Graphing Lines Topic 4.2.5

Graphing Lines 4.2.5 Topic California Standards: 6.0 Students graph a linear equation and compute the x- and y-intercepts (e.g. graph 2x + 6y = 4). They are also able to sketch the region defined by a linear inequality (e.g. they sketch the region defined by 2x + 6y < 4). 7.0 Students verify that a point lies on a line, given an equation of the line. Students are able to derive linear equations by using the point-slope formula. What it means for you: You’ll graph lines by first calculating the x- and y-intercepts. Key words: linear equation intercept

Topic 4.2.5 Graphing Lines In Topic 4.2.2 you learned how to graph a straight line by plotting two points. If you’re not given points on the line, it’s easiest to use the x- and y-intercepts. 2 4 6 y x –2 –4 –6 y-intercept x-intercept

Graphing Lines 4.2.5 Topic Graphing Lines by Computing the Intercepts The method below for plotting a straight-line graph is the same as in Topic 4.2.2. To graph the line, you plot two points — except this time you use the x-intercept and the y-intercept, then draw a straight line through them. 2 4 6 y x –2 –4 –6 y-intercept x-intercept

Graphing Lines 4.2.5 Topic Graphing a Line Find the x-intercept — let y = 0, then solve the equation for x. Find the y-intercept — let x = 0, then solve the equation for y. Draw a set of axes and plot the two intercepts. Draw a straight line through the points. Check your line by plotting a third point.

Topic 4.2.5 Graphing Lines Example 1 Draw the graph of 5x + 3y = 15 by computing the intercepts. Solution x-intercept y-intercept Let y = 0, then solve for x: Let x = 0, then solve for y: 5x + 3(0) = 15 5(0) + 3y = 15 5x + 0 = 15 0 + 3y = 15 5x = 15 3y = 15 x = 3 y = 5 So, (3, 0) is the x-intercept and (0, 5) is the y-intercept. Solution continues… Solution follows…

Topic 4.2.5 Graphing Lines Example 1 Draw the graph of 5x + 3y = 15 by computing the intercepts. Solution (continued) –2 2 4 6 8 y x x-intercept – (3, 0) y-intercept – (0, 5) Plot the intercepts… …then draw a straight line. Finally, check the line is correct by finding and plotting a third point. Solution continues…

Topic 4.2.5 Graphing Lines Example 1 Draw the graph of 5x + 3y = 15 by computing the intercepts. Solution (continued) –2 2 4 6 8 y x Check the line using x = 1. 5x + 3y = 15, so y = 5 – x 5 3 Þ y = 5 – 5 × 1 3 (1, 3 ) 1 3 15 – 5 3 Þ y = = = 3 10 1 (1 , 3 ) lies on the line — which means the line is correct. 1 3

Topic 4.2.5 Graphing Lines Finding the intercepts is the quickest way of finding two points. When you substitute 0 for y to solve for x, the y-term disappears, and vice versa — making the equations easier to solve.

Topic 4.2.5 Graphing Lines Example 2 Draw the graph of y = –x + 3 by computing the intercepts. Solution x-intercept y-intercept –2 2 4 6 y x 0 = –x + 3 y = –(0) + 3 x = 3 y = 3 So, (3, 0) is the x-intercept and (0, 3) is the y-intercept. Plot the intercepts… …then draw a straight line. Solution continues… Solution follows…

Topic 4.2.5 Graphing Lines Example 2 Draw the graph of y = –x + 3 by computing the intercepts. Solution (continued) Check the line using x = 1. –2 2 4 6 y x y = –x + 3 (1, 2) Þ y = –1 + 3 = 2 (1, 2) lies on the line — which means the line is correct.

Graphing Lines 4.2.5 Topic Guided Practice Draw the graphs of the following equations by computing the intercepts. 1. 5x + 2y = 10 2. –3x – 3y = 12 3. 6x – y = 3 4. –4x + 5y = 20 5. 2x + y = 3 6. –x – 8y = 2 –6 –4 –2 2 4 6 y x 5 4 6 1 2 3 Intercepts: (2, 0) and (0, 5) Intercepts: (–4, 0) and (0, –4) Intercepts: (0.5, 0) and (0, –3) Intercepts: (–5, 0) and (0, 4) Intercepts: (1.5, 0) and (0, 3) Intercepts: (–2, 0) and (0, –0.25) Solution follows…

Graphing Lines 4.2.5 Topic Guided Practice Draw the graphs of the following equations by computing the intercepts. 7. x + y = 10 8. x – y = 4 9. 2x – y = 4 10. x + 5y = 10 11. 3x + y = 9 12. x – 4y = 8 y x –8 –4 4 8 11 10 12 7 8 9 Intercepts: (10, 0) and (0, 10) Intercepts: (4, 0) and (0, –4) Intercepts: (2, 0) and (0, –4) Intercepts: (10, 0) and (0, 2) Intercepts: (3, 0) and (0, 9) Intercepts: (8, 0) and (0, –2) Solution follows…

Graphing Lines 4.2.5 Topic Guided Practice Draw the graphs of the following equations by computing the intercepts. 13. y = 2x + 4 14. y = 5x – 10 15. y = 3x + 9 16. y = x – 8 17. y = – x – 2 18. y = x + 4 8 16 24 32 y x –4 4 12 20 28 –8 –12 y x –8 –4 4 8 18 16 17 13 14 15 Intercepts: (–2, 0) and (0, 4) (2, 0) and (0, –10) (–3, 0) and (0, 9) 1 4 (32, 0) and (0, –8) 2 5 (–5, 0) and (0, –2) 4 5 (–5, 0) and (0, 4) Solution follows…

Graphing Lines 4.2.5 Topic Independent Practice Draw graphs of the lines using the x- and y-intercepts in Exercises 1–3. 1. x-intercept: (–3, 0) y-intercept: (0, 2) 2. x-intercept: (1, 0) y-intercept: (0, 6) 3. x-intercept: (4, 0) y-intercept: (0, –3) –6 –4 –2 2 4 6 y x 3 2 1 Solution follows…

Graphing Lines 4.2.5 Topic Independent Practice Draw graphs of the lines using the x- and y-intercepts in Exercises 4–6. 4. x-intercept: (–6, 0) y-intercept: (0, –4) 5. x-intercept: (–1, 0) y-intercept: (0, 7) 6. x-intercept: (2, 0) y-intercept: (0, –5) –6 –4 –2 2 4 6 y x 6 5 4 Solution follows…

Graphing Lines 4.2.5 Topic Independent Practice Draw the graphs of the equations in Exercises 7–12 by computing the intercepts. 7. x + y = 6 8. x + y = –4 9. x – y = –5 10. x – y = 7 11. 3x + y = 6 12. 2x + y = 8 y x –8 –4 4 8 11 10 12 7 8 9 Intercepts: (6, 0) and (0, 6) Intercepts: (–4, 0) and (0, –4) Intercepts: (–5, 0) and (0, 5) Intercepts: (7, 0) and (0, –7) Intercepts: (2, 0) and (0, 6) Intercepts: (4, 0) and (0, 8) Solution follows…

Graphing Lines 4.2.5 Topic Independent Practice Draw the graphs of the equations in Exercises 13–18 by computing the intercepts. 13. 2x – y = –4 14. 3x – y = –3 15. 4x + 3y = –12 16. 5x – 2y = 10 17. 6x – 3y = 24 18. 10x – 12y = 60 –6 –4 –2 2 4 6 y x y x –8 –4 4 8 17 18 16 13 14 15 Intercepts: (–2, 0) and (0, 4) Intercepts: (–1, 0) and (0, 3) Intercepts: (–3, 0) and (0, –4) Intercepts: (2, 0) and (0, –5) Intercepts: (4, 0) and (0, –8) Intercepts: (6, 0) and (0, –5) Solution follows…

Graphing Lines 4.2.5 Topic Independent Practice 19. Show that the graphs of x + y = 6 and –6x – 6y = –36 are the same. The intercepts of both graphs are (6, 0) and (0,6), so the graphs will be the same. 20. Explain why the graph of 5x + 8y = 0 cannot be drawn using the intercepts. Both intercepts are at (0, 0). Since you need two different points to draw a line, you can’t draw this line from just the intercepts. Solution follows…

Graphing Lines 4.2.5 Topic Round Up This Topic follows on neatly from Topic 4.2.2, where you graphed lines by plotting two points and joining them with a straight line. You can use any two points — the main reason for using the intercepts is that they’re usually easier to calculate.