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HOMEWORK 7
RICKY NG
Question 7.1. The point J = (2, 5) is in Quadrant I.
Question 7.2. The point L = (4, −5) is in Quadrant IV.
Question 7.3. The point B = (−4, −3) is in Quadrant III.
Question 7.4. The point A = (−3, 0) is on the x-axis.
Question 7.5. Given the following points A = (−3, 4), B = (0, 4), C = (1, 4), and
D = (3, 4), what is the equation of the line containing these points.
Solution. Note that all of these points have y-coordinates equal to 4. Hence, y is not
varying, and the equation is y = 4.
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Question 7.6. Graph the line y = −5.
Solution. y is fixed to be −5, so...
Question 7.7. Graph the line x = −3.
Solution. x is fixed to be −3:
Question 7.8. Graph both x = 5 and y = −2
Solution. Similar to the previous questions:
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Question 7.9. Let y = −2x + 5. Complete the table the plot these points, then graph
the line.
Solution. For x = −2, we have
y = −2(−2) + 5
= 4 + 5 = 9.
Proceed similarly for x = −1, 0, 1, 2, we get the following table:
x y = −2x + 5
−2
9
−1
7
0
5
1
3
2
1
Now plot the points and graph the line:
Question 7.10. Let y = 5x − 1. Complete the table and plot the points, then graph
the line.
Solution. For x = 2, we have
y = 5(2) − 1 = 9.
Likewise for x = 53 ,
3
y=5
− 1 = 3 − 1 = 2.
5
Now, given y = −1, to find x, we need to solve the equation:
5x − 1 = −1
5x = 0
x = 0.
3
Proceed similarly for y = −6:
5x − 1 = −6
5x = −5
x = −1.
And for y = 0,
5x − 1 = 0
5x = 1
1
x= .
5
So we get this table:
y = 5x − 1
9
−1
3
2
5
−1
−6
1
0
5
x
2
0
The line looks like this:
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