Graphing Inequalities in Two Variables
Objective • Graph inequalities in the coordinate plane.
Graphing an Inequality <, >, ≤, or ≥ 1. Solve the equation for y (if necessary). 2. Graph the equation as if it contained an = sign. 3. Draw the line solid if the inequality is ≤, or ≥ 4. Draw the line dashed if the inequality is < or > 5. Pick a point not on the line to use as a test point. The point (0, 0) is a good test point if it is not on the line. 6. If the point makes the inequality true, shade that side of the line. If the point does not make the inequality true, shade the opposite side of the line.
Graph the inequality x < 3
Graph y > x + 1 x -1 0 1 y 0 1 2 The boundary line for this graph is the line y = x + 1. since the boundary is not part of the graph, it is shown as a dashed line on the graph. To determine which halfplane is the graph of y > x + 1, test a point NOT on the boundary. For example, you can test my favorite point (0, 0) the origin. Since 0 > 0 + 1 is false, (0, 0) is not a solution of y > x + 1. Thus, the graph is all points in the half-plane that does NOT contain (0, 0). This graph is called an open half-plane since the boundary is not part of the graph.
Graph y ≤ x + 1 x -1 0 1 y 0 1 2 The boundary line for this graph is also the line y = x + 1. Since the inequality y ≤ x + 1 means y < x + 1 or y = x + 1, this boundary is part of the graph. Therefore, the boundary is shown as a solid line on the graph. The origin (0, 0) is part of the graph of y ≤ x + 1 since 0 ≤ 0 + 1 is true. Thus, the graph is all points in the half plane that contains the origin and the line y = x + 1. This graph is called a closed half plane.
Graph y > -4 x – 3 Graph the equation y = -4 x – 3. Draw it as a dashed line since this boundary is NOT part of the graph. The origin, (0, 0), is part of the graph since 0 >-4(0) – 3 is true. Thus, the graph is all points in the half plane that contains the origin. x -1 0 1 y 1 -3 -7 CHECK: Test a point on the other side of the boundary, say (-2, -2). Since -2 > -4(-2) – 3 or -2 > 5 is false, (-2, -2) is NOT part of the graph.
Ex. 2: Graph 15 x + 20 y ≤ 240 to answer the application at the beginning of the lesson. How many of each ticket can Mr. Harris purchase? • First solve for y in terms of x. 15 x + 20 y ≤ 240 – 15 x y ≤ 12 Then graph the equation as if it were equal to y as a solid line since the boundary is part of the graph. The origin (0, 0) is part of the graph since 15(0) + 20(0) ≤ 240 is true. Thus, the graph is all points in the half-plane that contains the origin.
Mr. Harris cannot buy fractional or negative numbers of tickets. • So, any point in the shaded region whose x- and y-coordinates are whole numbers is a possible solution. • For example, (5, 8) is a solution. This corresponds to buying five $15 tickets and eight $20 tickets for a total cost of 15(5) + 20(8) or $235.
Is the boundary included in the graphs of each inequality? 2 x + y ≥ 3 3 x – 2 y ≤ 1 5 x - 2 > 3 y
Included or not? • Signs with < or > are NOT included • Signs with ≤ or ≥ are included • If the graph is < or >, then it is said to be an open half plane since the boundary is not part of the graph. • If a graph is ≤ or ≥, then it is a closed halfplane because the boundary is part of the graph. • Greater than → shade above • Less than → shade below
Determine which (-2, 2), (4, -1), or (3, 1) are solutions to the inequality x + 2 y ≥ 3 -2 + 2(2) ≥ 3 -2 + 4 ≥ 3 2≥ 3 x + 2 y ≥ 3 3 + 2(1) ≥ 3 3+2≥ 3 5≥ 3 x + 2 y ≥ 3 4 + 2(-1) ≥ 3 4– 2≥ 3
Graph each inequality g(x)>0 and f(x)≥(⅓)x The overlap is where these two meet which is where your answer to both of these lies.
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