# Graph the boundary line the same as if the problem was a linear equation.  Pretend that there is an equal sign and use an appropriate method to graph.

## Presentation on theme: "Graph the boundary line the same as if the problem was a linear equation.  Pretend that there is an equal sign and use an appropriate method to graph."— Presentation transcript:

Graph the boundary line the same as if the problem was a linear equation.  Pretend that there is an equal sign and use an appropriate method to graph the line.

When you draw the boundary line, you need to have a way to indicate if the line is included or not in the final answer.

If the problem includes where it is equal, then you will have a solid boundary line. In other words, if you have, you will have a solid line for your boundary line.

This shows the boundary line for x + y < 6: (Note that this does not show the inequality part ) This shows the boundary line for x + y < 6: (Note that this does not show the inequality part )

If the problem does not include where it is equal, then you will use a dashed boundary line. In other words, if you have, you will have a dashed line for your boundary line.

This shows the boundary line for x + y < 6: (note that this does not show the inequality part) This shows the boundary line for x + y < 6: (note that this does not show the inequality part) In either case, you still graph the line the same. You just have to decide if you are needing a solid line or a dashed line.

 Use a solid line if your equation contains any part of an equal sign ( =, ≤, ≥ ) to show that points that fall on the line are include in the solution.  Use a dotted or dashed line if you have to show that the points on this line are not part of the solution area.

Pick a test point on either side of the boundary line and plug it into the original problem. This will help determine which side of the boundary line is the solution.

If you get a true statement when you plug in the test point in step 2, then you have found a solution. Shade the region that the test point is in. If you get a false statement when you plug in the test point in step 2, then you don’t have a solution. Shade in the region that is on the other side of the test point.

Step 1: Graph the boundary Line  Use the intercepts to graph the boundary line. When working with only the only boundary line, put an equal sign between the two sides to emphasize that you are working on the boundary line.

 What value is y on the x-intercept? If y = 0, x + y = 2 x + 0 = 2 x = 2 x-intercept is (2, 0)

 What is the value of x on the y-intercept? If x = 0, x + y = 2 0 + y = 2 y = 2 y-intercept is (0, 2).

xy(x,y) 20(2, 0) 02(0, 2) 11(1, 1) Since the original problem has a >, this means it DOES NOT include the boundary line.

 Putting it all together, we get the following boundary line for this problem:

Note how the boundary line separates it into two parts. An easy test point would be (0,0). Note that it is a point that is not on the boundary line. In fact, it is located below the boundary line.

 Let’s put (0, 0) into the original problem and see what happens: Replacing x and y with 0 x + y > 2 0 + 0 > 2 0 > 2

 Since our test point (0, 0) made our inequality FALSE, this means it is not a solution.  Since it has to be on one side or the other of the boundary line, and it is not below it, then our solution would lie above the boundary line. This means we will shade in the part that is above it.

Sketch x = -3 x = -3 Now pick a test point on one side of the dotted line like (0,0) BACK

 Take the point (0,0) and plug in the x value in x < -3 0 < -3 Since it’s false, shade the side opposite of (0,0).

x = -3 Shade the area with true solutions!

Sketch y = 4 y = 4 Now pick a point on one side of the solid line (0,0)

 Take the point (0,0) and plug in the y value in y ≤ 4 0 ≤ 4 Since it’s True, shade the side that (0,0) is on.

 Graph y ≤ 2x - 1 You try this one BACK

Sketch y = 2x - 1 y= 2x - 1 Now pick a test point on one side of the dotted line (-1,0)

 Take the point (-1,0) and plug in the values in y ≤ 2x - 1 y ≤ 2x - 1 0 ≤ 2(-1) -1 0 ≤ -3 Since it’s False, shade the opposite side of (-1,0).

x y Graph the following linear system of inequalities. Use the slope and y- intercept to plot the two lines. Draw in the line. For < use a dashed line. Pick a point and test it in the inequality. Shade the appropriate region. BACK

Sketch y = 2x - 1 y= 2x - 1 Now pick a point on one side of the dotted line - (-1,0)

 Take the point (-1,0) and plug in the values in y < 2x - 1 0 < 2(-1) -1 0 < -3 Since it’s False, shade the opposite side of (-1,0). on….

Find the answer of the following inequality.  y < x + 2  y > -1/2x + 5

Graph y = x + 2 Now pick a point on one side of the dotted line -(0,0)

 Substitute (0,0) in for x and y y < x + 2 0 < 0 + 2 0 < 2

Graph y = x + 2

Graph y = -1/2x + 5 Now pick a point on one side of the dotted line -(0,0)

 Substitute (0,0) in for x and y y > -1/2x + 5 0 > -1/2(0) + 5 0 > 5

Graph y = -1/2x + 5

Perform the (0,0) test. Insert 0 for x and 0 for y into the inequality and see if the statement remains true. Because the statement is true, (0,0) is included in the solution and covered in the graph. Key Skill Graph y < 2x + 4 Create a table, treat the inequality like an equation. x y 0 4 2 8 Graph the line, use a dashed line because the inequality is < and not <. 0 < 2(0) + 4 0 < 4, which is true.

Perform the (0,0) test. Insert 0 for x and 0 for y into the inequality and see if the statement remains true. Because the statement is true, (0,0) is included in the solution and covered in the graph. Key Skill Graph y > (1/3)x - 4 Create a table, treat the inequality like an equation. x y 0 -4 3 -3 Graph the line, use a solid line because the inequality is > and not >. 0 > (1/3)(0) -4 0 > -4, which is true. TRY THIS

Key Skill Graph y < 2x + 4 and y < 3 x y 0 2 4 8 0 < 4 true x y 0 3 2 3 0 < 3 true Fill in the region of intersection.

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