Graphing Equations by Hand

Chapter
3
Introduction to Graphing
3.1
Graphing Equations by Hand
1. To plot the point A(2, −4), move 2 units to the right, then move 4 units
down. The remaining points are plotted in a similar manner.
y
y
5
5
C(−3, 2)
2
−5
5
x
−5
5
−4
−5
B(−4, −3)
A(2, −4)
−5
A(2, −4)
3. To plot the point A(3, −2), move 3 units to the right, then move 2 units
down. The remaining points are plotted in a similar manner.
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CHAPTER 3. INTRODUCTION TO GRAPHING
128
y
y
5
5
C(−2, 3)
3
−5
−2 5
x
−5
x
5
A(3, −2)
A(3, −2)
B(−3, −4)
−5
−5
5. To plot the point A(−2, 2), move 2 units to the left, then move 2 units up.
The remaining points are plotted in a similar manner.
y
y
5
5
A(−2, 2)
A(−2, 2)
2
−5
−2
5
x
−5
5
C(2, −2)
B(−2, −3)
−5
x
−5
7. Start at the origin, move 4 units to the left then move 3 units up to reach
the point P .
y
5
P (−4, 3)
3
−5
−4
5
−5
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3.1. GRAPHING EQUATIONS BY HAND
129
9. Start at the origin, move 3 units to the right then move 4 units down to
reach the point P .
y
5
3
−5
5
x
−4
P (3, −4)
−5
11. Substitute (x, y) = (5, 20) into the equation y = 5x − 5.
y = 5x − 5
Original equation.
20 = 5(5) − 5
Substitute: 5 for x, 20 for y.
20 = 25 − 5
20 = 20
Multiply: 5(5) = 25.
Subtract: 25 − 5 = 20.
Because this last statement is a true statement, (5, 20) satisfies (is a solution
of) the equation y = 5x − 5.
For contrast, consider the point (8, 36).
y = 5x − 5
36 = 5(8) − 5
Original equation.
Substitute: 8 for x, 36 for y.
36 = 40 − 5
36 = 35
Multiply: 5(8) = 40.
Subtract: 40 − 5 = 35.
Note that this last statement is false. Hence, the pair (8, 36) is not a solution
of y = 5x − 5. In similar fashion, readers should also check that the remaining
two points are not solutions.
13. Substitute (x, y) = (1, 1) into the equation y = −5x + 6.
y = −5x + 6
Original equation.
1 = −5(1) + 6
1 = −5 + 6
Substitute: 1 for x, 1 for y.
Multiply: −5(1) = −5.
1=1
Add: −5 + 6 = 1.
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130
Because this last statement is a true statement, (1, 1) satisfies (is a solution of)
the equation y = −5x + 6.
For contrast, consider the point (2, −3).
y = −5x + 6
Original equation.
−3 = −5(2) + 6
−3 = −10 + 6
Substitute: 2 for x, −3 for y.
Multiply: −5(2) = −10.
−3 = −4
Add: −10 + 6 = −4.
Note that this last statement is false. Hence, the pair (2, −3) is not a solution
of y = −5x+6. In similar fashion, readers should also check that the remaining
two points are not solutions.
15. Substitute (x, y) = (7, 395) into the equation y = 8x2 + 3.
y = 8x2 + 3
2
Original equation.
395 = 8(7) + 3
Substitute: 7 for x, 395 for y.
395 = 8(49) + 3
Square: (7)2 = 49.
395 = 392 + 3
395 = 395
Multiply: 8(49) = 392.
Add: 392 + 3 = 395.
Because this last statement is a true statement, (7, 395) satisfies (is a solution
of) the equation y = 8x2 + 3.
For contrast, consider the point (1, 12).
y = 8x2 + 3
2
Original equation.
12 = 8(1) + 3
Substitute: 1 for x, 12 for y.
12 = 8(1) + 3
Square: (1)2 = 1.
12 = 8 + 3
12 = 11
Multiply: 8(1) = 8.
Add: 8 + 3 = 11.
Note that this last statement is false. Hence, the pair (1, 12) is not a solution
of y = 8x2 + 3. In similar fashion, readers should also check that the remaining
two points are not solutions.
17. Substitute (x, y) = (8, 400) into the equation y = 6x2 + 2x.
y = 6x2 + 2x
2
Original equation.
400 = 6(8) + 2(8)
Substitute: 8 for x, 400 for y.
400 = 6(64) + 2(8)
400 = 384 + 16
Square: (8)2 = 64.
Multiply: 6(64) = 384; 2(8) = 16
400 = 400
Add: 384 + 16 = 400.
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3.1. GRAPHING EQUATIONS BY HAND
131
Because this last statement is a true statement, (8, 400) satisfies (is a solution
of) the equation y = 6x2 + 2x.
For contrast, consider the point (2, 29).
y = 6x2 + 2x
Original equation.
2
29 = 6(2) + 2(2)
Substitute: 2 for x, 29 for y.
29 = 6(4) + 2(2)
29 = 24 + 4
Square: (2)2 = 4.
Multiply: 6(4) = 24; 2(2) = 4
29 = 28
Add: 24 + 4 = 28.
Note that this last statement is false. Hence, the pair (2, 29) is not a solution of
y = 6x2 + 2x. In similar fashion, readers should also check that the remaining
two points are not solutions.
19. First, complete the table of points that satisfy the equation y = 2x + 1.
y
y
y
y
y
y
y
x
−3
−2
−1
0
1
2
3
= 2(−3) + 1 = −5
= 2(−2) + 1 = −3
= 2(−1) + 1 = −1
= 2(0) + 1 = 1
= 2(1) + 1 = 3
= 2(2) + 1 = 5
= 2(3) + 1 = 7
y = 2x + 1
−5
−3
−1
1
3
5
7
(x, y)
(−3, −5)
(−2, −3)
(−1, −1)
(0, 1)
(1, 3)
(2, 5)
(3, 7)
Next, plot the points in the table, as shown in the image on the right. This
first image gives us enough evidence to believe that if we plotted all points
satisfying the equation y = 2x + 1, the result would be the graph on the right.
y
y
10
−10
10
10
−10
x
−10
y = 2x + 1
10
−10
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CHAPTER 3. INTRODUCTION TO GRAPHING
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21. First, complete the table of points that satisfy the equation y = |x − 5|.
y
y
y
y
y
y
y
x
2
3
4
5
6
7
8
= |2 − 5| = 3
= |3 − 5| = 2
= |4 − 5| = 1
= |5 − 5| = 0
= |6 − 5| = 1
= |7 − 5| = 2
= |8 − 5| = 3
y = |x − 5|
3
2
1
0
1
2
3
(x, y)
(2, 3)
(3, 2)
(4, 1)
(5, 0)
(6, 1)
(7, 2)
(8, 3)
Next, plot the points in the table, as shown in the image on the right. This
first image gives us enough evidence to believe that if we plotted all points
satisfying the equation y = |x − 5|, the result would be the graph on the right.
y
y
10
10
y = |x − 5|
−10
10
x
−10
−10
10
−10
23. First, complete the table of points that satisfy the equation y = (x + 1)2 .
y
y
y
y
y
y
y
2
= (−4 + 1) = 9
= (−3 + 1)2 = 4
= (−2 + 1)2 = 1
= (−1 + 1)2 = 0
= (0 + 1)2 = 1
= (1 + 1)2 = 4
= (2 + 1)2 = 9
x
−4
−3
−2
−1
0
1
2
y = (x + 1)2
9
4
1
0
1
4
9
(x, y)
(−4, 9)
(−3, 4)
(−2, 1)
(−1, 0)
(0, 1)
(1, 4)
(2, 9)
Next, plot the points in the table, as shown in the image on the left. This
first image gives us enough evidence to believe that if we plotted all points
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3.1. GRAPHING EQUATIONS BY HAND
133
satisfying the equation y = (x + 1)2 , the result would be the graph on the
right.
y
y = (x + 1)2
10
−10
10
x
y
10
−10
−10
10
−10
25. First, complete the table of points that satisfy the equation y = −x − 5.
y
y
y
y
y
y
y
= −(−3) − 5 = −2
= −(−2) − 5 = −3
= −(−1) − 5 = −4
= −(0) − 5 = −5
= −(1) − 5 = −6
= −(2) − 5 = −7
= −(3) − 5 = −8
x
−3
−2
−1
0
1
2
3
y = −x − 5
−2
−3
−4
−5
−6
−7
−8
(x, y)
(−3, −2)
(−2, −3)
(−1, −4)
(0, −5)
(1, −6)
(2, −7)
(3, −8)
Next, plot the points in the table, as shown in the image on the right. This
first image gives us enough evidence to believe that if we plotted all points
satisfying the equation y = −x − 5, the result would be the graph on the right.
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CHAPTER 3. INTRODUCTION TO GRAPHING
134
y
y
10
10
−10
10
x
−10
10
−10
−10
x
y = −x − 5
27. Enter the equation y = x2 − 6x + 5 in Y1 in the Y= menu, using the
following keystrokes.
∧
X,T,θ,n
2
−
6
×
X,T,θ,n
+
5
ENTER
Press 2ND WINDOW, then set TblStart = 0 and ΔTbl = 1. Press 2ND
GRAPH to produce the table.
Enter the data from your calculator into a table on your graph paper. Use
a ruler to construct the axes of your coordinate system, label and scale each
axis. Plot the points in the table, as shown in the image on the right.
y
10
x
y
(x, y)
0
1
2
3
4
5
6
5
0
−3
−4
−3
0
5
(0, 5)
(1, 0)
(2, −3)
(3, −4)
(4, −3)
(5, 0)
(6, 5)
−10
10
−10
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x
3.1. GRAPHING EQUATIONS BY HAND
135
The plotted points provide enough evidence to convince us that if we plotted
all points that satisfied the equation y = x2 −6x+5, we would get the following
result.
y
10
y = x2 − 6x + 5
−10
10
x
−10
29. Enter the equation y = −x2 + 2x + 3 in Y1 in the Y= menu using the
following keystrokes.
(-)
X,T,θ,n
∧
2
+
2
×
X,T,θ,n
+
3
ENTER
Press 2ND WINDOW, then set TblStart = −2 and ΔTbl = 1. Press 2ND
GRAPH to produce the table.
Enter the data from your calculator into a table on your graph paper. Use
a ruler to construct the axes of your coordinate system, label and scale each
axis. Plot the points in the table, as shown in the image on the right.
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CHAPTER 3. INTRODUCTION TO GRAPHING
136
y
10
x
y
(x, y)
−2
−1
0
1
2
3
4
−5
0
3
4
3
0
−5
(−2, −5)
(−1, 0)
(0, 3)
(1, 4)
(2, 3)
(3, 0)
(4, −5)
−10
10
x
−10
The plotted points provide enough evidence to convince us that if we plotted
all points that satisfied the equation y = −x2 +2x+3, we would get the following
result.
y
10
−10
10
x
y = −x2 + 2x + 3
−10
31. Enter the equation y = x3 − 4x2 − 7x + 10 in Y1 in the Y= menu using
the following keystrokes.
X,T,θ,n
∧
3
−
×
4
+
1
X,T,θ,n
0
∧
2
−
7
×
X,T,θ,n
ENTER
Press 2ND WINDOW, then set TblStart = −3 and ΔTbl = 1. Press 2ND
GRAPH to produce the table.
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3.1. GRAPHING EQUATIONS BY HAND
137
Enter the data from your calculator into a table on your graph paper. Use
a ruler to construct the axes of your coordinate system, label and scale each
axis. Plot the points in the table, as shown in the image on the right.
y
x
y
(x, y)
−3
−2
−1
0
1
2
3
4
5
6
−32
0
12
10
0
−12
−20
−18
0
40
(−3, −32)
(−2, 0)
(−1, 12)
(0, 10)
(1, 0)
(2, −12)
(3, −20)
(4, −18)
(5, 0)
(6, 40)
50
−10
10
−50
The plotted points provide enough evidence to convince us that if we plotted
all points that satisfied the equation y = x3 − 4x2 − 7x + 10, we would get the
following result.
y
50
y = x3 − 4x2 − 7x + 10
−10
10
x
−50
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CHAPTER 3. INTRODUCTION TO GRAPHING
138
33. Enter the equation y = −x3 + 4x2 + 7x − 10 in Y1 in the Y= menu using
the following keystrokes.
(-)
X,T,θ,n
∧
3
+
−
×
4
1
0
X,T,θ,n
∧
2
+
7
×
ENTER
Press 2ND WINDOW, then set TblStart = −3 and ΔTbl = 1. Press 2ND
GRAPH to produce the table.
Enter the data from your calculator into a table on your graph paper. Use
a ruler to construct the axes of your coordinate system, label and scale each
axis. Plot the points in the table, as shown in the image on the right.
x
y
(x, y)
−3
−2
−1
0
1
2
3
4
5
6
32
0
−12
−10
0
12
20
18
0
−40
(−3, 32)
(−2, 0)
(−1, −12)
(0, −10)
(1, 0)
(2, 12)
(3, 20)
(4, 18)
(5, 0)
(6, −40)
y
50
−10
10
−50
The plotted points provide enough evidence to convince us that if we plotted
all points that satisfied the equation y = −x3 + 4x2 + 7x − 10, we would get
the following result.
Second Edition: 2012-2013
x
X,T,θ,n
3.1. GRAPHING EQUATIONS BY HAND
139
y
50
−10
10
x
y = −x3 + 4x2 + 7x − 10
−50
√
35. First, enter the equation y = x + 5 into Y1 in the Y= menu using the
following keystrokes. The results are shown in the first image below.
2ND
√
X,T,θ,n
+
5)
ENTER
Next, select 2ND WINDOW and adjust the settings to match those in the
second image. below. Finally, select 2ND GRAPH to open the TABLE. Copy
the results from the third image below into the tables on your graph paper.
x
−5
−4
−3
−2
−1
0
1
2
y=
√
x+5
0
1
1.4
1.7
2
2.2
2.4
2.6
(x, y)
(−5, 0)
(−4, 1)
(−3, 1.4)
(−2, 1.7)
(−1, 2)
(0, 2.2)
(1, 2.4)
(2, 2.6)
x
3
4
5
6
7
8
9
10
y=
√
x+5
2.8
3
3.2
3.3
3.5
3.6
3.7
3.9
(x, y)
(3, 2.8)
(4, 3)
(5, 3.2)
(6, 3.3)
(7, 3.5)
(8, 3.6)
(9, 3.7)
(10, 3.9)
Next, plot the points in the tables, as shown in the image on the left. This
first image gives us enough√evidence to believe that if we plotted all points
satisfying the equation y = x + 5, the result would be the graph on the right.
Second Edition: 2012-2013