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CHAPTER 3 Graphs of Liner Equations Slide 2Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 3.1Graphs and Applications of Linear Equations 3.2More.

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Presentation on theme: "CHAPTER 3 Graphs of Liner Equations Slide 2Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 3.1Graphs and Applications of Linear Equations 3.2More."— Presentation transcript:

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2 CHAPTER 3 Graphs of Liner Equations Slide 2Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 3.1Graphs and Applications of Linear Equations 3.2More with Graphing and Intercepts 3.3Slope and Applications 3.4Equations of Lines 3.5Graphing Using the Slope and the y-Intercept 3.6Parallel and Perpendicular Lines 3.7Graphing Inequalities in Two Variables

3 OBJECTIVES 3.2 More with Graphing and Intercepts Slide 3Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. aFind the intercepts of a linear equation, and graph using intercepts. bGraph equations equivalent to those of the type x = a and y = b.

4 (0, b) (a, 0) y-intercept x-intercept The y-intercept is (0, b). To find b, let x = 0 and solve the original equation for y. The x-intercept is (a, 0). To find a, let y = 0 and solve the original equation for x. 3.2 More with Graphing and Intercepts Intercepts Slide 4Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

5 EXAMPLE Solution To find the y-intercept, we let x = 0 and solve for y: 5 0 + 2y = 10 2y = 10 y = 5 The y-intercept is (0, 5). Replacing x with 0 Replacing y with 0 3.2 More with Graphing and Intercepts a Find the intercepts of a linear equation, and graph using intercepts. AConsider 5x + 2y = 10. Find the intercepts. Then graph the equation using the intercepts. (continued) Slide 5Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

6 EXAMPLE To find the x-intercept, we let y = 0 and solve for x. 5x + 2 0 = 10 5x = 10 x = 2 The x-intercept is (2, 0). 3.2 More with Graphing and Intercepts a Find the intercepts of a linear equation, and graph using intercepts. AConsider 5x + 2y = 10. Find the intercepts. Then graph the equation using the intercepts. (continued) Slide 6Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

7 EXAMPLE We plot these points and draw the line, or graph. A third point should be used as a check. We substitute any convenient value for x and solve for y. If we let x = 4, then 5 4 + 2y = 10 20 + 2y = 10 2y = –10 y = –5 3.2 More with Graphing and Intercepts a Find the intercepts of a linear equation, and graph using intercepts. AGraph using intercepts. (continued) Slide 7Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

8 EXAMPLE 5x + 2y = 10 x-intercept (2, 0) y-intercept (0, 5) xy 05 20 4 55 3.2 More with Graphing and Intercepts a Find the intercepts of a linear equation, and graph using intercepts. AGraph using intercepts. Slide 8Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

9 EXAMPLE Solution : (0, 0) is both the x-intercept and y-intercept. Calculate two other points and complete the graph. xy 11 44 00 14 x-intercept y-intercept y = 4x (1, 4) (  1,  4) (0, 0) 3.2 More with Graphing and Intercepts a Find the intercepts of a linear equation, and graph using intercepts. BGraph y = 4x. Slide 9Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

10 EXAMPLE Choose any number for x. y must be 2. Solution: We regard the equation y = 2 as 0 x + y = 2. No matter what number we choose for x, we find that y must equal 2. xy(x, y) 02(0, 2) 42(4, 2) 44 2 (  4, 2) 3.2 More with Graphing and Intercepts b Graph equations equivalent to those of the type x = a and y = b. CGraph y = 2. (continued) Slide 10Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

11 EXAMPLE When we plot the ordered pairs (0, 2), (4, 2) and (  4, 2) and connect the points, we obtain a horizontal line. Any ordered pair of the form (x, 2) is a solution, so the line is parallel to the x-axis with y- intercept (0, 2). y = 2 (  4, 2) (0, 2) (4, 2) 3.2 More with Graphing and Intercepts b Graph equations equivalent to those of the type x = a and y = b. CGraph y = 2. Slide 11Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

12 EXAMPLE xy(x, y) 22 4 (  2, 4) 22 1 (  2, 1) 22 44(  2,  4) x must be  2. Solution: We regard the equation x =  2 as x + 0 y =  2. We make up a table with all  2 in the x-column. Any number can be used for y. 3.2 More with Graphing and Intercepts b Graph equations equivalent to those of the type x = a and y = b. D Graph x =  2. (continued) Slide 12Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

13 EXAMPLE Solution: When we plot the ordered pairs (  2, 4), (  2, 1), and (  2,  4) and connect them, we obtain a vertical line. Any ordered pair of the form (  2, y) is a solution. The line is parallel to the y-axis with x- intercept (  2, 0). x =  2 (  2,  4) (  2, 4) (  2, 1) 3.2 More with Graphing and Intercepts b Graph equations equivalent to those of the type x = a and y = b. D Graph x =  2. Slide 13Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

14 The graph of y = b is a horizontal line. The y-intercept is (0, b). The graph of x = a is a vertical line. The x-intercept is (a, 0). 3.2 More with Graphing and Intercepts Horizontal and Vertical Lines Slide 14Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.


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