supplemental worksheet problems to accompany: the algebra 2
TRANSCRIPT
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© 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.
The DVD is located at:
http://www.mathtutordvd.com/products/item9.cfm
Sample Videos For this DVD Are Located Here: http://www.mathtutordvd.com/public/department48.cfm
Page 1
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© 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−
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© 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−
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© 2008 Jason Gibson / MathTutorDVD.com The Algebra 2 Tutor Section 1 – Graphing Equations
5) Plot the point below on an x-y coordinate axis.
12,2
⎛ ⎞− −⎜ ⎟⎝ ⎠
6) Plot the point below on an x-y coordinate axis.
3 1,2 2
⎛ ⎞−⎜ ⎟⎝ ⎠
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© 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.
4y x= − x y
54-1
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© 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.
2xy =
x y
1-1-4
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© 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.
3 1y x= +
x y -2
0
1
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© 2008 Jason Gibson / MathTutorDVD.com The Algebra 2 Tutor Section 1 – Graphing Equations
10) Graph the equation by using the intercept method.
7x y+ =
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© 2008 Jason Gibson / MathTutorDVD.com The Algebra 2 Tutor Section 1 – Graphing Equations
11) Graph the equation by using the intercept method.
2x y− = −
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© 2008 Jason Gibson / MathTutorDVD.com The Algebra 2 Tutor Section 1 – Graphing Equations
12) Graph the equation by using the intercept method.
3 1x y+ = −
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© 2008 Jason Gibson / MathTutorDVD.com The Algebra 2 Tutor Section 1 – Graphing Equations
13) Graph the equation by using the intercept method.
3 2x y 6− =
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© 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.
( )5, 2
Begin.
We start by drawing the x-y coordinate axes.
Since our point is (5,2), we move along the x-axis 5 units, then move up the y-axis by 2 units. This is our point. Ans: See graph at left.
x
y
x
y
x
y
x
y
(5,2)
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© 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.
( )1, 3−
Begin.
We start by drawing the x-y coordinate axes.
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. Ans: See graph at left.
x
y
x
y
x
y
x
y
(1,-3)
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© 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.
( )3,2−
Begin.
We start by drawing the x-y coordinate axes.
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. Ans: See graph at left.
x
y
x
y
x
y
x
y
(-3,2)
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© 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.
( )4,5−
Begin.
We start by drawing the x-y coordinate axes.
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. Ans: See graph at left.
x
y
x
y
x
y
x
y
(-4,5)
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© 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.
12,2
⎛ ⎞− −⎜ ⎟⎝ ⎠
Begin.
We start by drawing the x-y coordinate axes.
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. Ans: See graph at left.
x
y
x
y
x
y
x
y
(-2, -1/2)
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© 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.
3 1,2 2
⎛ ⎞−⎜ ⎟⎝ ⎠
Begin.
We start by drawing the x-y coordinate axes.
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. Ans: See graph at left.
x
y
x
y
x
y
x
y
(3/2, -1/2)
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© 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.
4y x= −
Begin.
Plug in x = 5...
5 41
yy= −=
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’.
Plug in x = 4...
4 40
yy= −=
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’.
Plug in x = -1...
1 45
yy= − −= −
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).
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x y 54-1
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© 2008 Jason Gibson / MathTutorDVD.com 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. After you have plotted the points, draw a straight line through them. Ans: See graph at left.
x y 54-1
10-5
x
y
x
y
(5, 1)
(4, 0)
(-1, -5)
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© 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.
2xy =
Begin.
Plug in x = 1...
12
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 first point and find the corresponding value for ‘y’.
Plug in x = -1...
12
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 second point and find the corresponding value for ‘y’.
Plug in x = -4...
422
y
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).
x y 1-1-4
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© 2008 Jason Gibson / MathTutorDVD.com The Algebra 2 Tutor Section 1 – Graphing Equations
© 2008 Jason Gibson / MathTutorDVD.com The Algebra 2 Tutor Section 1 – Graphing Equations
Put the calculated values for ‘y’ into the table.
x y 1
-1
-4
½
- ½
-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.
After you have plotted the points, draw a straight line through them.
Ans: See graph at left.
x
y
x
y
(1, 1/2)(-1, -1/2)
(-4, -2)
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© 2008 Jason Gibson / MathTutorDVD.com The Algebra 2 Tutor Section 1 – Graphing Equations
Question Answer
9) Complete the table of values for the equation below then graph the equation using the points.
Begin.
3 1y x= +
x y -2
0
1
orresponding alues of ‘y’.
int and find the corresponding value for ‘y’.
We need to plug the values of ‘x’ into the equation and solve for the cv We do this for the first po( )
Plug in x = -2...
3 2 16 15
yyy
= − +
= − += −
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’.
( )
Plug in = 0...
3 0 10 11
yyy
= +
= +=
x
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).
( )
Plug in x = 1...
3 1 13 14
yyy
= +
= +=
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© 2008 Jason Gibson / MathTutorDVD.com The Algebra 2 Tutor Section 1 – Graphing Equations
Put the calculated values for ‘y’ into the
ble.
ta
x y -2 -5
0
1
1
4
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.
Page 23
After you have plotted the points, draw a straight line through them. Ans: See graph at left.
x
y
x
y
(1, 4)
(0, 1)
(-2, -5)
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© 2008 Jason Gibson / MathTutorDVD.com The Algebra 2 Tutor Section 1 – Graphing Equations
Question
Answer
10) Graph the equation by using the intercept method. Begin.
7x y+ =
Substitute y=0 and calculate ‘x’.
0 and alculate ‘y’.
Plug in = 0...
0 77
x yy
y
+ =+ ==
x
7
Any line is completely defined by two points. We can easily find two points onany line by: - Substitute x=0 and calculate ‘y’. - We begin by substituting x=c
( )0, 7
We write down the point that we found in the previous step. This is called the y-intercept, because it is the point where the line intercepts the y-axis.
Plug in y = 0...
70 77
x yxx
+ =+ ==
We now substitute y=0 and calculate the corresponding ‘x’ value.
( )7,0
Write down the point that we found in the previous step. This is called the x-intercept, because it is the point where the line intercepts the x-axis. (continued on next page)
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© 2008 Jason Gibson / MathTutorDVD.com The Algebra 2 Tutor Section 1 – Graphing Equations
Question
Answer
two intercept points that we have found and draw a line through them.
Ans: See graph at left.
Plot the
x
y
x
y
(0, 7)
(7, 0)
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© 2008 Jason Gibson / MathTutorDVD.com The Algebra 2 Tutor Section 1 – Graphing Equations
Question
Answer
11) Graph the equation by using the intercept method.
2x y− = −
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’.
Plug in = 0...
20 2
22
x yy
yy
− =− =
− = −=
x
−−
We write down the point that we found in the previous step. This is called the y-intercept, because it is the point where the line intercepts the y-axis.
( )0, 2
We now substitute y=0 and calculate the corresponding ‘x’ value.
Plug in y = 0...
20 2
2
x yxx
− = −− = −= −
Write down the point that we found in the previous step. This is called the x-intercept, because it is the point where the line intercepts the x-axis. (continued on next page)
( )2,0−
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© 2008 Jason Gibson / MathTutorDVD.com The Algebra 2 Tutor Section 1 – Graphing Equations
Question
Answer
two intercept points that we have found and draw a line through them. Ans: See graph at left.
Plot the
yy
(0, 2)(-2, 0)
xx
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© 2008 Jason Gibson / MathTutorDVD.com The Algebra 2 Tutor Section 1 – Graphing Equations
Question
Answer
12) Graph the equation by using the intercept method.
3 1x y+ = −
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’.
( )
Plug in = 0...
3 13 0 1
1
x yy
y
+ = −
+ = −
= −
x
We write down the point that we found in the previous step. This is called the y-intercept, because it is the point where the line intercepts the y-axis.
( )0, 1 −
Plug in y = 0...
3 13 0 13 1
13
x yxx
x
+ = −+ = −= −
= −
We now substitute y=0 and calculate the corresponding ‘x’ value.
1 ,03
⎛ ⎞−⎜ ⎟⎝ ⎠
Write down the point that we found in the previous step. This is called the x-intercept, because it is the point where the line intercepts the x-axis. (continued on next page)
Page 28
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© 2008 Jason Gibson / MathTutorDVD.com The Algebra 2 Tutor Section 1 – Graphing Equations
Question
Answer
two intercept points that we have found and draw a line through them. Ans: See graph at left.
Plot the
yy
(-1/3,0)
Page 29
xx
(0, -1)
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© 2008 Jason Gibson / MathTutorDVD.com The Algebra 2 Tutor Section 1 – Graphing Equations
Question
Answer
13) Graph the equation by using the intercept method.
63 2x y− =
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’.
( )
Plug in = 0...
3 2 63 0 2 6
2 63
x yy
yy
− =
− =
− == −
x
We write down the point that we found in the previous step. This is called the y-intercept, because it is the point where the line intercepts the y-axis.
( )0, 3 −
We now substitute y=0 and calculate the corresponding ‘x’ value.
( )
Plug in y = 0...
3 2 63 2 0 63 6
2
x yxx
x
− =
− =
==
Write down the point that we found in the previous step. This is called the x-intercept, because it is the point where the line intercepts the x-axis. (continued on next page)
( )2,0
Page 30
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© 2008 Jason Gibson / MathTutorDVD.com The Algebra 2 Tutor Section 1 – Graphing Equations
Question
Answer
two intercept points that we have found and draw a line through them. Ans: See graph at left.
Plot the
yy
Page 31
xx(2, 0)
(0,-3)