11 Graphs of Linear Equations, and Inequalities, in Two Variables

11.2 Graphing Linear Equations in Two Variables Objectives 1. Graph linear equations by plotting ordered pairs. 2. Find intercepts. 3. Graph linear equations of the form Ax + By = 0. 4. Graph linear equations of the form y = k or x = k. 5. Use a linear equation to model data.

Graph by Plotting Ordered Pairs The graph of any linear equation in two variables is a straight line. Example 1 Graph the linear equation y = 2x – 1. Note that although this equation is not of the form Ax + By = C, it could be. Therefore, it is linear. To graph it, we will first find two points by letting x = 0 and then y = 0. If x = 0, then If y = 0, then y = 2(0) – 1 0 = 2x – 1 + 1 + 1 y = – 1 1 = 2x So, we have the ordered pair (0,–1). ½ = x So, we have the ordered pair (½,0).

Graph by Plotting Ordered Pairs Example 1 (concluded) Graph the linear equation y = 2x – 1. Now we will find a third point (just as a check) by letting x = 1. When we graph, all three points, (0,–1), (½,0), and (1,1), should lie on the same straight line. If x = 1, then y = 2(1) – 1 y = 1 So, we have the ordered pair (1,1).

Find Intercepts Finding Intercepts To find the x-intercept, let y = 0 in the given equation and solve for x. Then (x, 0) is the x-intercept. To find the y-intercept, let x = 0 in the given equation and solve for y. Then (0, y) is the y-intercept.

Find Intercepts Example 2 Find the intercepts for the graph of x + y = 2. Then draw the graph. Plotting the intercepts gives the graph. To find the y-intercept, let x = 0; to find the x-intercept, let y = 0. 0 + y = 2 x + 0 = 2 y = 2 x = 2 The y-intercept is (0, 2). The x-intercept is (2, 0).

Graph Linear Equations of the Form Ax + By = 0 Example 4 Graph the linear equation –6x + 2y = 0. First, find the intercepts. –6(0) + 2y = 0 –6x + 2(0) = 0 Since the x and y intercepts are the same (the origin), choose a different value for x or y. 2y = 0 –6x = 0 y = 0 x = 0 The y-intercept is (0, 0). The x-intercept is (0, 0).

Graph Linear Equations of the Form Ax + By = 0 Example 4 (concluded) Graph the linear equation –6x + 2y = 0. Let x = 1. –6(1) + 2y = 0 –6 + 2y = 0 +6 +6 2y = 6 y = 3 A second point is (1, 3).

Graph Linear Equations of the Form Ax + By = 0 Line through the Origin The graph of a linear equation of the form Ax + By = 0 where A and B are nonzero real numbers, passes through the origin (0,0). 9

Graph Linear Equations of the Form y = k or x = k Example 5 (a) Graph y = –2. The expanded version of this linear equation would be 0 · x + y = –2. Here, the y-coordinate is unaffected by the value of the x-coordinate. Whatever x-value we choose, the y-value will be –2. Thus, we could plot the points (–1, –2), (2,–2), (4,–2), etc. Note that this is the graph of a horizontal line with y-intercept (0,–2).

Graph Linear Equations of the Form y = k or x = k Example 6 Graph x – 1 = 0. Add 1 to each side of the equation. x = 1. The x-coordinate is unaffected by the value of the y-coordinate. Thus, we could plot the points (1, –3), (1, 0), (1, 2), etc. Note that this is the graph of a vertical line with no y-intercept.

Graph Linear Equations of the Form y = k or x = k Horizontal and Vertical Lines The graph of the linear equation y = k, where k is a real number, is the horizontal line with y-intercept (0, k) and no x-intercept. The graph of the linear equation x = k, where k is a real number, is the vertical line with x-intercept (k, 0) and no y-intercept.

Use a Linear Equation to Model Data Example 7 Bob has owned and managed Bob’s Bagels for the past 5 years and has kept track of his costs over that time. Based on his figures, Bob has determined that his total monthly costs can be modeled by C = 0.75x + 2500, where x is the number of bagels that Bob sells that month. Use Bob’s cost equation to determine his costs if he sells 1000 bagels next month, 4000 bagels next month. C = 0.75(1000) + 2500 C = 0.75(4000) + 2500 C = $3250 C = $5500

Use a Linear Equation to Model Data Example 7 (concluded) (b) Write the information from part (a) as two ordered pairs and use them to graph Bob’s cost equation. From part (a) we have (1000, 3250) and (4000, 5500). Note that we did not extend the graph to the left beyond the vertical axis. That area would correspond to a negative number of bagels, which does not make sense.