Transcript The x- and y-Intercepts

The x- and y-Intercepts
Topic 4.2.4
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Topic
4.2.4
The x- and y-Intercepts
California Standard:
What it means for you:
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).
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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Topic
4.2.4
The x- and y-Intercepts
y
The intercepts of a graph
are the points where the
graph crosses the axes.
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2
–4
–2
0
x
0
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–2
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This Topic is all about how to calculate them.
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Topic
4.2.4
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 Ax + By = C is the point
at which the graph of Ax + By = C crosses the x-axis.
y-axis
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The x-intercept
is here (–1, 0)
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–2
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0
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x-axis
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Topic
4.2.4
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.
y-axis
(–1, 0)
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–4
–2
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x-axis
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Topic
The x- and y-Intercepts
4.2.4
Example
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Find the x-intercept of the line 3x – 4y = 18.
Solution
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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Solution follows…
Topic
The x- and y-Intercepts
4.2.4
Example
2
Find the x-intercept of the line 2x + y = 6.
Solution
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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Solution follows…
Topic
4.2.4
The x- and y-Intercepts
Guided Practice
In Exercises 1–8, find the x-intercept.
1. x + y = 5
x + 0 = 5 x = 5 (5, 0)
2. 3x + y = 18
3x + 0 = 15 x = 6 (6, 0)
3. 5x – 2y = –10
5x – 2(0) = –10 x = –2 (–2, 0)
4. 3x – 8y = –21
3x – 8(0) = –21 x = –7 (–7, 0)
5. 4x – 9y = 16
4x – 9(0) = 16 x = 4 (4, 0)
6.15x – 8y = 5
15x – 8(0) = 5 x =
7. 6x – 10y = –8
8. 14x – 6y = 0
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( , 0)
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6x – 10(0) = –8 x = – (– , 0)
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14x – 6(0) = 0 x = 0 (0, 0)
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Solution follows…
Topic
4.2.4
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 Ax + By = C is the point
at which the graph of Ax + By = C crosses the y-axis.
y-axis
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The y-intercept
here is (0, 3)
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–4
–2
–0
0
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x-axis
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Topic
4.2.4
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.
y-axis
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(0, 3)
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–4
–2
–0
0
2
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x-axis
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Topic
The x- and y-Intercepts
4.2.4
Example
3
Find the y-intercept of the line –2x – 3y = –9.
Solution
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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Solution follows…
Topic
The x- and y-Intercepts
4.2.4
Example
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Find the y-intercept of the line 3x + 4y = 24.
Solution
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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Solution follows…
Topic
4.2.4
The x- and y-Intercepts
Guided Practice
In Exercises 9–16, find the y-intercept.
9. 4x – 6y = 24
4(0) – 6y = 24 y = –4 (0, –4)
10. 5x + 8y = 24
5(0) + 8y = 24 y = 3 (0, 3)
11. 8x + 11y = –22
8(0) + 11y = –22 y = –2 (0, –2)
12. 9x + 4y = 48
9(0) + 4y = 48 y = 12 (0, 12)
13. 6x – 7y = –28
6(0) – 7y = –28 y = 4 (0, 4)
14. 10x – 12y = 6
10(0) – 12y = 6 y = –0.5 (0, –0.5)
15. 3x + 15y = –3
3(0) + 15y = –3 y = –0.2 (0, –0.2)
16. 14x – 5y = 0
14(0) – 5y = 0 y = 0 (0, 0)
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Solution follows…
Topic
4.2.4
The x- and y-Intercepts
Independent Practice
1. Define the x-intercept.
The point at which the graph of a line crosses the y-axis.
2. Define the y-intercept.
The point at which the graph of a line crosses the x-axis.
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Solution follows…
Topic
4.2.4
The x- and y-Intercepts
Independent Practice
Find the x- and y-intercepts of the following lines:
3. x + y = 9
(9, 0); (0, 9)
4. x – y = 7
(7, 0); (0, –7)
5. –x – 2y = 4
(–4, 0); (0, –2)
6. x – 3y = 9
(9, 0); (0, –3)
7. 3x – 4y = 24
(8, 0); (0, –6)
8. –2x + 3y = 12
(–6, 0); (0, 4)
9. –5x – 4y = 20
(–4, 0); (0, –5)
10. –0.2x + 0.3y = 1
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(–5, 0); (0, 3 )
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12. – x – y = 6
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(–12, 0); (0, –9)
11. 0.25x – 0.2y = 2
(8, 0); (0, –10)
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Solution follows…
Topic
4.2.4
The x- and y-Intercepts
Independent Practice
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13. ( g, 0) is the x-intercept of the line –10x – 3y = 12.
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Find the value of g. g = –2
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14. (0, k) is the y-intercept of the line 2x – 15y = –3.
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Find the value of k. k = 1
15. The point (–3, b) lies on the line 2y – x = 8.
Find the value of b. b = 2.5
16. Find the x-intercept of the line in Exercise 15.
(–8, 0)
17. Another line has x-intercept (4, 0) and equation
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2y + kx = 20. Find the value of k. k = 5
Solution follows…
Topic
The x- and y-Intercepts
4.2.4
Independent Practice
In Exercises 18-22, use the graph below to help you reach
your answer.
18. Find the x- and y-intercepts of line n.
y
(1, 0); (0, –4)
19. Find the x-intercept of line p.
(–3, 0)
20. Find the y-intercept of line r.
(0, 2)
21. Explain why line p does not have
a y-intercept.
Line p is vertical and never crosses the y-axis.
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p
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n
2
0
–6 –4 –2 0
–2
r
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x
–4
–6
22. Explain why line r does not have
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an x-intercept. Line r is horizontal and never crosses the x-axis.Solution follows…
Topic
4.2.4
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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