1. Copyright © 2010 Pearson Education, Inc. All rights reserved. 3.2 – Slide 1

2. Copyright © 2010 Pearson Education, Inc. All rights reserved. 3.2 – Slide 2 Graphs of Linear Equations and Inequalities; Functions Chapter 3

3. Copyright © 2010 Pearson Education, Inc. All rights reserved. 3.2 – Slide 3 3.2 Graphing Linear Equations in Two Variables

4. Copyright © 2010 Pearson Education, Inc. All rights reserved. 3.2 – Slide 4 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. 3.2 Graphing Linear Equations in Two Variables

5. Copyright © 2010 Pearson Education, Inc. All rights reserved. 3.2 – Slide 5 y = 2(0) – 1 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. Graphing by Plotting Ordered Pairs If x = 0, then 3.2 Graphing Linear Equations in Two Variables The graph of any linear equation in two variables is a straight line. y = – 1 So, we have the ordered pair (0,–1). 0 = 2x – 1 If y = 0, then 1 = 2x So, we have the ordered pair (½,0). + 1 ½ = x

6. Copyright © 2010 Pearson Education, Inc. All rights reserved. 3.2 – Slide 6 y = 2(1) – 1 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. Graphing by Plotting Ordered Pairs If x = 1, then 3.2 Graphing Linear Equations in Two Variables y = 1 So, we have the ordered pair (1,1). When we graph, all three points, (0,–1), (½,0), and (1,1), should lie on the same straight line.

7. Copyright © 2010 Pearson Education, Inc. All rights reserved. 3.2 – Slide 7 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. Finding Intercepts 3.2 Graphing Linear Equations in Two Variables

8. Copyright © 2010 Pearson Education, Inc. All rights reserved. 3.2 – Slide 8 0 + y = 2 Example 2 Find the intercepts for the graph of x + y = 2. Then draw the graph. To find the y-intercept, let x = 0; to find the x-intercept, let y = 0. Finding Intercepts 3.2 Graphing Linear Equations in Two Variables y = 2 The y-intercept is (0,2). x + 0 = 2 x = 2 The x-intercept is (2,0). Plotting the intercepts gives the graph.

9. Copyright © 2010 Pearson Education, Inc. All rights reserved. 3.2 – Slide 9 –6(0) + 2y = 0 Example 3 Graph the linear equation –6x + 2y = 0. First, find the intercepts. Graphing Linear Equations of the Form Ax + By = 0 3.2 Graphing Linear Equations in Two Variables 2y = 0 The y-intercept is (0,0). The x-intercept is (0,0). Since the x and y intercepts are the same (the origin), choose a different value for x or y. y = 0 –6x + 2(0) = 0 –6x = 0 x = 0

10. Copyright © 2010 Pearson Education, Inc. All rights reserved. 3.2 – Slide 10 –6(1) + 2y = 0 Example 3 (concluded) Graph the linear equation –6x + 2y = 0. Let x = 1. Graphing Linear Equations of the Form Ax + By = 0 3.2 Graphing Linear Equations in Two Variables 2y = 6 A second point is (1,3). y = 3 –6 + 2y = 0 +6

11. Copyright © 2010 Pearson Education, Inc. All rights reserved. 3.2 – Slide 11 Line through the Origin If A and B are nonnzero real numbers, the graph of the linear equation of the form Ax + By = 0 passes through the origin (0,0). Graphing Linear Equations of the Form Ax + By = 0 3.2 Graphing Linear Equations in Two Variables

12. Copyright © 2010 Pearson Education, Inc. All rights reserved. 3.2 – Slide 12 Note that this is the graph of a horizontal line with y-intercept (0,–2). Example 4 (a) Graph y = –2. Graphing Linear Equations of the Form y = k or x = k 3.2 Graphing Linear Equations in Two Variables 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.

13. Copyright © 2010 Pearson Education, Inc. All rights reserved. 3.2 – Slide 13 Example 4 (continued) (b) Graph x = 1. Graphing Linear Equations of the Form y = k and x = k 3.2 Graphing Linear Equations in Two Variables The expanded version of this linear equation would be x + 0 · y = 1. Here, the x- coordinate is unaffected by the value of the y-coordinate. Whatever y-value we choose, the x-value will be 1. 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.

14. Copyright © 2010 Pearson Education, Inc. All rights reserved. 3.2 – Slide 14 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. 3.2 Graphing Linear Equations in Two Variables Graphing Linear Equations of the Form y = k and x = k

15. Copyright © 2010 Pearson Education, Inc. All rights reserved. 3.2 – Slide 15 Example 5 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. (a)Use Bob’s cost equation to determine his costs if he sells 1000 bagels next month, 4000 bagels next month. Using a Linear Equation to Model Data 3.2 Graphing Linear Equations in Two Variables C = 0.75(1000) + 2500 C = $3250 C = 0.75(4000) + 2500 C = $5500

16. Copyright © 2010 Pearson Education, Inc. All rights reserved. 3.2 – Slide 16 Example 5 (concluded) (b) Write the information from part (a) as two ordered pairs and use them to graph Bob’s cost equation. Using a Linear Equation to Model Data 3.2 Graphing Linear Equations in Two Variables 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.