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1 Topic 4.2.4 The x - and y -Intercepts. 2 Topic 4.2.4 The x - and y -Intercepts California Standard: 6.0 Students graph a linear equation and compute.

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Presentation on theme: "1 Topic 4.2.4 The x - and y -Intercepts. 2 Topic 4.2.4 The x - and y -Intercepts California Standard: 6.0 Students graph a linear equation and compute."— Presentation transcript:

1 1 Topic The x - and y -Intercepts

2 2 Topic The x - and y -Intercepts California Standard: 6.0 Students graph a linear equation and compute the x - and y -intercepts (e.g. graph 2 x + 6 y = 4). They are also able to sketch the region defined by a linear inequality (e.g. they sketch the region defined by 2 x + 6 y < 4). What it means for you: You’ll learn about x - and y -intercepts and how to compute them from the equation of a line. Key Words: intercept linear equation

3 3 Topic The x - and y -Intercepts The intercepts of a graph are the points where the graph crosses the axes. This Topic is all about how to calculate them. –4– –4 y x

4 4 Topic The x - and y -Intercepts The x -Intercept is Where the Graph Crosses the x -Axis The x -axis on a graph is the horizontal line through the origin. Every point on it has a y -coordinate of 0. That means that all points on the x -axis are of the form ( x, 0). The x -intercept of the graph of A x + B y = C is the point at which the graph of A x + B y = C crosses the x -axis. –4– y -axis x -axis The x -intercept is here (–1, 0)

5 5 Topic The x - and y -Intercepts Computing the x -Intercept Using “ y = 0” Since you know that the x -intercept has a y -coordinate of 0, you can find the x -coordinate by letting y = 0 in the equation of the line. –4– y -axis x -axis (–1, 0)

6 6 The x - and y -Intercepts Example 1 Find the x -intercept of the line 3 x – 4 y = 18. Solution follows… Solution Topic Let y = 0, then solve for x : 3 x – 4 y = 18 3 x – 4(0) = 18 3 x – 0 = 18 3 x = 18 x = 6 So (6, 0) is the x -intercept of 3 x – 4 y = 18.

7 7 The x - and y -Intercepts Example 2 Find the x -intercept of the line 2 x + y = 6. Solution follows… Solution Topic Let y = 0, then solve for x : 2 x + y = 6 2 x + 0 = 6 2 x = 6 x = 3 So (3, 0) is the x -intercept of 2 x + y = 6.

8 8 Topic The x - and y -Intercepts Guided Practice Solution follows… In Exercises 1–8, find the x -intercept. 1. x + y = x + y = x – 2 y = – x – 8 y = – x – 9 y = x – 8 y = x – 10 y = – x – 6 y = 0 x + 0 = 5  x = 5  (5, 0) 3 x + 0 = 15  x = 6  (6, 0) 5 x – 2(0) = –10  x = –2  (–2, 0) 3 x – 8(0) = –21  x = –7  (–7, 0) 4 x – 9(0) = 16  x = 4  (4, 0) 15 x – 8(0) = 5  x =  (, 0) x – 10(0) = –8  x = –  (–, 0) x – 6(0) = 0  x = 0  (0, 0)

9 9 Topic The x - and y -Intercepts The y -Intercept is Where the Graph Crosses the y -Axis The y -axis on a graph is the vertical line through the origin. Every point on it has an x -coordinate of 0. That means that all points on the y -axis are of the form (0, y ). The y -intercept of the graph of A x + B y = C is the point at which the graph of A x + B y = C crosses the y -axis. –4– y -axis x -axis –0 The y -intercept here is (0, 3)

10 10 Topic The x - and y -Intercepts Computing the y -Intercept Using “ x = 0” Since the y -intercept has an x -coordinate of 0, find the y -coordinate by letting x = 0 in the equation of the line. –4– y -axis x -axis –0 (0, 3)

11 11 The x - and y -Intercepts Example 3 Find the y -intercept of the line –2 x – 3 y = –9. Solution follows… Solution Topic Let x = 0, then solve for y : –2 x – 3 y = –9 –2(0) – 3 y = –9 0 – 3 y = –9 –3 y = –9 y = 3 So (0, 3) is the y -intercept of –2 x – 3 y = –9.

12 12 The x - and y -Intercepts Example 4 Find the y -intercept of the line 3 x + 4 y = 24. Solution follows… Solution Topic Let x = 0, then solve for y : 3 x + 4 y = 24 3(0) + 4 y = y = 24 4 y = 24 y = 6 So (0, 6) is the y -intercept of 3 x + 4 y = 24.

13 13 Topic The x - and y -Intercepts Guided Practice Solution follows… In Exercises 9–16, find the y -intercept x – 6 y = x + 8 y = x + 11 y = – x + 4 y = x – 7 y = – x – 12 y = x + 15 y = – x – 5 y = 0 4(0) – 6 y = 24  y = –4  (0, –4) 5(0) + 8 y = 24  y = 3  (0, 3) 8(0) + 11 y = –22  y = –2  (0, –2) 9(0) + 4 y = 48  y = 12  (0, 12) 6(0) – 7 y = –28  y = 4  (0, 4) 10(0) – 12 y = 6  y = –0.5  (0, –0.5) 3(0) + 15 y = –3  y = –0.2  (0, –0.2) 14(0) – 5 y = 0  y = 0  (0, 0)

14 14 Topic The x - and y -Intercepts Independent Practice Solution follows… 1. Define the x -intercept. 2. Define the y -intercept. The point at which the graph of a line crosses the x -axis. The point at which the graph of a line crosses the y -axis.

15 15 Topic The x - and y -Intercepts Independent Practice Solution follows… Find the x - and y -intercepts of the following lines: 3. x + y = 9 4. x – y = 7 5. – x – 2 y = 4 6. x – 3 y = x – 4 y = –2 x + 3 y = –5 x – 4 y = –0.2 x y = x – 0.2 y = – x – y = 6 (9, 0); (0, 9)(7, 0); (0, –7) (–4, 0); (0, –2) (9, 0); (0, –3) (8, 0); (0, –6) (–6, 0); (0, 4) (–4, 0); (0, –5) (8, 0); (0, –10) (–12, 0); (0, –9) (–5, 0); (0, 3 ) 1 3

16 ( g, 0) is the x -intercept of the line –10 x – 3 y = 12. Find the value of g. 14. (0, k ) is the y -intercept of the line 2 x – 15 y = –3. Find the value of k. 15. The point (–3, b ) lies on the line 2 y – x = 8. Find the value of b. 16. Find the x -intercept of the line in Exercise Another line has x -intercept (4, 0) and equation 2 y + kx = 20. Find the value of k Topic The x - and y -Intercepts Independent Practice Solution follows… g = –2 k = 1 b = 2.5 (–8, 0) k = 5

17 17 Topic The x - and y -Intercepts Independent Practice Solution follows… In Exercises 18-22, use the graph below to help you reach your answer. 18. Find the x - and y -intercepts of line n. 19. Find the x -intercept of line p. 20. Find the y -intercept of line r. 21. Explain why line p does not have a y -intercept. 22. Explain why line r does not have an x -intercept. (1, 0); (0, –4) (–3, 0) (0, 2) Line p is vertical and never crosses the y -axis. Line r is horizontal and never crosses the x -axis. –6–4– –4 –6 y x p n r

18 18 Topic The x - and y -Intercepts Round Up Make sure you get the method the right way around — to find the x -intercept, put y = 0 and solve for x, and to find the y -intercept, put x = 0 and solve for y. In the next Topic you’ll see that the intercepts are really useful when you’re graphing lines from the line equation.


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