1 Topic 4.2.4 The x - and y -Intercepts. 2 Topic 4.2.4 The x - and y -Intercepts California...

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Topic 4.2.4Topic 4.2.4

The x- and y-InterceptsThe x- and y-Intercepts

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Topic4.2.4


California Standard: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).

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

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Topic4.2.4


The intercepts of a graph are the points where the graph crosses the axes.

This Topic is all about how to calculate them.

–4 –2 0 2 4

4

2

0

–2

–4

y

x

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Topic4.2.4


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 Ax + By = C is the point at which the graph of Ax + By = C crosses the x-axis.

–4 –2 0 2 4

2

0

–2

y-axis

x-axis

The x-intercept is here (–1, 0)

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Topic4.2.4


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 –2 0 2 4

2

0

–2

y-axis

x-axis

(–1, 0)

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Example 1

Find the x-intercept of the line 3x – 4y = 18.

Solution follows…

Solution

Topic4.2.4

Let y = 0, then solve for x:

3x – 4y = 18

3x – 4(0) = 18

3x – 0 = 18

3x = 18

x = 6

So (6, 0) is the x-intercept of 3x – 4y = 18.

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Example 2

Find the x-intercept of the line 2x + y = 6.

Solution follows…

Solution

Topic4.2.4

Let y = 0, then solve for x:

2x + y = 6

2x + 0 = 6

2x = 6

x = 3

So (3, 0) is the x-intercept of 2x + y = 6.

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Topic4.2.4


Guided Practice

Solution follows…

In Exercises 1–8, find the x-intercept.

1. x + y = 5

2. 3x + y = 18

3. 5x – 2y = –10

4. 3x – 8y = –21

5. 4x – 9y = 16

6.15x – 8y = 5

7. 6x – 10y = –8

8. 14x – 6y = 0

x + 0 = 5 x = 5 (5, 0)

3x + 0 = 15 x = 6 (6, 0)

5x – 2(0) = –10 x = –2 (–2, 0)

3x – 8(0) = –21 x = –7 (–7, 0)

4x – 9(0) = 16 x = 4 (4, 0)

15x – 8(0) = 5 x = ( , 0)1

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1

3

6x – 10(0) = –8 x = – (– , 0)4

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4

3

14x – 6(0) = 0 x = 0 (0, 0)

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Topic4.2.4


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 Ax + By = C is the point at which the graph of Ax + By = C crosses the y-axis.

–4 –2 0 2 4

6

4

2

y-axis

x-axis–0

The y-intercept here is (0, 3)

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Topic4.2.4


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 –2 0 2 4

6

4

2

y-axis

x-axis–0

(0, 3)

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Example 3

Find the y-intercept of the line –2x – 3y = –9.

Solution follows…

Solution

Topic4.2.4

Let x = 0, then solve for y:

–2x – 3y = –9

–2(0) – 3y = –9

0 – 3y = –9

–3y = –9

y = 3

So (0, 3) is the y-intercept of –2x – 3y = –9.

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Example 4

Find the y-intercept of the line 3x + 4y = 24.

Solution follows…

Solution

Topic4.2.4

Let x = 0, then solve for y:

3x + 4y = 24

3(0) + 4y = 24

0 + 4y = 24

4y = 24

y = 6

So (0, 6) is the y-intercept of 3x + 4y = 24.

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Topic4.2.4


Guided Practice

Solution follows…

In Exercises 9–16, find the y-intercept.

9. 4x – 6y = 24

10. 5x + 8y = 24

11. 8x + 11y = –22

12. 9x + 4y = 48

13. 6x – 7y = –28

14. 10x – 12y = 6

15. 3x + 15y = –3

16. 14x – 5y = 0

4(0) – 6y = 24 y = –4 (0, –4)

5(0) + 8y = 24 y = 3 (0, 3)

8(0) + 11y = –22 y = –2 (0, –2)

9(0) + 4y = 48 y = 12 (0, 12)

6(0) – 7y = –28 y = 4 (0, 4)

10(0) – 12y = 6 y = –0.5 (0, –0.5)

3(0) + 15y = –3 y = –0.2 (0, –0.2)

14(0) – 5y = 0 y = 0 (0, 0)

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Topic4.2.4


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.

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Topic4.2.4



Solution follows…

Find the x- and y-intercepts of the following lines:

3. x + y = 9 4. x – y = 7

5. –x – 2y = 4 6. x – 3y = 9

7. 3x – 4y = 24 8. –2x + 3y = 12

9. –5x – 4y = 20 10. –0.2x + 0.3y = 1

11. 0.25x – 0.2y = 2 12. – 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)1

2

2

3

(–5, 0); (0, 3 )1

3

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13. ( g, 0) is the x-intercept of the line –10x – 3y = 12.

Find the value of g.

14. (0, k) is the y-intercept of the line 2x – 15y = –3.

Find the value of k.

15. The point (–3, b) lies on the line 2y – x = 8.Find the value of b.

16. Find the x-intercept of the line in Exercise 15.

17. Another line has x-intercept (4, 0) and equation 2y + kx = 20. Find the value of k.

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5

1

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Topic4.2.4



Solution follows…

g = –2

k = 1

b = 2.5

(–8, 0)

k = 5

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Topic4.2.4



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 –2 0 2 4 6

6

4

2

0

–2

–4

–6

y

x

pn

r

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Topic4.2.4


Round UpRound 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.

1 Topic 4.2.4 The x - and y -Intercepts. 2 Topic 4.2.4 The x - and y -Intercepts California...

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Transcript of 1 Topic 4.2.4 The x - and y -Intercepts. 2 Topic 4.2.4 The x - and y -Intercepts California...