the inequality <
  • Step 2: Test the boundary using a point not on the line, the easiest point to test is (0,0). Always check that is not on the line, or else the test will be useless.
  •  Test (0,0)
     0 < −[1/3](0)−2
    y < −[1/3]x−2 0 < 0−2
      0 < −2
      Not True, shade away from (0,0) since (0,0) is not part of the solution set
Graph y > [3/2]x + 1
  • Step 1 : Graph the boundary line using the slope, m = [3/2] = [3/2] and the y - intercept (0,1). Starting at (0,1), go Up 3 and Right 2. If you don't prefer this method, you may create a table of values.
  • The boundary line must be a dashed line becasue you have the inequality >
  • Step 2 : Test the boundary using a point not on the line, the easiest point to test is (0,0). Always check that is not on the line, or else the test will be useless.
  •  Test (0,0)
     0 > [3/2](0)+1
    y < [3/2]x+1 y > 0+1
      0 > 1
      Not True, shade away from (0,0) since (0,0) is not part of the solution set
Graph 5x + 2y > 10
  • Step 1: Graph the boundary line by finding the x - and y - intercept. Boundary line will be a dashed line.
  •   X-Intercept Y-Intercept
      Set y = 0 Set x = 0
    5x + 2y = 10    
  •   X-Intercept Y-Intercept
      Set y = 0 Set x = 0
    5x + 2y = 10 5x + 2(0) = 10 5(0)+2y = 0
      5x = 10 2y = 10
      x = 2 y = 5
      (2,0) (0,5)
  • Step 2: Test the boundary using a point not on the line, the easiest point to test is (0,0). Always check that is not on the line, or else the test will be useless.
  •  Test (0,0)
     5(0)+2(0) > 10
    5x+2y > 10 0 > 10
      Not True, shade away from (0,0) since (0,0) is not part of the solution set
Graph 2x + 3y < 6
  • Step 1: Graph the boundary line by finding the x - and y - intercept. Boundary line will be a dashed line.
  •   X-Intercept Y-Intercept
      Set y = 0 Set x = 0
    2x+3y=6    
  •   X-Intercept Y-Intercept
      Set y = 0 Set x = 0
    2x+3y=6 2x + 3(0) = 6 2(0) + 3y = 6
      2x = 6 3y = 6
      x = 3 y = 2
      (3,0) (0,2)
  • Step 2: Test the boundary using a point not on the line, the easiest point to test is (0,0). Always check that is not on the line, or else the test will be useless.
  •  Test (0,0)
     2(0) +3(0) < 6
    2x+3y=6 0 < 6
      True, shade towards (0,0) since (0,0) is part of the solution set
Graph − 4x + y > 4
  • Step 1: Graph the boundary line by finding the x - and y - intercept. Boundary line will be a dashed line.
  •   X-Intercept Y-Intercept
      Set y = 0 Set x = 0
    -4x+y=4    
  •   X-Intercept Y-Intercept
      Set y = 0 Set x = 0
    -4x+y=4 -4x+0=4 -4(0)+y=4
      -4x = 4 y=4
      x = -1 (0,4)
      (-1,0)  
  • Step 2: Test the boundary using a point not on the line, the easiest point to test is (0,0). Always check that is not on the line, or else the test will be useless.
  •   Test (0,0)
     −4x+y > 4
    −4x+y=4 −4(0)+0 > 4
      0 > 4
      Not True, shade away (0,0) since (0,0) is not part of the solution set
Graph − 2x + 5y < 10
  • Step 1: Graph the boundary line by finding the x - and y - intercept. Boundary line will be a dashed line.
  •   X-Intercept Y-Intercept
      Set y = 0 Set x = 0
    -2x+5y=10    
  •   X-Intercept Y-Intercept
      Set y = 0 Set x = 0
    -2x+5y=10 -2x+5(0)=10 -2(0)+5y=10
      -2x = 10 5y = 10
      x = -5 y = 2
      (-5,0) (0,2)
  • Step 2: Test the boundary using a point not on the line, the easiest point to test is (0,0). Always check that is not on the line, or else the test will be useless.
  •   Test (0,0)
     −2(0)+5(0) < 10
    −2x+5y < 10 0+0 < 6
      0 < 6
      True, shade towards (0,0) since (0,0) is part of the solution set
Graph y > |x − 3| + 3
  • Step 1: Graph the boundary line by completing a table of values.
  • x f(x) = - x-3 - + 3
    0  
    1  
    2  
    3  
    4  
    5  
    6  
  • x f(x) = |x−3| + 3
    0 = |0 − 3| + 3 = 3 + 3 = 6
    1 = |1 − 3| + 3 = 2 + 3 = 5
    2 = |2 − 3| + 3 = 1 + 3 = 4
    3 = |3 − 3| + 3 = 0 + 3 = 3
    4 = |4 − 3| + 3 = 1 + 3 = 4
    5 = |5 − 3| + 3 = 2 + 3 = 5
    6 = |6 − 3| + 3 = 3 + 3 = 6
  • Step 2: Test the boundary using a point not on the line, the easiest point to test is (0,0). Always check that is not on the line, or else the test will be useless.
  •   Test (0,0)
     0 > |0−3|+3
    y > |x−3|+3 0 > 3+3
      0 > 6
      Not True, (0,0) is not part of the solution set
      Shade within the boundaries of the function
Graph y < |x − 4| − 2
  • Step 1: Graph the boundary line by completing a table of values.
  • x f(x) = y < |x − 4| − 2
    1  
    2  
    3  
    4  
    5  
    6  
    7  
  • x f(x) = y < |x − 4| − 2
    1 = |1 − 4| − 2 = 3 − 2 = 1
    2 = |2 − 4| − 2 = 2 − 2 = 0
    3 = |3 − 4| − 2 = 1 − 2 = − 1
    4 = |4 − 4| − 2 = 0 − 2 = − 2
    5 = |5 − 4| − 2 = 1 − 2 = − 1
    6 = |6 − 4| − 2 = 2 − 2 = 0
    7 = |7 − 4| − 2 = 3 − 2 = 1
  • Step 2: Test the boundary using a point not on the line, the easiest point to test is (0,0). Always check that is not on the line, or else the test will be useless.
  •   Test (0,0)
     0 < |0 − 4| − 2
    y < |x−4|−2 0 < 4 − 2
      0 < 2
      True, (0,0) is part of the solution set
      Shade outside the boundaries of the function

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Graphing Inequalities

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Graphing Linear Inequalities 0:07
    • Shaded Region
    • Using Test Points
    • Graph Corresponding Linear Function
    • Dashed or Solid Lines
    • Use Test Point
    • Example: Linear Inequality
  • Graphing Absolute Value Inequalities 4:50
    • Graph Corresponding Equations
    • Use Test Point
    • Example: Absolute Value Inequality
  • Example 1: Linear Inequality 9:17
  • Example 2: Linear Inequality 11:56
  • Example 3: Linear Inequality 14:29
  • Example 4: Absolute Value Inequality 17:06

Transcription: Graphing Inequalities

Welcome to Educator.com.0000

In today's lesson, we are going to be graphing inequalities.0002

Recall that, in order to graph a linear inequality, you have to take a series of steps.0008

And the graph of a linear inequality is a shaded region in the coordinate plane.0014

And those shaded regions are known as half-planes; the boundary of the region, or the half-plane, is a straight line,0019

which is the graph of the corresponding linear function.0027

The region, or the half-plane, containing the solution set can be determined by using a test point.0032

OK, so the steps in order to graph a linear inequality are to first graph the corresponding linear function; and that is a straight line.0040

And that line is dashed if the inequality you are working with is a strict inequality (less than or greater than).0058

The line is solid if you are dealing with less than or equal to, or greater than or equal to; and we will talk in a minute about what that means.0071

After you have graphed the line, then you end up with the coordinate plane divided into two half-planes.0079

You use a test point to determine the region containing the solution set.0097

So, I am going to go ahead and do an example now: but if you need more review on this, check out the Algebra I video on this topic.0104

All right, the easiest way to understand is to look at an example: let's look at the inequality y - 3x < -2.0112

My first step is to graph the corresponding linear function, which would be y - 3x = -2.0123

So, I am going to go ahead and find some x-values and some y-values--some coordinate pairs on this line.0132

When x is 0, y is -2; so that gives me the y-intercept.0140

When x is 1, this is going to give me -3; add 3 to both sides--y will be 1.0145

When x is 2, -3 times 2 is -6; add that to both sides, and y is 4.0155

All right, I have enough to graph the line: x is 0, y is -2--the y-intercept; then I have another point at (1,1) and another point at (2,4).0164

Because this is a strict inequality, I am going to use a dashed line.0177

This line divides the coordinate plane into two half-planes: an upper and lower half-plane.0187

One of these half-planes contains the solution set for the inequality; the way I figure out which one is to use a test point.0193

Pick a point that is not on the boundary line, and that is easy to work with.0201

(0,0) is very easy to work with; so I am going to take my test point, and I am going to go back to my inequality and plug it in.0206

I am going to see what happens: this gives me 0 - 0 < -2; 0 is less than -2--that is not true.0216

What this means is that this is not part of the solution set.0230

If this is not part of the solution set, it is in the half-plan that is not contained in the solution set.0234

Therefore, the solution set is actually in the lower half-plane.0240

And you could verify that by checking a point down here.0245

The idea is to graph the corresponding linear function to find the boundary line.0249

Now, I mentioned that this is a strict inequality, so I am using a dashed line.0253

And what that means is that the boundary line is not part of the solution set.0256

If it were a solid line (because I had less than or equal to), then the boundary line would be part of the solution set.0262

But actually, in this, the solution set is just below that boundary line.0268

I determined which half-plane was correct by using a test point.0272

Since the test point ended up giving me an inequality that is not true, that is not valid,0276

I knew that that test point did not lie in the half-plane of the solution set, so I shaded in the lower half-plane.0281

Graphing absolute value inequalities: the procedure for these inequalities is similar to the procedure for linear inequalities.0290

So, we are going to use the same technique.0298

Recall that the technique involved first graphing the line for the corresponding equation.0302

Graph the corresponding equation: in this case, it is going to be an absolute value equation.0309

And then, use a test point to determine the half-plane containing the solution set for the inequality.0318

All right, let's take an example: y ≤ |x-3|.0338

The corresponding equation would be y = |x-3|, so I need to graph that to find my boundary.0347

So, some values for x and y: when x is -2, -2 minus 3 is -5; the absolute value is 5.0357

When x is -1, this becomes -4; the absolute value is 4; when x is 0, this becomes -3; the absolute value is 3.0368

Let's jump up to 3: 3 minus 3 is 0--the absolute value is 0.0381

4 minus 3 is 1; the absolute value is 1; 5 minus 3 is 2--the absolute value is 2.0389

OK, plotting these points: first, (-2,5), (-1,4), (0,3), (3,0), (4,1), (5,2); OK, I have my typical v-shaped graph of absolute value.0399

And you notice that I made this a solid line; and the reason that is a solid line is because I had less than or equal to.0439

So, this boundary is actually going to be part of the solution set.0447

OK, so I graphed the corresponding equation, and I determined that it is a solid line.0451

Now, I need to determine if the solution set is down here, or if it is above the boundary; is it above or below the boundary line?0455

So, I am going to use a test point of (0,0): that is going to be my test point.0462

I go back to the inequality, and I see what happens when I put my test point values in there.0468

This gives me 0 ≤ |-3|; that would say that 0 is less than or equal to 3; that is true.0477

Since that is true, what that tells me is that this is part of the solution set.0489

(0,0) is part of the solution set, and that means that this lower half-plane where (0,0) is, my test point, contains the solution set.0495

So, as I mentioned, this is very similar to the technique for graphing linear inequalities.0505

First, graph the corresponding equation; and use a solid line if it is less than or greater than or equal to.0510

Use a dashed line if it is a strict inequality (just greater than or less than).0520

So then, once you have your boundary line, find a test point that is off the boundary line, and not so close to it as to cause any confusion.0525

And choose one that is simple: (0,0), (1,1)--very easy to work with.0533

Plug that into the absolute value inequality; and if you come out with a true inequality,0538

like this one, you know that you are in the correct area with the solution set.0544

If you come out with a statement or inequality that is not valid, you shade in the other region.0548

OK, the first example: we have an inequality that we are asked to graph: y ≤ 2x - 3.0557

My first step is going to be to graph the corresponding linear equation, in order to determine the boundary.0567

Some values for x and y: when x is 0, y is -3; when x is 1, 1 times 2 is 2, minus 3 is -1.0581

When x is 2, 2 times 2 is 4, minus 3 is 1.0594

OK, so plotting this out: (0,-3), (1,-1), (2,1); I have plotted this boundary line, and I look up here,0599

and I see that it is less than or equal to; so I am going to make this a solid line.0615

It is going to be a solid line: this line will be part of the solution set.0621

So, first I graph the boundary; next, I am going to look at a test point.0626

OK, I can again use this very simple test point of (0,0), the origin.0638

Going back to the inequality to look at my test point, y ≤ 2x -3; the test point is (0,0), so y is 0; x is 0.0646

This is saying that 0 is less than or equal to -3; that is not true--this is not in the solution set.0664

Well, the origin is up here; this means that this whole area is not part of the solution set.0678

So, I am going to go to the lower half-plane and shade that in to indicate that this shaded region, including the boundary line, is the solution set.0684

Again, the first step is to take the corresponding linear equation and graph it.0692

The second step: find a test point that is easy to work with and away from the boundary line, and plug that in.0698

If you come up with an inequality that is not true, like this, shade the other half.0705

If it is true, shade the half containing the test point.0711

Example 2: We are graphing another inequality, 2x + y > 6.0716

First, I am going to find the corresponding equation: 2x + y = 6.0724

I am going to keep this simple: I am going to use the intercept method in order to graph this.0733

I am going to let x equal 0 to find the y-intercept: when x is 0, y is 6.0738

Then, I am going to let y equal 0 to find the x-intercept: when y is 0, then I am going to end up with 2x + 0 = 6, or 2x = 6; x = 3.0745

So, I have my x and y intercepts; it is going to give me...when x is 0, y is 6, right about there; when x is 3, y is 0.0759

OK, then I look, and I have a strict inequality, so I am going to have a dashed line;0771

this line is going to be dashed, and is not going to be part of the solution set.0777

So, I found my boundary line; and that is my first step; my second step is going to be to work with my test point.0788

The test point is right here: it is the origin, (0,0).0798

Going back and looking at the inequality: 2 times 0, plus 0, is greater than 6; 0 plus 0 is greater than 6.0803

Well, 0 is not greater than 6, so this is not true--not in the solution set.0817

The origin is not part of the solution set, so the lower half-plane is not the correct half-plane; I am going to shade, instead, this upper half-plane.0830

Again, graph the linear equation corresponding to the inequality.0839

Since this is strict inequality, plot that line using a dashed line, since this boundary is not part of the solution set.0844

Find a test point; the origin is a really good one, as long as it is not on the boundary line or very close to the boundary line.0853

Substitute (0,0) into the inequality; and you come up with an inequality that is not true.0859

Example 3: Graph 3x - 4 > y.0870

First, graph the boundary line: the corresponding equation is 3x - 4 = y.0876

To graph that, I am going to need some x and y values; when x is 0, y is -4; when x is 1, that gives me 3 - 4 = y, so y equals -1.0889

One more value: when x is 2, y is 2; OK.0908

So, I have enough information to plot my line: when x is 0, y is -4; (1,-1), (2,2).0918

The next issue: strict inequality--that means this is a dashed line.0935

OK, so now, I have this boundary; I have an upper and a lower half-plane; and I need to figure out where my solution set is.0948

The origin is well away from the boundary line, so I can use that as my test point.0956

So, test point is (0,0): the inequality is 3x - 4 > y; and this gives me 0 - 4 > 0, which comes out to -4 > 0.0963

And that is, of course, not true; this is not in the solution set, because this inequality did not hold true.0986

Therefore, I am going to go to the other half-plane, and I am going to shade that in.0997

I graphed the boundary line using a dashed line, since it is a strict inequality.1007

And then, I took a test point, substituted those values into the inequality, and determined that the origin (0,0) is not part of the solution set.1011

So, my other half-plane contains the solution set.1022

Example 4: we are working with an absolute value inequality where y is less than the absolute value of x - 2, plus 1.1027

I am starting out with just thinking that I expect this to look like a v, because absolute value graphs are v-shaped.1036

I am graphing the corresponding equation, y = |x - 2| + 1.1046

And with absolute value inequalities or equations, I like to get more values,1051

because I want to make sure that I don't end up missing half of my v.1055

So, when x is -2, that is going to give me -4; the absolute value of that is 4; plus one is 5.1061

-1 minus 2 is -3; the absolute value is 3, plus one is 4.1070

0 minus 2 is -2; the absolute value is 2, plus one is 3.1077

Now, some positive numbers: 1 minus 2 is -1; the absolute value is 1; plus one is 2.1085

2 minus 2 is 0, plus one is 1; 3 minus 2 is 1; the absolute value is 1, plus one is 2.1094

4 minus 2 is 2; the absolute value is 2, plus one is 3; and now I have enough values to work with.1111

So, (-2,5), (-1,4), (0,3), (1,2), (2,1), (3,2)...so you see the v forming, so I know I have enough points,1119

because I have both parts of my v...(4,3).1145

It's a strict inequality, so I am using a dashed line.1150

And you see that it is the typical v shape, but it is shifted over.1161

And I have a -2 here, and so this is actually shifted over to the right by 2; and plus 1, so it is shifted up by 1.1165

I found the boundary; again, I am fortunate--I can easily use (0,0) as the test point--the origin.1173

Going back to the inequality: this is going to be 0 < |0 - 2| + 1.1187

So, that is 0 < |-2| (the absolute value of -2 is 2) + 1, so this gives me 0 < 3.1201

0 actually is less than 3; therefore, the origin is part of the solution set, so this is a true statement, and the origin is part of the solution set.1219

(0,0) is part of the solution set, and that means that I am in the correct half-plane, and I am going to go ahead and shade that in.1240

OK, so we handled this just with a very similar technique to linear inequalities.1257

Graphing out the absolute value equation, I got my v-shaped graph with a dashed line for the strict inequality.1263

I used a test point, (0,0), substituted that into the inequality, and found that I had a true statement--a true inequality, 0 < 3.1270

Since that is true, I knew that the origin lay within the region that contains the solution set.1279

So, I went ahead and shaded that region to note that my solution set is located there.1289

That concludes this lesson of Educator.com; see you next lesson!1297