© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
Supplemental Worksheet Problems To Accompany:
The Algebra 2 Tutor
Section 1 – Graphing Equations
Please watch Section 1 of this DVD before working these problems.
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Page 1
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
1) Plot the point below on an x-y coordinate axis.
( 5, 2 )
2) Plot the point below on an x-y coordinate axis.
(1, −3)
Page 2
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
3) Plot the point below on an x-y coordinate axis.
( −3, 2 )
4) Plot the point below on an x-y coordinate axis.
( −4,5)
Page 3
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
5) Plot the point below on an x-y coordinate axis.
1⎞
⎛
⎜ −2, − ⎟
2⎠
⎝
6) Plot the point below on an x-y coordinate axis.
⎛3 1⎞
⎜ ,− ⎟
⎝2 2⎠
Page 4
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
7) Complete the table of values for the equation below then graph the equation using
the points.
y = x−4
x
5
4
-1
y
Page 5
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
8) Complete the table of values for the equation below then graph the equation using
the points.
y=
x
1
-1
-4
x
2
y
Page 6
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
9) Complete the table of values for the equation below then graph the equation using
the points.
y = 3x + 1
x
-2
y
0
1
Page 7
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
10) Graph the equation by using the intercept method.
x+ y =7
Page 8
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
11) Graph the equation by using the intercept method.
x − y = −2
Page 9
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
12) Graph the equation by using the intercept method.
3x + y = −1
Page 10
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
13) Graph the equation by using the intercept method.
3x − 2 y = 6
Page 11
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
Question
Answer
1) Plot the point below on an x-y
coordinate axis.
Begin.
( 5, 2 )
y
We start by drawing the x-y coordinate
axes.
x
Since our point is (5,2), we move along
the x-axis 5 units, then move up the yaxis by 2 units. This is our point.
y
(5,2)
Ans: See graph at left.
x
Page 12
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
Question
Answer
2) Plot the point below on an x-y
coordinate axis.
Begin.
(1, −3)
y
We start by drawing the x-y coordinate
axes.
x
Since our point is (1,-3), we move along
the x-axis 1 units, then move down the
y-axis by 3 units. This is our point.
y
Ans: See graph at left.
x
(1,-3)
Page 13
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
Question
Answer
3) Plot the point below on an x-y
coordinate axis.
Begin.
( −3, 2 )
y
We start by drawing the x-y coordinate
axes.
x
Since our point is (-3,2), we move to the
left along the x-axis 3 units, then move up
the y-axis by 2 units. This is our point.
y
(-3,2)
Ans: See graph at left.
x
Page 14
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
Question
Answer
4) Plot the point below on an x-y
coordinate axis.
Begin.
( −4,5)
y
We start by drawing the x-y coordinate
axes.
x
Since our point is (-4,5), we move to the
left along the x-axis 4 units, then move up
the y-axis by 5 units. This is our point.
y
(-4,5)
Ans: See graph at left.
x
Page 15
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
Question
Answer
5) Plot the point below on an x-y
coordinate axis.
Begin.
1⎞
⎛
⎜ −2, − ⎟
2⎠
⎝
y
We start by drawing the x-y coordinate
axes.
x
Since our point is (-2, -1/2), we move to
the left along the x-axis 2 units, then
move down the y-axis by ½ of a unit.
This is our point.
y
Ans: See graph at left.
x
(-2, -1/2)
Page 16
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
Question
Answer
6) Plot the point below on an x-y
coordinate axis.
Begin.
⎛3 1⎞
⎜ ,− ⎟
⎝2 2⎠
y
We start by drawing the x-y coordinate
axes.
x
Since our point is (3/2, -1/2), we move to
the right along the x-axis by 3/2 units
(note that 3/2 = 1.5), then move down the
y-axis by ½ of a unit. This is our point.
y
Ans: See graph at left.
x
(3/2, -1/2)
Page 17
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
Question
Answer
7) Complete the table of values for the
equation below then graph the equation
using the points.
Begin.
y = x−4
x
5
4
-1
y
Plug in x = 5...
We need to plug the values of ‘x’ into the
equation and solve for the corresponding
values of ‘y’.
y = 5−4
y =1
We do this for the first point and find the
corresponding value for ‘y’.
Plug in x = 4...
y = 4−4
y=0
Plug in x = -1...
y = −1 − 4
y = −5
We need to plug the values of ‘x’ into the
equation and solve for the corresponding
values of ‘y’.
We do this for the second point and find
the corresponding value for ‘y’.
We need to plug the values of ‘x’ into the
equation and solve for the corresponding
values of ‘y’.
We do this for the third point and find the
corresponding value for ‘y’.
(continued on next page).
Page 18
© 2008 Jason Gibson / MathTutorDVD.com
x
5
4
-1
The Algebra 2 Tutor
Section 1 – Graphing Equations
Put the calculated values for ‘y’ into the
table.
y
1
0
-5
Now we need to plot the points. For each
point, we travel along ‘x’ by the amount
given in the ‘x’ column, then travel up or
down in the ‘y’ column as shown in the
table. We put a dot for each point.
y
(5, 1)
After you have plotted the points, draw a
straight line through them.
x
(4, 0)
Ans: See graph at left.
(-1, -5)
Page 19
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
Question
Answer
8) Complete the table of values for the
equation below then graph the equation
using the points.
y=
x
1
-1
-4
x
2
y
Plug in x = 1...
y=
Begin.
1
2
We do this for the first point and find the
corresponding value for ‘y’.
Plug in x = -1...
y=−
We need to plug the values of ‘x’ into the
equation and solve for the corresponding
values of ‘y’.
1
2
Plug in x = -4...
4
2
y = −2
We need to plug the values of ‘x’ into the
equation and solve for the corresponding
values of ‘y’.
We do this for the second point and find
the corresponding value for ‘y’.
We need to plug the values of ‘x’ into the
equation and solve for the corresponding
values of ‘y’.
y=−
We do this for the third point and find the
corresponding value for ‘y’.
(continued on next page).
Page 20
© 2008 Jason Gibson / MathTutorDVD.com
x
1
The Algebra 2 Tutor
Section 1 – Graphing Equations
Put the calculated values for ‘y’ into the
table.
y
½
-1
-½
-4
-2
Now we need to plot the points. For each
point, we travel along ‘x’ by the amount
given in the ‘x’ column, then travel up or
down in the ‘y’ column as shown in the
table. We put a dot for each point.
y
(-1, -1/2)
(1, 1/2)
After you have plotted the points, draw a
straight line through them.
x
Ans: See graph at left.
(-4, -2)
Page 21
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
Question
9) Complete the table of values for the
equation below then graph the equation
using the points.
Answer
Begin.
y = 3x + 1
x
-2
y
0
1
Plug in x = -2...
y = 3 ( −2 ) + 1
y = −6 + 1
y = −5
Plug in x = 0...
y = 3 ( 0) + 1
We need to plug the values of ‘x’ into the
equation and solve for the corresponding
values of ‘y’.
We do this for the first point and find the
corresponding value for ‘y’.
We need to plug the values of ‘x’ into the
equation and solve for the corresponding
values of ‘y’.
y = 0 +1
y =1
We do this for the second point and find
the corresponding value for ‘y’.
Plug in x = 1...
We need to plug the values of ‘x’ into the
equation and solve for the corresponding
values of ‘y’.
y = 3 (1) + 1
y = 3 +1
y=4
We do this for the third point and find the
corresponding value for ‘y’.
(continued on next page).
Page 22
© 2008 Jason Gibson / MathTutorDVD.com
x
-2
y
-5
0
1
1
4
The Algebra 2 Tutor
Section 1 – Graphing Equations
Put the calculated values for ‘y’ into the
table.
Now we need to plot the points. For each
point, we travel along ‘x’ by the amount
given in the ‘x’ column, then travel up or
down in the ‘y’ column as shown in the
table. We put a dot for each point.
y
(1, 4)
(0, 1)
After you have plotted the points, draw a
straight line through them.
x
Ans: See graph at left.
(-2, -5)
Page 23
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
Question
Answer
10) Graph the equation by using the intercept
method.
x+ y =7
Begin.
Plug in x = 0...
Any line is completely defined by two
points. We can easily find two points on
any line by:
x+ y =7
0+ y = 7
- Substitute x=0 and calculate ‘y’.
- Substitute y=0 and calculate ‘x’.
y=7
We begin by substituting x=0 and
calculate ‘y’.
( 0, 7 )
We write down the point that we found in
the previous step. This is called the yintercept, because it is the point where
the line intercepts the y-axis.
Plug in y = 0...
We now substitute y=0 and calculate the
corresponding ‘x’ value.
x+ y =7
x+0=7
x=7
( 7, 0 )
Write down the point that we found in the
previous step. This is called the xintercept, because it is the point where
the line intercepts the x-axis.
(continued on next page)
Page 24
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
Question
Answer
y
Plot the two intercept points that we have
found and draw a line through them.
(0, 7)
Ans: See graph at left.
(7, 0)
x
Page 25
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
Question
Answer
11) Graph the equation by using the intercept
method.
x − y = −2
Plug in x = 0...
x − y = −2
0 − y = −2
− y = −2
y=2
Begin.
Any line is completely defined by two
points. We can easily find two points on
any line by:
- Substitute x=0 and calculate ‘y’.
- Substitute y=0 and calculate ‘x’.
We begin by substituting x=0 and
calculate ‘y’.
( 0, 2 )
We write down the point that we found in
the previous step. This is called the yintercept, because it is the point where
the line intercepts the y-axis.
Plug in y = 0...
We now substitute y=0 and calculate the
corresponding ‘x’ value.
x − y = −2
x − 0 = −2
x = −2
( −2, 0 )
Write down the point that we found in the
previous step. This is called the xintercept, because it is the point where
the line intercepts the x-axis.
(continued on next page)
Page 26
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
Question
Answer
y
Plot the two intercept points that we have
found and draw a line through them.
Ans: See graph at left.
(0, 2)
(-2, 0)
x
Page 27
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
Question
Answer
12) Graph the equation by using the intercept
method.
3x + y = −1
Plug in x = 0...
3x + y = −1
3 ( 0 ) + y = −1
y = −1
Begin.
Any line is completely defined by two
points. We can easily find two points on
any line by:
- Substitute x=0 and calculate ‘y’.
- Substitute y=0 and calculate ‘x’.
We begin by substituting x=0 and
calculate ‘y’.
( 0, −1)
We write down the point that we found in
the previous step. This is called the yintercept, because it is the point where
the line intercepts the y-axis.
Plug in y = 0...
We now substitute y=0 and calculate the
corresponding ‘x’ value.
3x + y = −1
3x + 0 = −1
3x = −1
1
x=−
3
⎛ 1 ⎞
⎜ − ,0⎟
⎝ 3 ⎠
Write down the point that we found in the
previous step. This is called the xintercept, because it is the point where
the line intercepts the x-axis.
(continued on next page)
Page 28
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
Question
Answer
y
Plot the two intercept points that we have
found and draw a line through them.
Ans: See graph at left.
(-1/3,0)
x
(0, -1)
Page 29
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
Question
Answer
13) Graph the equation by using the intercept
method.
3x − 2 y = 6
Plug in x = 0...
3x − 2 y = 6
Begin.
Any line is completely defined by two
points. We can easily find two points on
any line by:
3(0) − 2 y = 6
- Substitute x=0 and calculate ‘y’.
- Substitute y=0 and calculate ‘x’.
−2 y = 6
y = −3
We begin by substituting x=0 and
calculate ‘y’.
( 0, −3)
We write down the point that we found in
the previous step. This is called the yintercept, because it is the point where
the line intercepts the y-axis.
Plug in y = 0...
We now substitute y=0 and calculate the
corresponding ‘x’ value.
3x − 2 y = 6
3x − 2 ( 0 ) = 6
3x = 6
x=2
( 2, 0 )
Write down the point that we found in the
previous step. This is called the xintercept, because it is the point where
the line intercepts the x-axis.
(continued on next page)
Page 30
© 2008 Jason Gibson / MathTutorDVD.com
The Algebra 2 Tutor
Section 1 – Graphing Equations
Question
Answer
y
Plot the two intercept points that we have
found and draw a line through them.
Ans: See graph at left.
(2, 0)
x
(0,-3)
Page 31