Lesson 1: Graphing Points on a Coordinate Plane Equations/Graphin… · ... Graph the following...
Transcript of Lesson 1: Graphing Points on a Coordinate Plane Equations/Graphin… · ... Graph the following...
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 1: Graphing Points on a Coordinate Plane
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 1: Graphing Points on a Coordinate Plane
Directions: Graph the following points on the coordinate plane. (0,0) Y axis
Example: Z. (-4, 6) x axis
x y
A. (2,4) F. (-2,-8)
B. (-4, 1) G. (1,8)
C. (-5, -2) H. (-2,4)
D. (0, -7) I. (5, -2)
E. (8, 0) J. (4, -4)
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Part 1: Identify the coordinates of the points graphed on the grid. (1 point each)
1. Point A
2. Point B
3. Point C
Part 2: Graph the following points on the graph. Label your points. (1 point each)
1. Point D: (-5, 8)
2. Point E: (2, -4)
3. Point F: (10, 0)
4. Point G: (-9, -1)
A
B C
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 1: Graphing Points on a Coordinate Plane
Answer Key
Directions: Graph the following points on the coordinate plane. (0,0) Y axis
Example: Z. (-4, 6) x axis
x y
A. (2,4) F. (-2,-8)
B. (-4, 1) G. (1,8)
C. (-5, -2) H. (-2,4)
D. (0, -7) I. (5, -2)
E. (8, 0) J. (4, -4)
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Part 1: Identify the coordinates of the points graphed on the grid. (1 point each)
1. Point A: (6,3)
2. Point B: (-7, -5)
3. Point C: (0, -4)
Part 2: Graph the following points on the graph. Label your points. (1 point each)
1. Point D: (-5, 8)
2. Point E: (2, -4)
3. Point F: (10, 0)
4. Point G: (-9, -1)
A
B C
D
E
F
G
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 2: Using a Table of Values
Example 1
y = -4x – 8
x -4x – 8 y
-4( ) – 8
-4( ) – 8
-4( ) – 8
Example 2
y = -2/3x + 1
x -2/3x + 1 y
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 2: Graphing Linear Equations Using a Table of Values
Directions: Use a table of values to plot points and graph the line of the following linear equations.
1. y = 2x -4
x 2x - 4 y
Ordered Pairs:
2. y = -x + 3
x -x + 3 y
Ordered Pairs:
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
3. y = -3x – 5
x -3x - 5 y
Ordered Pairs:
4. y = ½x + 4
x ½x + 4 y
Ordered Pairs:
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
5. y = -¾x - 2
x -¾x - 2 y
Ordered Pairs:
6. y = ⅕x - 5
x ⅕x - 5 y
Ordered Pairs:
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Part 2: Which points are solutions to the equations? Justify your answer algebraically
and graphically.
7. y = 4x – 1 (3,12) (2,7) (-1,-5)
Solutions:
8. y = -1/2x + 2 (-2,1) (2,1) (-4,0)
Solutions:
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Part 1: Complete the table of values. (3 points) Then
graph the line on the grid. (1 point) (4 total)
1. y = -1/2x + 2
x y
Ordered Pairs:
Part 2: Which points are solutions to the
given equation? Justify your answer
algebraically and by graphing. (3 points)
2. y = 3x - 2 (3,7) (1,-1) (-2,-4)
Solutions:
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 2: Graphing Using a Table of Values
Answer Key
1. y = 2x -4
x 2x - 4 y
0 2(0) -4 -4
1 2(1) -4 -2
2 2(2)-4 0
Ordered Pairs: (0,-4) (1,-2) (2,0)
2. y = -x + 3
x -x + 3 y
0 0+3 3
1 -1 +3 2
2 -2 +3 1
Ordered Pairs: (0,3) (1,2) (2,1)
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
3. y = -3x – 5
x -3x - 5 y
0 -3(0) -5 -5
1 -3(1)-5 -8
-1 -3(-1) -5 -2
Ordered Pairs: (0,-5) (1,-8) (-1,-2)
4. y = ½x + 4
x ½x + 4 y
0 ½(0) + 4 4
2 ½(2) + 4 5
4 ½(4) + 4 6
Ordered Pairs: (0,4) (2,5) (4,6)
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
5. y = -¾x - 2
x -¾x - 2 y
0 y = -¾(0) - 2
-2
4 y = -¾(4) - 2
-5
8 y = -¾(8) - 2
-8
Ordered Pairs: (0,-2) (4,-5) (8,-8)
6. y = ⅕x - 5
x ⅕x - 5 y
0 ⅕(0) - 5 -5
5 ⅕(5) - 5 -4
10 ⅕(10) - 5 -3
Ordered Pairs: (0,-5) (5,-4) (10,-3)
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Part 2: Which points are solutions to the equations? Justify your answer algebraically and
graphically.
7. y = 4x -1 (3,12) (2,7) (-1,-5)
4x -1
3 4(3) -1 11
2 4(2) -1 7
-1 4(-1) -1 -5
Solutions:
(2,7) (-1,-5)
8. y = -½x + 2 (-2, 1) (2,1) (-4,0)
-½x + 2
-2 -½(-2) + 2 3
2 -½(2)+ 2 1
-4 -½(-4) + 2 4
Solutions:
(2,1)
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
(Answers in the table of values may
vary depending on what x values you
choose. )
1. y = -1/2x + 2
x -1/2x + 2 y
0 -1/2(0) + 2 2
2 -1/2(2) +2 1
-2 -1/2(-2) +2 3
Ordered Pairs:
(0,2) (2, 1) (-2, 3)
Part 2: Which points are solutions to the
given equation? Justify your answer
algebraically and by graphing. (3 points)
2. y = 3x - 2 (3,7) (1,-1) (-2,-4)
X Y = 3x-2 Y
3 Y = 3(3) – 2 7
1 Y = 3(1) – 2 1
-2 Y = 3(-2) – 2 -8
The point (3,7) is the only point that is a
solution to this equation.
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 3: Calculating Slope
Example 1
Rise = ______________ Run = ______________ Slope = rise run __________
Formula for Slope
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Example 2
Example 3
Rise = _________ Run = _________ Slope = rise run _________
Rise = _________ Run = _________ Slope = rise run _________
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Example 4
Special Notes:
A horizontal line always has a
slope of _____________because
_____________________________
A vertical line has an ___________
slope because ________________
_____________________________
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 3: Calculating Slope
Directions: Calculate the slope for each graph.
1.
2.
3.
4.
Rise = ________ Slope =
Run = _________
Slope = rise
run
Rise = ________ Slope =
Run = _________
Rise = ________ Slope =
Run = _________
Rise = ________ Slope =
Run = _________
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
5.
6.
7.
8. John calculated that the slope of the
following line was 3/2. Is he correct?
Explain why or why not!
Rise = ________ Slope =
Run = _________
Rise = ________ Slope =
Run = _________
Rise = ________ Slope =
Run = _________
Explanation:
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Part 1: Calculate the slope of each line. (1 point each)
1. 2.
Part 2: Graph the points. Then calculate the slope of the line. (2 points each)
1. (2, 5) & (-2, 0) 2. (-3, 8) & (2, -3)
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 3: Calculating Slope – Answer Key
Directions: Calculate the slope for each graph.
1.
2.
3.
4.
1 = run
3 = rise
2 = run
1 = run
1 = run
1 = rise
Slope = Rise = 1 SLOPE= 1
Run 1
Slope = Rise = 3 SLOPE= 3
Run 1
- 4 = rise (down)
Slope = Rise = -4 SLOPE = -4
Run 1
1 = rise
Slope = Rise = 1 SLOPE= 1/2
Run 2
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
5.
6.
7. 8. John calculated that the slope of the
following line was 3/2. Is he correct?
Explain why or why not!
3 = run
-2 = rise (down)
Slope = Rise = -2 SLOPE= -2/3
Run 3
4 = rise
3 = run
Slope = Rise = 4 SLOPE= 4/3
Run 3
Slope = Rise = 0 SLOPE= 0
Run 1
For all horizontal lines, the slope is 0.
The slope is not 3/2, it is -3/2. The line is
falling from left to right, so the slope is
negative. There is a rise (fall) of -3 and a run
of 2. Therefore, the slope is -3/2.
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
1. 2.
Rise: 5 The slope is 5/6. Rise = -1 The slope is -1/4 Run 6 Run = 4
Part 2: Graph the points. Then calculate the slope of the line. (2 points each)
1. (2, 5) & (-2, 0) 2. (-3, 8) & (2, -3)
Rise = 5 The slope is 5/4. Rise = -11 The slope is -11/5
Run = 4 Run = 5
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
**BONUS LESSON** - Predicting Slope
There are a couple of rules that you can use to predict the slope of a line. These rules will
also help you to check your answers as you calculate slope!
For each graph below, determine whether the slope of the line will be positive or negative.
You read a graph from left to right just as you do
a book. If the line is falling (from left to right),
then the slope is negative.
The slope of this line is negative!
If the line is rising (from left to right), then the
slope of the line is positive.
The slope of this line is positive!
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Predicting Slope – Answer Key
For each graph below, determine whether the slope of the line will be positive or negative.
Negative Positive Negative
Notice that the sign (negative or positive) only dictates whether the line is rising or falling. It
has no impact on the steepness of the line. In order to identify the steepness of the slope,
we will look at the absolute value of the slope. (By absolute value, I am referring to the
number without regards to it’s’ sign). Take a look at the following examples.
Notice that in the first graph, the slopes of the lines are less than 1. The slopes of the lines in
the second graph are 1 or more. Notice how the slope of 1 and 2 are greater than the slopes
of ½ or ¼. The larger the absolute value of the slope, the greater (steeper) the slope of the
line.
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 4: Graphing Slope
Example 1
Example 2
1. Start with a point on (0,-4).
2. Draw a line with a slope of
¾.
1. Start with a point on (0,3).
2. Draw a line with a slope of
-2.
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Example 3
Example 4
1. Start with a point on (0,8).
2. Draw a line with a slope of
3/5.
Part 1:
1. Start with a point on (0,-2).
2. Draw a line with a slope of 0.
Part 2:
3. Start with a point on (4,5).
4. Draw a line with an undefined
slope through this point.
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 4: Graphing Slope
Directions: Start at the point given to you on the graph. Use the slope given to plot the next 2
points and then draw a line to create your graph.
1. Slope = 3 2. Slope = -2
3. Slope = 2/3 4. Slope = -1/2
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
5. Slope = 4/3 6. Slope = 0
7. Slope = -1/3 8. Slope = Undefined
Quiz #1 is coming up next! You will need to be able to:
• graph an equation using a table of values
• calculate slope
• graph slope.
**Hint: Create yourself a study guide to help you prepare for the quiz.
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Directions: Start at the point given to you on the graph.
Use the slope given to plot the next 2 points
and then draw a line to create your graph. (2 points each)
1. Slope = 1/3 2. Slope = -4
3. Slope = -3/4 4. Slope = 0
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 4: Graphing Slope – Answer Key
1. Slope = 3 2. Slope = -2
Rise = 3 (up)
Run 1 (right)
Rise = -2 (down)
Run 1 (right)
Rise = 2 (up)
Run 3 (right)
3. Slope = 2/3 4. Slope = -1/2
Rise = -1 (down)
Run 2 (right)
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
5. Slope = 4/3 6. Slope = 0
7. Slope = -1/3 8. Slope = Undefined
Rise = 4 (up)
Run 3 (right)
Rise = 0 **A slope of 0 always results
Run 1 in a horizontal line.
Rise = -1 (down)
Run 3 (right)
A slope that’s undefined always
results in a vertical line.
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
3. Slope = -3/4 4. Slope = 0
Rise 1
Run 3
Rise -4 OR Rise 4
Run 1 Run -1
Rise -3 Or Rise 3
Run 4 Run -4
A line with a slope of 0, is always a
horizontal line.
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Quiz # 1
1. Which line on the graph has a slope of -2/3?
2. What is the slope of the line that passes 3. Start with a point on (-4,8). Draw a
through (-2, 4) (5, 8)? line with a slope of -2.
A. Line A
B. Line B
C. Line C
D. Line D
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
4. Start with a point on the origin. Draw a line with a slope of 3/5.
5. Complete the table of values for the following equation. Then find the slope of the line.
y = 1/4x - 2
x y
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Quiz # 1 – Answer Key
1. Which line on the graph has a slope of -2/3? (1 point)
2. What is the slope of the line that passes 3. Start with a point on (-4,8). Draw a
through (-2, 4) (5, 8)? (2 points) line with a slope of -2. (2 points)
A. Line A
B. Line B
C. Line C
D. Line D
B
You can eliminate answers C and
A because they are positive slopes.
(When read from left to right, they
rise).
Letters D and B are negative
slopes, but only letter B rises -2
and runs 3.
Rise = 4
Run = 7
Slope = rise/run
Slope = 4/7
(-4, 8)
A slope of -2 means that you count
down 2 and right 1.
Slope = rise/run
Slope = -2/1
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
4. Start with a point on the origin. Draw a line with a slope of 3/5. (2 points)
5. Complete the table of values for the following equation. Then find the slope of the line.
y = 1/4x – 2 (3 points)
x y = 1/4x - 2
y
0 y = 1/4x - 2
y = ¼(0) – 2
-2
4 y = 1/4x - 2
y = ¼(4) – 2
-1
-4 y = 1/4x - 2
y = ¼(-4) -2
-3
Origin (0,0)
A slope of 3/5 means that you count up 3 and
right 5.
Slope = rise/run
Slope = 3/5
Ordered Pairs
(0, -2)
(4, -1)
(-4, -3)
This quiz is worth 10 points.
The slope of the line is
¼. You rise 1 and run
4 to get from one
point to the next
point.
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 5: Graphing Using Slope Intercept Form
Example 1 Example 2
Slope = ________ Y-intercept = __________
Slope = ________ Y-intercept = __________
Notes
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Example 3
Example 4
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 5: Using Slope Intercept Form to Graph Equations
Directions: Graph each equation using the following directions:
Example:
1. Circle the y-intercept in the equation.
Then plot this point on your graph.
y = 1/2x +2
2. Put a square around the slope in the
equation.
Then use the slope to plot your next point.
y = ½ x +2
3. Draw a line through your two points.
(0,2)
Rise = 1 (up 1)
Run 2 (right 2)
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
1. y = 2x +2 2. y = -3x +1
3. y = 1/3x - 4 4. y = -2/5x - 2
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
5. y = -3/4x 6. y = x +5
7. y = -2 8. y = -4x
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
9. Which graph represents the equation: y = -2/3x – 7?
A. Line A
B. Line B
C. Line C
D. Line D
10. Graph the equation: y = 3x -8
on the grid. Label this line A.
Identify three solutions for this
equation.
11. Graph the equation: y = -x +10
on the grid. Label this line B.
Identify three solutions for this
equation.
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
1. Which line on the graph represents the equation: y = -3/4x – 2? (1 point)
A. Line A
B. Line B
C. Line C
D. Line D
Part 2: Identify the slope and y-intercept in each equation. Then graph the equation on the grid.
(3 points each)
1. y = 3x -6 2. y = -1/4x
Slope: _______ Slope: __________
Y-intercept:_________ Y-intercept:___________
A
B
C
D
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 5: Using Slope Intercept Form - Answer Key
1. y = 2x + 2
3. y = 1/3x - 4 4. y = -2/5x - 2
2. y = -3x +1
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
5. y = -3/4x 6. y= x +5
7. y = -2 8. y = -4x
**The y-intercept is (0,0). When no
number is written in the y-intercept’s
place, it is assumed to be 0.
**The slope is 1. When there is no
coefficient (slope), the slope is assumed
to be 1. (1 times x is still x)
**There is no variable term in this
equation; therefore the slope is 0.
Anytime the slope is 0, the line is
horizontal. Therefore, we have a
horizontal line through the y-intercept.
**The y-intercept is (0,0). When no
number is written in the y-intercept’s
place, it is assumed to be 0.
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
9. Which graph represents the equation: y = -2/3x – 7?
A. Line A
B. Line B
C. Line C
D. Line D
10. Graph the equation: y = 3x -8
on the grid. Label this line A.
Identify three solutions for this
equation.
11. Graph the equation: y = -x +10
On the grid. Label this line B.
Identify three solutions for this
equation.
Since the equation has a slope of
-2/3, you can eliminate C and D
because both lines have a
positive slope (rising from left to
right.)
The equation y = -2/3x – 7 has a
y-intercept of -7. Therefore, letter
B, must be the correct answer
because the line passes through
the point (0, -7).
Letter B has a slope of -2/3. y-intercept = -7
Possible solutions:
(4,4) (3,1) (2,-2) (1,-5) (0,-8)
(Or any point on the line)
Possible solutions:
(0,10) (1,9) (2,8) (3,7) (4,6)
(Or any point on the line)
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
1. Which line on the graph represents the equation: y = -3/4x – 2? (1 point)
A. Line A
B. Line B
C. Line C
D. Line D
Part 2: Identify the slope and y-intercept in each equation. Then graph the equation on the grid.
(3 points each)
1. y = 3x -6 2. y = -1/4x +0
Slope: 3 Slope: -1/4
Y-intercept: -6 Y-intercept: 0 (this is the origin)
A
B
C
D
Line B has a y-intercept of -2. The
slope is -3/4, so if you count down
three and right 4, you will end up on
the next point on the line. The slope is
negative, so you can immediately
eliminate letter D since its slope is
positive.
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 6: Finding Slope Given Two Points
Example 1 Example 2
Find the slope of the line that passes through
(-3,0) and (3,6)
Find the slope of the line that passes through
(4, -4) and (6,7)
Formula for Finding Slope Given Two Points
Using two points: (x1, y1) (x2, y2)
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 6: Finding Slope Given Two Points
Part 1: Find the slope of a line given two points.
1. (2,7) (-5,-3)
2. (10, 3) (9,2) 3. Find the slope of a line that passes
through the origin and the point (-6,2) .
4. Which number represents the slope of the line that contains the following points:
(-2, 8) & (7, -10)
A. -2/3 B. 2 C. -2 D. 2/3
Graph the line for the two points above (-2, 8) & (7, -10) to check your answer. Was your answer
correct? Explain.
Formula For Finding Slope:
Using Two Points: (x1, y1) & (x2, y2)
y2 – y1
x2 - x1
Explain Your Answer:
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
5. Line A passes through the points (2, -7) & ( -3,5). Line B passes through the points
(10, -8) & (22, 9). Which line has a greater slope? Explain your answer.
6. Find the slope of the line that passes through the points (2,3) & (2,-5). Describe what this line
would look like when graphed. Make a sketch of the line.
7. Find the slope of the line that passes through the points (8,4) & (-7,4). Describe what this line
would look like when graphed. Make a sketch of the line.
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
1. Using the slope formula, find the slope of the line that contains the points: (-3,6) & (5,10) (2 points) 2. The three points are vertices of a triangle. Find the slope of each side of the triangle. (6 points) A (-5,1) B (-2, 5) C (1, -4) AB_______ AC_______ BC_______
3. Find the missing value using the given slope and points. (1 point)
1. m = 3/2 and (x, 4) (5,7)
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 6: Finding Slope Given Two Points – Answer Key
Part 1: Find the slope of a line given two points.
1. (2, 7) (-5,-3)
x1 y1 x2 y2
y2 – y1 = -3 – 7 = -10 = 10
x2 – x1 -5 -2 -7 7
Slope = 10/7
2. (10, 3) (9, 2)
x1 y1 x2 y2
y2 – y1 = 2 - 3 = -1= 1
x2 – x1 9 – 10 -1
Slope = 1
4. Which number represents the slope of the line that contains the following points:
(-2, 8) & (7, -10) y2 – y1 = -10 – 8 = -18 = -2
x2 – x1 7 – (-2) 9
A. -2/3 B. 2 C. -2 D. 2/3
b. Graph the line for the two points above (-2, 8) & (7, -10) to check your answer. Was your
answer correct? Explain.
Formula For Finding Slope:
Using Two Points: (x1, y1) & (x2, y2)
y2 – y1
x2 - x1
Explain Your Answer: (Your answer may vary based on
your answer above.)
My answer was correct because the slope is -2. From each
point on the line, if I count a rise of 2 and a run of -1 (to the
left), I will find the next point on the line. Therefore, the
slope is -2.
3. Find the slope of the line that passes
through the origin and the point (-6,2)
(0,0) (-6, 2)
x1 y1 x2 y2
y2 – y1 = 2 – 0 = 2 = -1
x2 – x1 -6 -0 -6 3
Slope = -1/3
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
5. Line A passes through the points (2, -7) & ( -3,5). Line B passes through the points
(10, -8) & (22, 9). Which line has a greater slope? Explain your answer.
Line A: y2 – y1 = 5 – (-7) = 12 = -2 &2/5
x2 – x1 -3 – 2 -5
Line B: y2 – y1 = 9 – (-8) = 17 = 1 & 5/12
x2 – x1 22 – 10 12
Line A has the greater slope. Line A’s slope is – 12/5 and line B’s slope is 17/12. 12/5 is greater
than 17/12. When you are determining the steepness of the slope, disregard the negative signs.
Take the absolute value of the number and the larger the number the greater the slope. The
positive and negative signs only determine whether the line rises or falls.
6. Find the slope of the line that passes through the points (2,3) & (2,-5). Describe what this line
would look like when graphed. Make a sketch of the line.
y2 – y1 = -5 – 3 = -8
x2 – x1 2 -2 0
Since the slope has a 0 in the denominator, the slope of the
line is undefined. That means that this is a vertical line.
7. Find the slope of the line that passes through the points (8,4) & (-7,4). Describe what this line
would look like when graphed. Make a sketch of the line.
y2 – y1 = 4-4 = 0
x2 – x1 -7 -8 -15
Since this slope has a 0 in the numerator, the slope
of the line is 0. This is a horizontal line through (0,4).
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
1. Using the slope formula, find the slope of the line that contains the points: (-3,6) & (5,10) (2 points) y2 – y1 10 – 6 = 4 = 1 x2 - x1 5-(-3) 8 2 The slope of the line is 1/2 2. The three points are vertices of a triangle. Find the slope of each side of the triangle. (6 points) A (-5,1) B (-2, 5) C (1, -4) AB 4/3 AC -5/6 BC -3
3. Find the missing value using the given slope and points. (1 point)
1. m = 3/2 and (x, 4) (5,7)
y2 – y1 7 – 4 = 3 x2 - x1 5 – x 2
x = 3 because 5 – 3 = 2. Since the slope is 3/2, we know that the denominator must equal 3 once
the coordinates are substituted into the slope formula.
Line segment AB
A (-5,1) B (-2,5)
y2 – y1 5 – 1 = 4 x2 - x1 -2 – (-5) 3
Slope of AB = 4/3
Line segment AC
A (-5,1) C (1, -4)
y2 – y1 -4 - 1 = -5 x2 - x1 1 – (-5) 6
Slope of AC= 5/6
Line segment BC
B (-2,5) C (1, -4)
y2 – y1 -4 –5 = -9 = -3 x2 - x1 1-(-2) 3
Slope of BC = -3
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 7: Rate of Change
Example 1
Formula for Finding Slope Given Two Points
Using two points: (x1, y1) (x2, y2)
In real world problems, slope is related to how something changes (usually over time). In this
case, slope is referred to as rate of change. Therefore, in real world problems when you are
asked to find the rate of change, you are actually finding the slope. The same formulas that
you use for slope will also be used for rate of change.
1. What is the average rate of change
between hours 4 and 9?
2. What is the average rate of change
between hours 9 and 11? What most likely
occurred during this time?
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Example 2
Jerry purchased a house in 1994 for $230,000. In 2008, he sold the house for $378,000. What is
the average rate of change in the value of the house per year?
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 7: Rate of Change Practice
The following graph shows the weekly price of a stock over a ten week period. Use the graph to
answer the following questions:
1a. How much was the stock worth at the very beginning (week 0)? ___________
Write this as an ordered pair. (# of week, Price of stock) (0, ____)
b. How much was the stock worth during week 3? ___________
Write this as an ordered pair. (# of week, Price of stock) (3, ____)
c. What is the average rate of change during this time frame (week 0 to week 3)?
Write your two ordered pairs: (___ , ____) (___ , ____). Find the slope. (Slope and rate of
Slope = y2 – y1 change are synonymous)
x2 - x1
The rate of change from week 0 to week 3 is ________________. This means
that_________________________________________________________________________
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
2. What is the average rate of change from week 3 to week 5?
Week 3 ordered pair: ___________ Week 5 ordered pair _____________
Rate of Change = slope = y2 – y1
x2 - x1
The rate of change from week 3 to week 5 is _____________.
6b. Explain what the rate of change means during this time frame.
3. Find the rate of change for the company’s final downfall, between weeks 7 and 10.
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
4. As part of a New Year’s Resolution plan, Brenda started a weight loss program on January 1,
2009. On January 1, 2009 she weighed 185 pounds. Six months later she recorded a weight of
140 pounds. Find the average monthly rate of change of her weight in pounds per month.
Ordered Pairs: ___________________ ___________________________
January weight June weight
5. Jonathan was driving from Maryland to Virginia for vacation. He started his trip at 5:00am. By
7:00 am he had travelled 120 miles. By 9:00 am he had travelled 210 miles. He stopped for
breakfast between 9 and 10:00 am. By 12:00 pm he had reached his destination and had travelled
480 miles.
a. Create a graph to illustrate this situation. Don’t forget to label your axis and title your graph.
b. Find the average rate of change in miles per hour between 5:00 am and 9:00 am.
c. Find the average rate of change in miles per hour for the entire trip.
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
6. Michael bought 573 shares of stock for $730 in 1995. When he sold the stock in 2007, the stock
was valued at $1735. What is the average rate of change of the price of the stock per year?
7. Claudia bought her first brand new car in 2004 for $23,458. She sold that car in 2010 for
$12,010. What is the average rate of change of the price of the car per year?
8. In the first year that Patricia started her career as a teacher, she made $26,800 per year. She is
now in her 17th year of teaching and her annual salary is $62,000 per year. What is the average
rate of change in dollars per year?
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
1. Josie tracks the balance in her savings account
very carefully. She opened an account and in
January she deposited $480. In February she was
able to deposit $130 dollars more into her account.
In March, she did not work much, so she made no
deposits or withdrawals. In April she deposited
another $440. In May she bought a computer, so
she withdrew $500.
A. Create a graph to represent the balance in Josie’s savings account. (3 points)
B. What is the average rate of change between January and April? (2 points)
C. What is the average rate of change from the time she opened the account to the end of
May? (2 points)
D. What is the average rate of change between February and March? How do you know?
(2 points)
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 7: Rate of Change – Answer Key The following graph shows the weekly price of a stock over a ten week period. Use the graph to
answer the following
questions:
1a. How much was the stock worth at the very beginning (week 0)? $12
Write this as an ordered pair. (# of week, Price of stock) (0, 12)
b. How much was the stock worth during week 3? $19
Write this as an ordered pair. (# of week, Price of stock) (3, 19)
c. What is the average rate of change during this time frame (week 0 to week 3)?
Write your two ordered pairs: (0, 12) (3, 19) Find the slope. (Slope and rate of
Slope = y2 – y1 = 19 - 12 = 7 change are synonymous)
x2 - x1 3 -0 3
The rate of change from week 0 to week 3 is 7/3. This means that between weeks 0 and 3, the
average price of the stock rose $2.33 cents per week. (Since we are talking about money, I
converted the fraction 7/3 to a decimal. 7/3 = 2 – 1/3 which is 2.33333…)
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
2. What is the average rate of change from week 3 to week 5?
Week 3 ordered pair: (3, 19) Week 5 ordered pair (5, 8)
Rate of Change = slope = y2 – y1 = 8 – 19 = -11
x2 - x1 5 – 3 2
The rate of change from week 3 to week 5 is -11/2.
2b. Explain what the rate of change means during this time frame.
During the time period between weeks 3 and 5, the average price of the stock fell $5.50 per week. I
changed 11/2 to a decimal since we were talking about money. 11/2 = 5.50. The price of the stock
fell because the slope is negative.
3. Find the rate of change for the company’s final downfall, between weeks 7 and 10.
Week 7 ordered pair (7, 13) Week 10 ordered pair (10, 2) x1 y1 x2 y2
y2 – y1 = 2 – 13 = -11
x2 - x1 10 - 7 3
The rate of change between weeks 7 and 10 is -11/3
or -3.67. The average price of the stock fell $3.67 per
week between the weeks of 7 and 10.
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
4. As part of a New Year’s Resolution plan, Brenda started a weight loss program on January
1, 2009. On January 1, 2009 she weighed 185 pounds. Six months later she recorded a
weight of 140 pounds. Find the average monthly rate of change of her weight in pounds per
month.
Ordered Pairs: (1, 185) (6, 140)
January weight June weight
y2 – y1 = 140- 185 = -45 = -9
x2 - x1 6-1 5
5. Jonathan was driving from Maryland to Virginia for vacation. He started his trip at
5:00am. By 7:00 am he had travelled 120 miles. By 9:00 am he had travelled 210 miles. He
stopped for breakfast between 9 and 10:00 am. By 12:00 pm he had reached his destination
and had travelled 480 miles.
a. Create a graph to illustrate this situation. Don’t forget to label your axis and title your
graph.
b. Find the average rate of change in miles per hour between 5:00 am and 9:00 am.
(5, 0) (9, 210)
y2 – y1 = 210-0 = 210 = 52.5
x2 - x1 9-5 4
Brenda’s average rate of change is -9. This means
that she lost an average of 9 pounds per month
between January and June.
The average rate of change between 5 and 9 am is 52.5
miles per hour.
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
c. Find the average rate of change in miles per hour for the entire trip.
(5, 0) (12, 480)
y2 – y1 = 480-0 = 480 = 68.6
x2 - x1 12-5 7
6. Michael bought 573 shares of stock for $730 in 1995. When he sold the stock in 2007, the stock
was valued at $1735. What is the average rate of change of the price of the stock per year?
7. Claudia bought her first brand new car in 2004 for $23,458. She sold that car in 2010 for
$12,010. What is the average rate of change of the price of the car per year?
8. In the first year that Patricia started her career as a teacher, she made $26,800 per year. She is
now in her 17th year of teaching and her annual salary is $62,000 per year. What is the average
rate of change in dollars per year?
The average rate of change over the entire trip
is 68.6 miles per hour.
(year, price of stock)
(1995, 730) (2007, 1735)
y2 – y1 = 1735 – 730 = 1005 = 83.75
x2 - x1 2007-1995 12
The stock rose an average of $83.75 per year.
(year, price of car)
(2004, 23,458) (2010, 12,010)
y2 – y1 = 12010-23458 = -11448 = -1908
x2 - x1 2010-2004 6
The car’s value decreased in value by $1908 a year.
(year, salary)
(1, 26800) (17, 62000)
y2 – y1 = 62000-26800 = 35200 = 2200
x2 - x1 17-1 16
Patricia’s salary has an average rate of change
of $2200 per year. Therefore, her salary
increases by about $2200 per year.
**Note: The number of shares (573) is
irrelevant information in this problem. It is not
needed to find the rate of change.
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
1. Josie tracks the balance in her savings account
very carefully. She opened an account on
December 31st and in January she deposited $480.
In February she was able to deposit $130 dollars
more into her account. In March, she did not work
much, so she made no deposits or withdrawals. In
April she deposited another $440. In May she
bought a computer, so she withdrew $500.
A. Create a graph to represent the balance in Josie’s savings account. (3 points)
B. What is the average rate of change between January and April? (2 points)
January (1, 480) April (4, 1050)
y2 – y1 1050 – 480 = 570 = $190 x2 - x1 4-1 3
The rate of change between January and April is $190 per month.
C. What is the average rate of change from the time she opened the account to the end of
May? (2 points)
Open account (0,0) May (5,550)
y2 – y1 550-0 = 550 = $110 x2 - x1 5-0 5
The rate of change from the time she opened the account to the end of May is $110 per
month.
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
D. What is the average rate of change between February and March? How do you know?
(2 points)
Between February and March, the rate of change is 0 because she did not make any
deposits or withdrawals. Her account balance remained the same. The graph shows a
horizontal line which also indicates a slope of 0.
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 8: Graphing Standard Form Equations – Lesson 1 of 2
A Close Look at Intercepts
The y-intercept is on the ________________. The y-intercepts all have _______________
___________________________________.
The x-intercept is on the ________________. The x-intercepts all have _______________
___________________________________.
Finding Intercepts
To find the y-intercept: ________________________
To find the x-intercept: ________________________
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Example 1
Example 2
Graph the line:
4x – 3y = -24
X-intercept: (Let y = 0)
Y-intercept: (Let x = 0)
Graph the line:
-2x - 9y = 18
X-intercept: (Let y = 0)
Y-intercept: (Let x = 0)
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Example 3
Example 4
Mary is preparing for a Super Bowl Party. She has $200 to spend on pizza and wings. One pizza
costs $10. One order of wings costs $5. The equation representing x pizzas and y orders of
wings is: 10x + 5y = 200
• Graph the equation on the grid. Let x = the number of pizzas and y = the number of orders
of wings.
• Using your graph, list one realistic option for buying pizza and wings.
Find the x-intercept of the line: y = -3x – 2.
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 8: Graphing Standard Form Equations
1. Which line has an x intercept of 8 and a y intercept of 5?
2.. Find the x intercept of the line: y = -3x -1
A. -1 C. -1/3
B. -3 D. 1
3. Find the x intercept of the line: y = 2x +4
HINT:
Let y = 0 to find the x-intercept.
A. Line A
B. Line B
C. Line C
D. None
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
4. The slope of a line is ½. The y- intercept is -3. Find the x-intercept.
5. Given the equation: 3x +6y = 9, find the x and y intercepts.
6. Graph the following equations on the grid provided.
A. -2x +8y = 16 B. x – 4y = 8
X intercept = __________ X intercept = __________
Y intercept = __________ Y intercept = __________
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
7. John has started a summer business mowing grass. He charges $10 per hour, but had to spend
$50 on equipment. The equation, y = 10x – 50 is used to determine the total profit after mowing x
lawns.
• Graph the equation on the grid. Let x = the number of hours worked. Let y = the profit.
• Use your graph to determine the amount of profit for 10 hours. Justify your answer.
• Explain why the y-intercept is a negative number in this problem.
Hint: Since the numbers in the equation are so
large, it will be easier to find the x and y
intercepts.
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
8. Jerry spent $30 on ice cream and cookies for his children’s party. Ice cream costs $2 a cone and
cookies cost $1.50 a piece. The equation representing x ice cream cones and y cookies is
2x +1.50y = $30.
• Graph the equation on the grid. Let x = number of ice cream cones and y = number of
cookies.
• Suppose Jerry bought 6 ice cream cones, how many cookies can he buy for $30? Explain
your answer.
9. John is supplying hot dogs and sodas for the high school football game. He has $450 to spend.
The hot dogs cost $3 per pack of 6. The sodas cost $0.50 a piece. The equation representing x
hot dogs and y sodas is: 3x +.50y = 450.
• Graph the equation on the grid.
Let x = the number of hot dogs and
y = the number of sodas.
• Using your graph, list three different realistic options for buying hot dogs and sodas.
(For example, John could buy ____ packages of hot dogs and ____ sodas for $450).
Quiz #2 is next! You will need to be able to graph equations written in slope intercept form, find the
slope of a line given 2 points, and calculate rate of change. Don’t forget to create a study guide.
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
1. The slope of a line is 5 and the y- intercept is -4. Find the x-intercept.
(2 points)
For problems 2-3, identify the x and y-intercepts. Then graph the equations on the grid (3 points ea)
2. 3x – 2y = 18 3. x + 3y = 9
x-intercept: __________ x-intercept: ________
y-intercept: __________ y-intercept: ________
4. Kerry is purchasing food and drinks for an annual Halloween party. She has a budget of $250
and she must provide appetizers and drinks for 40 people. Appetizers average about $9 per box
and sodas average $.90 a 2 liter bottle. Let x represent the number of boxes of appetizers and let y
= the number of 2 liter bottles of soda. The equation that represents this situation is: 9x + .90y =
250. (3 points)
• Graph the equation on the grid.
Let x = the number of appetizers and
y = the number of 2 liters of soda.
• Using your graph, list two different realistic options for buying appetizers and sodas.
(For example, Kerry could buy ____ packages of appetizers and ____ sodas for $250).
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 8: Graphing Standard Form Equations– Answer Key
1. Which line has an x intercept of 8 and a y intercept of 5?
2.. Find the x intercept of the line: y = -3x -1
X intercept: Let y = 0
y = -3x - 1
0 = -3x -1
0+1 = -3x -1 + 1
1 = -3x
1/-3 = -3x/-3
-1/3 = x
A. -1 C. -1/3
B. -3 D. 1
A. Line A
B. Line B
C. Line C
D. None
B
HINT:
You can use the same method to find
the x intercept even if the equation is
written in slope intercept form.
Let y = 0 to find the x-intercept.
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
3. Find the x intercept of the line: y = 2x +4
Let y = 0
Y = 2x + 4
0 = 2x + 4 Substitute 0 for y.
0 – 4 = 2x + 4 -4 Subtract 4 from both sides.
-4 = 2x
-4/2 = 2x /2 Divide by 2 on both sides.
-2 = x The X intercept is -2
4. The slope of a line is ½. The y- intercept is -3. Find the x-intercept.
Since I know the slope and y intercept, I can write an equation in slope intercept form:
y = mx +b m = slope (1/2) b = y intercept (-3)
y = 1/2x -3 Write the equation in slope intercept form.
0 = 1/2x – 3 Let y = 0 to find the x-intercept
0 +3 = 1/2x – 3 + 3 Add 3 to both sides.
3 = 1/2x
(2)3 = (2)1/2x Multiply by 2 on both sides.
6 = x The x intercept is 6
5. Given the equation: 3x +6y = 9, find the x and y intercepts.
X intercept: Let y = 0 Y Intercept: Let x = 0
3x +6y = 9 3x +6y = 9
3x +6(0) = 9 3(0) +6y = 9
3x = 9 6y = 9
3 3 6 6
x = 3 X intercept = 3 y = 3/2 Y intercept = 3/2
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
6. Graph the following equations on the grid provided.
A. -2x +8y = 16 B. x – 4y = 8
X Intercept: y=0 Y Intercept: x=0 X intercept: y=0 y intercept: x=0
(Eliminates y term) (Eliminates x term) x = 8
-2x = 16 8y = 16 -4y = 8 -2 -2 8 8 -4 -4 X = -8 y = 2 y = -2
X intercept = ___-8_______ X intercept = ____8______
Y intercept = ____2______ Y intercept = _____-2_____
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
7. John has started a summer business mowing grass. He charges $10 per hour, but had to spend
$50 on equipment. The equation, y = 10x – 50 is used to determine the total profit after mowing x
lawns.
• Graph the equation on the grid. Let x = the number of hours worked. Let y = the profit.
• Use your graph to determine the amount of profit for 10 hours. Justify your answer.
• Explain why the y-intercept is a negative number in this problem.
Since the equation is written in slope intercept form, we
know the y intercept is -50. Since this is a large number
and we know that we cannot use a scale of 1 on the
graph, it would be easiest to find the x intercept.
Y = 10x-50
0 = 10x – 50 Let y = 0
0 + 50 = 10x – 50 + 50
50 = 10x
50/10 = 10x/10
5 = x
X intercept = 5 Y intercept = -50
**You need to establish your scale for the x and y axis.
Since the x intercept is 5, and you only need to
determine a profit for 10 hours, you can use a scale of 1.
Since the y intercept is -50, we can use a scale of 10.
• According to the graph, the amount of profit for 10 hours would be $50.
• Justify:
y = 10x-50 Original Equation
y = 10(10) – 50 Substitute 50 hours for x
y = 50 y is the profit, so our profit is $50.
• The y-intercept is a negative number in this problem because before John starts
working (0 hours), he has lost money. Since he had to buy equipment for $50, he is in
the negative. He would have to work 5 hours before breaking even at $0 and then he
will start making a profit.
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
8. Jerry spent $30 on ice cream and cookies for his children’s party. Ice cream costs $2 a cone and
cookies cost $1.50 a piece. The equation representing x ice cream cones and y cookies is
2x +1.50y = $30.
• Graph the equation on the grid. Let x = number of ice cream cones and y = number of
cookies.
• Suppose Jerry bought 6 ice cream cones, how many cookies can he buy for $30? Explain
your answer.
Since this equation is already written in standard form,
it is easiest to find the x and y intercepts in order to
graph this equation on the grid.
2x + 1.50y = 30
X Intercept: y = 0 Y Intercept: x =
0
2x + 1.50y = 30 2x + 1.50y = 30
2x + 1.50(0) = 30 2(0) +1.50y = 30
2x = 30 1.50y = 30
2 2 1.50 1.50
x = 15 y = 20
X intercept = 15 Y Intercept = 20
**I must choose a scale that will fit on my graph. I
chose to count by 2’s because then I can maximize the
space in my graph. This graph has endpoints at the x
and y intercept since it represents Jerry’s spending to
$30. It doesn’t go on until infinity like other graphs.
According to the graph, if Jerry bought 6 ice cream cones, he could buy 12 cookies for $30. The point on the
line with an x coordinate of 6 is (6, 12). That means 6 ice cream cones and 12 cookies can be purchased for
$30. Let’s prove it by substituting into the original equation.
2x + 1.50y = 30
2(6) +1.50(12) =
12 + 18 = 30
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
9. John is supplying hot dogs and sodas for the high school football game. He has $450 to spend.
The hot dogs cost $3 per pack of 6. The sodas cost $0.50 a piece. The equation representing x
hot dogs and y sodas is: 3x +.50y = 450.
• Graph the equation on the grid.
Let x = the number of hot dogs and
y = the number of sodas.
• Using your graph, list three different realistic options for buying hot dogs and sodas.
(For example, John could buy ____ packages of hot dogs and ____ sodas for $450).
Since this equation is already written in
standard form, it would be easiest to graph
using the x and y intercepts. We also know
that our endpoints on the graph are the x and
y intercept since John’s spending is limited to
$450.
3x + .50y = 450
X intercept: y = 0 Y intercept: x = 0
3x + .50y = 450 3x +.50y = 450
3x +.50(0) = 450 3(0) +.50y = 450
3x = 450 .50y = 450
3 3 .50 .50
x = 150 y = 900
I need to choose a scale for my graph based
on the x and y intercept. I have 10 spaces on
the horizontal axis and it must reach 150. I
am going to use a scale of 20 on the x axis.
The y axis must reach 900. So, I will need to
use a scale of 100.
With $450, John could buy:
50 packages of hot dogs and 600 sodas.
3x +.50y = 450
3(50) +.50(600) = 450
60 packages of hot dogs and 540 sodas.
3x +.50y = 450
3(60) +.50(540) = 450
80 packages of hot dogs and 420 sodas.
3x +.50y = 450
3(80) +.50(420) = 450
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
1. The slope of a line is 5 and the y- intercept is -4. Find the x-intercept. (2 points)
Slope = 5 Y-int = -4. Equation in slope intercept form: y = 5x – 4
If we are finding the x-intercept, then we will let y = 0.
Y = 5x – 4
0 = 5x – 4 let y = 0
0+4 = 5x – 4+ 4 Add 4 to both sides
4 = 5x Simplify
4/5 = 5x/5 Divide by 5 on both sides
4/5 = x Simplify
The x-intercept is equal to 4/5.
For problems 2-3, identify the x and y-intercepts. Then graph the equations on the grid. (3 points
ea)
2. 3x – 2y = 18 3. x + 3y = 9
x-intercept: 6 x-intercept: 9
y-intercept: -9 y-intercept: 3
X-intercept:
Let y = 0
3x – 2(0) = 18
3x = 18
3x/3 = 18/3
x= 6
Y-intercept:
Let x = 0
3(0)– 2y = 18
-2y = 18
-2y/-2 = 18/-2
x= -9
X-intercept:
Let y = 0
x +3(0) = 9
x = 9
Y-intercept:
Let x = 0
(0)+ 3y = 9
3y = 9
3y/3 = 9/3
x= 3
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
4. Kerry is purchasing food and drinks for an annual Halloween party. She has a budget of $250
and she must provide appetizers and drinks for 40 people. Appetizers average about $9 per box
and sodas average $.90 a 2 liter bottle. Let x represent the number of boxes of appetizers and let y
= the number of 2 liter bottles of soda. The equation that represents this situation is: 9x + .90y =
250.
(3 points)
• Graph the equation on the grid.
Let x = the number of appetizers and
y = the number of 2 liters of soda.
• Using your graph, list two different realistic options for buying appetizers and sodas.
(For example, Kerry could buy ____ packages of appetizers and ____ sodas for $250).
The x-intercept is 27.8 and the y-intercept is 277.8. Kerry could buy 21 boxes of appetizers
and 60 2-liters of soda. She could also buy 24 boxes of appetizers and 30 2-liters of soda.
(Answers may vary)
Find the x and y intercepts in order to graph the
equation on the grid.
X-intercept:
Let y = 0
9x+.9(0) = 250
9x = 250
9x/9 = 250/9
x= 27.8
Y-intercept:
Let x = 0
9(0)+.9y = 250
.9y = 250
.9y/.9 = 250/.9
y= 277.8
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Graphing Equations - Quiz #2
1. Graph the following 3 equations on the graph.
Label each line on the graph.
y = 4x -6
y = 2/3x + 2
y = -5x + 8
2. Use the slope formula to find the slope of the line that passes through the
points (4, 6) & (-2, -1)
3. Lindsey has been tracking her savings account in order to keep track of her spending habits.
The first month account balance was $672. Eight months later her account balance was
$389. What is the average rate of change in dollars per month?
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Graphing Equations - Quiz #2 (Answer Key)
1. Graph the following 3 equations on the graph.
Label each line on the graph.
y = 4x -6 (2 points)
y = 2/3x + 2 (2 points)
y = -5x + 8 (2 points)
2. Use the slope formula to find the slope of the line that passes through the
points (4, 6) & (-2, -1) (2 points)
3. Lindsey has been tracking her savings account in order to keep track of her spending habits.
The first month account balance was $672. Eight months later her account balance was
$389. What is the average rate of change in dollars per month? (2 points)
y2 – y1 = -1 - 6 = -7 = 7/6
x2 - x1 -2 - 4 -6
The slope of the line is 7/6
(month, balance)
(1, 672) (8, 389)
y2 – y1 = 389 - 672 = -283 = -40.43
x2 - x1 8-1 7
Lindsey’s rate of change is -40.43. Therefore, on average, she spent about $40.43 per month.
This quiz is worth 10 points.
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 9: Graphing Standard Form Equations – Lesson 2 of 2
Example 1
Method 2: Rewriting the equation in slope intercept form
Use the same strategies that were used for solving equations:
1. ______________________________________________________________
2. ______________________________________________________________
Your goal is to solve for _______________.
Graph the following equations:
6x – 8y = -16
x – y = -9
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 9: Graphing an Equation in Standard Form (2 of 2)
Directions: First rewrite each equation in slope intercept form. Then identify the slope and y-
intercept. Last, graph your equation on the grid.
1. -4x – y = -2
Slope = ____ y-intercept = ____
2. -2x – 3y = 6
Slope = ____ y-intercept = ____
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
3. -4x – y = -2
Slope = ____ y-intercept = ____
4. -3x +9y = -9
Slope = ____ y-intercept = ____
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
5. Which one of the following equations shows 12x – 3y = 6 in slope intercept form?
A. 3y = 12x -6
B. y = 4x -2
C. –y = -4x + 2
D. y = 4x +2
6. Write an equation (in slope intercept form) that is equivalent to: 8x -2y = 12
7. Which one of the following equations shows -3y = 6 – 12x in slope intercept form?
A. -3y = 6 – 12x
B. y = 6 – 12x
C. y = 4x – 2
D. y = -4x + 2
8. Given the following equation: 3x – 5y = 10, identify the slope of the line.
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
9. Given the following equation: 3x – 8y = 16, identify the slope and y-intercept of the line.
10. Are the following equations equivalent? Justify your answer.
2x – 4y = 8 & y = -1/2x -2
1. Write an equation in slope intercept form that is equivalent to: 2x – 5y = 12
(2 points)
2. Given the equation: 4x + 3y = 8. Identify the slope and y-intercept of the line. (2 points)
3. Are the following equations equivalent? Explain your answer, then justify by graphing each line
on the grid below. (4 points)
2x – 3y = 15
4x = 6y + 36
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Lesson 9: Graphing an Equation in Standard Form – Answer Key
Directions: For each problem below, identify the slope and the y-intercept.
1. 3x +2y = -4
3x – 3x +2y = -4 – 3x Subtract 3x
2y = -4 – 3x Divide by 2
2 2 2
y = -2 - 3x Simplify
2
y = -3/2x -2 Switch terms around
Slope = -3/2 y-intercept = -2
2. -2x – 3y = 6
-2x +2x – 3y = 6 +2x Add 2x
-3y = 6 +2x Divide by -3
-3 -3 -3
y = -2 – 2/3x Simplify
y = -2/3x – 2 Switch terms around
Slope = -2/3 y-intercept = -2
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
3. -4x – y = -2
-4x +4x –y = -2 +4x Add 4x
-y = -2 +4x Divide by -1
-1 -1 -1
y = 2 – 4x Simplify
y = -4x +2 Switch terms around
Slope = -4 y-intercept = 2
4. -3x +9y = -9
-3x +3x +9y = -9 +3x Add 3x
9y = -9 + 3x Divide by 9
9 9 9
y = -1 + 1/3x Simplify
y = 1/3 x – 1 Switch terms around
Slope = 1/3 y-intercept = -1
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
5. Which one of the following equations shows 12x – 3y = 6 in slope intercept form?
A. 3y = 12x -6
B. y = 4x -2
C. –y = -4x + 2
D. y = 4x +2
6. Write an equation (in slope intercept form) that is equivalent to: 8x -2y = 12
7. Which one of the following equations shows -3y = 6 – 12x in slope intercept form?
A. -3y = 6 – 12x
B. y = 6 – 12x
C. y = 4x – 2
D. y = -4x + 2
8. Given the following equation: 3x – 5y = 10, identify the slope of the line.
You can eliminate A and C immediately because slope
intercept form must by y = . The y cannot have a coefficient,
nor can it be negative. Eliminating answers that do not
make sense is a great test taking strategy!
12x – 3y = 6
12x -12x -3y = 6 – 12x
-3y = -12x +6
-3 -3 -3
y = 4x -2 - The answer is B
8x -8x – 2y = 12 -8x Subtract 8x from both sides
-2y = -8x + 12 Simplify & reverse the terms on the right hand side.
-2y = -8x + 12 Divide all terms by -2 -2 -2 -2
y = 4x – 6 The equation written in slope intercept form.
You can eliminate letter A, because slope intercept form must be solved for y.
This equation is almost in slope intercept form. We must get y by itself on the left
hand side; therefore, we need to get rid of the coefficient of -3.
-3y = 6 – 12x Divide all terms by -3
-3 -3 -3
y = -2 + 4x OR y = 4x – 2 Equation written in slope intercept form.
In order to find the slope, we must rewrite the equation in slope intercept form.
3x -3x -5y = 10 -3x Subtract 3x from both sides
-5y = -3x + 10 Simplify & reverse the terms on the right hand side.
-5y = -3x + 10 Divide all terms by -5
-5 -5 -5
y = 3/5x - 2 Equation written in slope intercept form.
The slope of the line is 3/5. (3/5 is the coefficient of x, therefore, it is the slope)
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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
9. Given the following equation: 3x – 8y = 16, identify the slope and y-intercept of the line.
10. Are the following equations equivalent? Justify your answer.
2x – 4y = 8 & y = -1/2x -2
3x -3x – 8y = 16 – 3x Subtract 3x from both sides.
-8y = -3x + 16 Simplify & reverse the terms on the right hand side.
-8y = -3x + 16 Divide all terms by -8
-8 -8 -8
y = 3/8x – 2 Equation written in slope intercept form.
The slope of the line is 3/8 and the y-intercept is -2.
In order to determine if the equations are equivalent, you must rewrite the standard form in slope intercept
form. If the equations are equivalent, then they will be the exact same equation when written in slope
intercept form.
Let’s rewrite the standard form equation in slope intercept form.
2x – 4y = 8
2x -2x – 4y = 8 -2x Subtract 2x from both sides.
-4y = -2x + 8 Simplify & reverse the terms on the right hand side.
-4y = -2x + 8 Divide all terms by -4
-4 -4 -4
y = 1/2x - 2 Equation written in slope intercept form.
The equations are not equivalent.
y = 1/2x -2 & y = -1/2x – 2 differ because equation #1 has a positive slope and equation #2 has a
negative slope.
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
1. Write an equation in slope intercept form that is equivalent to: 2x – 5y = 12 (2 points)
In order to write the equation in slope intercept form, I must solve for y.
2x – 5y = 12 Original equation
2x – 2x – 5y = 12 – 2x Subtract 2x from both sides
-5y = -2x + 12 Simplify and rewrite in correct format
-5y/-5 = -2x/-5 + 12/-5 Divide ALL terms by -5
y = 2/5x – 12/5 Simplify
The equation in slope intercept form is: y = 2/5x – 12/5
2. Given the equation: 4x + 3y = 8. Identify the slope and y-intercept of the line. (2 points)
In order to identify the slope and y-intercept of the line, we must rewrite the equation in slope
intercept form. Therefore, we must solve for y.
4x + 3y = 8 Original equation
4x – 4x + 3y = 8 – 4x Subtract 4x from both sides
3y = -4x + 8 Simplify and rewrite in correct format
3y/3 = -4x/3 + 8/3 Divide ALL terms by 3
y = -4/3x + 8/3 Simplify
The equation in slope intercept form is: y = -4/3x + 8/3. Therefore, the slope is -4/3 and the y-
intercept is 8/3.
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
3. Are the following equations equivalent? Explain your answer, then justify by graphing each line
on the grid below. (4 points)
2x – 3y = 15
4x = 6y + 36
We can tell if the equations are equivalent by rewriting both
in slope intercept form:
2x – 3y = 15 Original equation
2x-2x – 3y = 15 – 2x Subtract 2x from both sides
-3y = -2x + 15 Simplify
-3y/-3 = -2x/-3 + 15/-3 Divide All terms by -3
y = 2/3x – 5 Equation in slope intercept form
4x = 6y +36 Original equation
4x – 36= 6y + 36 -36 Subtract 36 from both sides
4x – 36 = 6y Simplify
4x/6 – 36/6 = 6y/6 Divide All terms by 6
2/3x – 6 = y Equation in slope intercept form
y = 2/3x - 6
These equations are not equivalent because when written in slope intercept form, they are
not the same exact equation. The graph shows parallel lines, which means that the
equations are not equivalent. They have the same slope but different y-intercepts and this is
why they are parallel.
Copyright © 2009-2012 Karin Hutchinson – Algebra-class.com
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Unit 2: Unit 2: Unit 2: Unit 2: Graphing EquationsGraphing EquationsGraphing EquationsGraphing Equations
Lesson 10: Graphing Absolute Value Equations
Example 1
Example 2
Graph: y = |x|
Graph: y = |x| +2
Give two solutions to this
equation.
Copyright © 2009-2012 Karin Hutchinson – Algebra-class.com
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Unit 2: Unit 2: Unit 2: Unit 2: Graphing EquationsGraphing EquationsGraphing EquationsGraphing Equations
Example 3
Example 4
Graph: y = -|x-2| +1
Analyze the following graphs.
Y = |1/2x| (red graph)
Y = |2x| (green graph)
Y = |4x| (yellow graph)
Y = -|4x| (blue graph)
What do you notice about the
size of the graphs in regards
to the coefficient of x?
Copyright © 2009-2012 Karin Hutchinson – Algebra-class.com
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Unit 2: Unit 2: Unit 2: Unit 2: Graphing EquationsGraphing EquationsGraphing EquationsGraphing Equations
Lesson 10: Graphing Absolute Value Equations and Inequalities – Practice Problems
Part 1: Graph the following absolute value equations and inequalities on the grid.
1 y= |x+2|
2. y = |x-1| +2 Identify two solutions.
3. y = |3x|+1
4. y = |2x+2|
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Unit 2: Unit 2: Unit 2: Unit 2: Graphing EquationsGraphing EquationsGraphing EquationsGraphing Equations
5. y = |x-1| -3
6. y = - |x| +4
7. y = -|x-1| Identify two solutions.
8. y = |1/2x| -3
Copyright © 2009-2012 Karin Hutchinson – Algebra-class.com
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Unit 2: Unit 2: Unit 2: Unit 2: Graphing EquationsGraphing EquationsGraphing EquationsGraphing Equations
1. Graph the following inequality on
the grid. (3 points)
Y = |2x-4| -2
2. Identify two solutions. (2 points)
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Unit 2: Unit 2: Unit 2: Unit 2: Graphing EquationsGraphing EquationsGraphing EquationsGraphing Equations
Lesson 10: Graphing Absolute Value Equations and Inequalities –Answer Key
Part 1: Graph the following absolute value equations and inequalities on the grid.
1. y= |x+2|
First create a table of values with positive and
negative x values. Then plot the points and
draw your V shaped graph.
2. y = |x-1| +2 Identify two solutions.
First create a table of values with positive and negative x
values. Then plot the points and draw your V shaped
graph.
Two Solutions are: (-3, 6) and (0,3). These are points
on the graph. You can choose any two points on the
graph: (Hint: any point in the table of values is a
solution.)
Copyright © 2009-2012 Karin Hutchinson – Algebra-class.com
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Unit 2: Unit 2: Unit 2: Unit 2: Graphing EquationsGraphing EquationsGraphing EquationsGraphing Equations
3. y = |3x|+1
First create a table of values with positive and
negative x values. Then plot the points and draw
your V shaped graph.
4. y = |2x+2|
First create a table of values with positive and
negative x values. Then plot the points and draw
your V shaped graph.
Copyright © 2009-2012 Karin Hutchinson – Algebra-class.com
Algebra 1Algebra 1Algebra 1Algebra 1
Unit 2: Unit 2: Unit 2: Unit 2: Graphing EquationsGraphing EquationsGraphing EquationsGraphing Equations
5. y= |x-1| -3
First create a table of values with positive and
negative x values. Then plot the points and draw
your V shaped graph.
6. y = - |x| +4
First create a table of values with positive and
negative x values. Then plot the points and draw
your upside down V shaped graph. (This graph is
upside down since we have a negative absolute
value.
Copyright © 2009-2012 Karin Hutchinson – Algebra-class.com
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Unit 2: Unit 2: Unit 2: Unit 2: Graphing EquationsGraphing EquationsGraphing EquationsGraphing Equations
7. y = -|x-1| Identify two solutions.
First create a table of values with positive and
negative x values. Then plot the points and draw
your upside down V shaped graph.
Two solutions are: (-3,-4) and (-1,-2) (Any
point in the table of values is a solution).
8. y = |1/2x| -3
First create a table of values with positive and
negative x values. Then plot the points and draw
your V shaped graph.
Copyright © 2009-2012 Karin Hutchinson – Algebra-class.com
Algebra 1Algebra 1Algebra 1Algebra 1
Unit 2: Unit 2: Unit 2: Unit 2: Graphing EquationsGraphing EquationsGraphing EquationsGraphing Equations
1. Graph the following inequality on
the grid. (3 points)
Y= |2x-4| -2
2. Identify two solutions. (2 points)
First create a table of values with positive and negative
x values. Then plot the points and draw your V shaped
graph.
Two solutions: (0,2) & (3,0) (Any point in the table of
values is a solution)
Graphs and table of values were created using a free calculator on rentcalculators.org
Unit 2: Graphing Equations
Copyright © 2011 Karin Hutchinson – Algebra-class.com
Graphing Equations Chapter Test Review
Part 1: Calculate the slope of the following lines: (Lesson 3)
2. Find the slope of a line that has a 3. Find the slope of the line that
y-intercept of -4 and contains the point (3, -2) contains the following points: (-5, 4) (2, -4)
Unit 2: Graphing Equations
Copyright © 2011 Karin Hutchinson – Algebra-class.com
Part 2: Graphing Slope and Slope Intercept Form (Lessons 4 and 5)
4. Write the equation for the line represented on the graph:
5. Graph the following equations on the grid:
a. y = -3x+ 2 b. y = 1/3 x – 8
Unit 2: Graphing Equations
Copyright © 2011 Karin Hutchinson – Algebra-class.com
Part 3: Finding Slope Given Two Points and Rate of Change (Lessons 6 & 7)
6. Find the slope of the line that contains the following points: (-6, 9)(-4, -4)
7. A line has a y-intercept of 8 and passes through the point (-2, -9). Without graphing the line,
determine the slope of the line.
8. In Jessica’s first year in her career, she made $39000. In her sixth year, she makes $52500.
What is Jessica’s average rate of change over her six years in her career.
9. Jesse is investing money in the stock market. His initial investment when he opened the account
was $50. By the end of January he had gained $250. By the end of February he had lost $150. At
the end of March he remained steady with no gains or losses. By the end of April he gained $440.
a. Create a graph to show the total amounts in Jesse’s account each month.
Unit 2: Graphing Equations
Copyright © 2011 Karin Hutchinson – Algebra-class.com
b. What was Jesse’s total amount at the end of April?
c. What was the rate of change between his initial deposit and his amount at the end of April?
d. What was the rate of change between February and march? Explain how you determined
your answer.
Part 4: Graphing Standard Form Equations (Lessons 8 & 9)
10. Identify the x and y intercepts for the following equation: 4x – 3y = 24
11. Write an equation that is equivalent to each of the following equations:
a. 6x + 2y = -14 2. 3x – 2y = -12
12. Graph the following equations:
a. 2x + 4y = -16 b. -6x – 4y = -24
Unit 2: Graphing Equations
Copyright © 2011 Karin Hutchinson – Algebra-class.com
13. Mel is ordering t-shirts for the lacrosse team. Short sleeve t-shirts cost $10 and long sleeve t-
shirts cost $17. The total bill came to $293.00. The equation that represents x short sleeve t-shirts
and y long sleeve t-shirts is:
10x+17y = 293
• Graph the equation on the grid. Let x = the number of short sleeve t-shirts
y = the number of long sleeve t-shirts
• If 14 people ordered short sleeve t-shirts, how many long sleeve t-shirts were ordered? Justify
your answer mathematically.
Part 5: Graphing Absolute Value Equations
1. y= |x+2 |
2. y= - |x+2| -1 Identify two solutions.
Unit 2: Graphing Equations
Copyright © 2011 Karin Hutchinson – Algebra-class.com
Graphing Equations Chapter Test Review – Answer Key
Part 1: Calculate the slope of the following lines: (Lesson 3)
2. Find the slope of a line that has a 3. Find the slope of the line that
y-intercept of -4 and contains the point (3, -2) contains the following points: (-5, 4) (2, -4)
Slope = rise Slope = 10 = 5 Run 4 2 The slope is 5/2.
Slope = rise Slope = -6 = - 3 Run 2 The slope is -3.
Slope = rise Slope = 2 Run 3 The slope of this line is 2/3.
Slope = rise Slope = 8 Run -7 The slope of this line is -8/7.
Unit 2: Graphing Equations
Copyright © 2011 Karin Hutchinson – Algebra-class.com
Part 2: Graphing Slope and Slope Intercept Form (Lessons 4 and 5)
4. Write the equation for the line represented on the graph:
5. Graph the following equations on the grid:
a. y = -3x+ 2 b. y = 1/3 x – 8
To write the equation in slope
intercept form, we need to know the
slope and y-intercept. Y = mx +b
Slope (m) = -1 Y-int (b) = -2
Y = -x - 2
y-int
To write the equation in slope
intercept form, we need to know the
slope and y-intercept. Y = mx +b
Slope (m) = 1/2 Y-int (b) = 4
Y = 1/2x + 4
We can start by plotting the y-intercept
which is y = 2. From that point, graph
the slope. Slope is a rise of -3 and run
of 1.
We can start by plotting the y-intercept
which is y = -8. From that point, graph
the slope. Slope is a rise of 1 and run
of 3.
Unit 2: Graphing Equations
Copyright © 2011 Karin Hutchinson – Algebra-class.com
Part 3: Finding Slope Given Two Points and Rate of Change (Lessons 6 & 7)
6. Find the slope of the line that contains the following points: (-6, 9)(-4, -4)
7. A line has a y-intercept of 8 and passes through the point (-2, -9). Without graphing the line,
determine the slope of the line.
8. In Jessica’s first year in her career, she made $39000. In her sixth year, she makes $52500.
What is Jessica’s average rate of change over her six years in her career.
Use the slope formula to calculate the slope when given two points.
y2 – y1 -4 – 9 = -13 or -6.5
x2 - x1 -4 –(-6) 2
The slope of the line is: -13/2 or -6.5
In order to determine the slope of the line, we must have two points. We know one point is (-2,-9)
and we know the y-intercept is 8. The y-intercept can also be written as (0,8) because the y-
intercept is the point on the y-axis, so we know the x-coordinate is 0. Now we have two points.
(0,8) (-2,-9) Now we can use the slope formula to find the slope.
y2 – y1 = -9-8 = -17 or 8.5
x2 - x1 -2 – 0 -2
The slope of the line is 17/2 or 8.5
If we write two ordered pairs for this problem, then we can determine the rate of change, which is
also the slope. Therefore, we can use the slope formula.
Let x = the number of years & y = her salary
First year - $39000 can be written as: (1, 39000)
Sixth year - $52500 can be written as: (6, 52500)
Now we can use the slope formula to determine the rate of change.
y2 – y1 = 52500 – 39000 = 13500 = 2700
x2 - x1 6-1 5
Jessica’s rate of change is $2700 per year.
Unit 2: Graphing Equations
Copyright © 2011 Karin Hutchinson – Algebra-class.com
9. Jesse is investing money in the stock market. His initial investment when he opened the account
was $50. By the end of January he had gained $250. By the end of February he had lost $150. At
the end of March he remained steady with no gains or losses. By the end of April he gained $440.
a. Create a graph to show the total amounts in Jesse’s account each month.
b. What was Jesse’s total amount at the end of April?
c. What was the rate of change between his initial deposit and his amount at the end of April?
d. What was the rate of change between February and March? Explain how you determined your
answer.
At the end of April Jesse’s total was $590. (50+250-150+0+440 = 590)
Let’s write two ordered pairs for this information.
Initial Deposit(month 0) - $50 is (0, 50) April (4th month) – 590 is (4, 590)
Let x = the month Let y = the total amount)
Now we can use our slope formula to find the rate of change.
y2 – y1 = 590 – 50 = 540 = 135
x2 - x1 4-0 4
The rate of change between the initial deposit and the end of April is $135.
The rate of change between February and March is 0. The problem states that at the end of March is
account remained steady with no gains or losses. The ordered pairs would be: (2, 150) (3, 150)
Unit 2: Graphing Equations
Copyright © 2011 Karin Hutchinson – Algebra-class.com
Part 4: Graphing Standard Form Equations (Lessons 8 & 9)
10. Identify the x and y intercepts for the following equation: 4x – 3y = 24
11. Write an equation that is equivalent to each of the following equations:
a. 6x + 2y = -14 2. 3x – 2y = -12
To identify the x intercept, we will let y = 0
4x – 3(0) = 24 Substitute 0 for y
4x = 24 Simplify
4x/4 = 24/4 Divide by 4
X = 6
The x-intercept is 6.
To identify the y-intercept, we will let x = 0
4(0) – 3y = 24 Substitute 0 for x
-3y = 24 Simplify
-3y/-3 = 24/-3 Divide by -3
Y = -8
The y-intercept is -8
I need to write an equation in slope intercept
form in order for the equation to be
equivalent. Therefore, we will solve for y.
6x – 6x + 2y = -14 – 6x Subtract 6x
2y = -6x – 14 Simplify
2y/2 = -6x/2- 14/2 Divide by 2
Y = -3x – 7
Y = -3x – 7 is an equivalent equation.
I need to write an equation in slope intercept form in
order for the equation to be equivalent. Therefore, we
will solve for y.
3x – 3x – 2y = - 12 – 3x Subtract 3x
-2y = -3x – 12 Simplify
-2y/-2 = -3x/-2 – 12/-2 Divide by -2
Y = 3/2x + 6
Y = 3/2x + 6 is an equivalent equation.
Unit 2: Graphing Equations
Copyright © 2011 Karin Hutchinson – Algebra-class.com
12. Graph the following equations:
a. 2x + 4y = -16 b. -6x – 4y = -24
Since this equation is written in standard
form, you can find the x and y intercepts.
(You may also rewrite it in slope intercept
form). We will find the intercepts.
X-intercept (Let y = 0) Y-int (Let x=0)
2x +4(0) = -16 2(0) +4y = -16
2x = -16 4y = -16
2x/2 = -16/2 4y/4=-16/4
X = -8 Y = -4
**If you rewrote the equation in slope
intercept form, it would be:
Y = -1/2x 4
Since this equation is written in standard form,
you can find the x and y intercepts. (You may
also rewrite it in slope intercept form). We will
find the intercepts.
X-intercept (Let y = 0) Y-int (Let x=0)
-6x – 4(0) = -24 -6(0) – 4y = -24
-6x = -24 -4y = -24
-6x/-6 = -24/-6 -4y/-4 = -24/-4
X = 4 y = 6
**If you rewrote the equation in slope intercept
form, it would be:
Y = -3/2 + 6
Unit 2: Graphing Equations
Copyright © 2011 Karin Hutchinson – Algebra-class.com
13. Mel is ordering t-shirts for the lacrosse team. Short sleeve t-shirts cost $10 and long sleeve t-
shirts cost $17. The total bill came to $293.00. The equation that represents x short sleeve t-shirts
and y long sleeve t-shirts is:
10x+17y = 293
• Graph the equation on the grid. Let x = the number of short sleeve t-shirts
y = the number of long sleeve t-shirts
• If 14 people ordered short sleeve t-shirts, how many long sleeve t-shirts were ordered? Justify
your answer mathematically.
Since this equation is written in standard form, I can find the x and y intercepts in order to graph. Since
we are dealing with larger numbers, this is definitely the easiest way to graph this equation.
10x + 17y = 293
X int: (Let y = 0) Y int: (Let x = 0)
10x + 17(0) = 293 10(0) + 17y = 293
10x = 293 17y = 293
10x/10 = 293/10 17y/17 = 293/17
X = 29.3 y = 17.2
According to the graph, if 14 people ordered short sleeve t-shirts, then 9 long sleeve t-shirts were ordered.
Justify: 10(14) + 17(9) = 293
140 + 153 = 293
293= 293
Unit 2: Graphing Equations
Copyright © 2011 Karin Hutchinson – Algebra-class.com
Part 5: Graphing Absolute Value Equations and Inequalities
**Graphs and table of values were created using a free calculator on rentcalculators.org**
1. y= |x+2 |
2. y = -|x+2| -1 Identify two solutions.
Two solutions are: (-5, -4) and (-4, -3)
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Graphing Equations Chapter Test
1. Which line on the graph has a slope of 2/3?
2. Which equation is represented on the graph?
A. Line A
B. Line B
C. Line C
D. Line D
A. y = 4x – 6
B. y = -4x – 6
C. y = 4x +6
D. y = 6x +4
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
3. Find the slope of the line that has a y-intercept of 3 and contains the point (6,-3).
Patti is tracking the price of gas over a 10 week period. See the graph below to answer questions 4
– 6.
4. What is the y-intercept in this problem? What does it mean in the context of this problem?
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
5. What is the rate of change between week 3 and week 6?
6. What is the rate of change between week 0 and Week 3? Explain what the rate of change means
in the context of this problem.
7. What is the x intercept for the equation: 3x + 5y = 27?
8. Judy bought 10 shares of a stock on February 1, 2008 for $350. On October 1, 2008 she sold the
stock for $210. Find Judy’s average monthly rate of change on the stock.
9. Graph the following equation on the grid. Identify the slope and the y intercept.
2x +5y = -15
Slope = ___________
Y intercept = ___________
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
10. Which equation is equivalent to: 4x +2y = -12
A. y = -2x – 6 C. y = -4x - 12
B. y = 2x – 6 D. y = -2x + 6
11. Identify the graph that represents the function: y = |x+3|
A. B. C. D.
12. Joe is traveling to his sister’s house for Thanksgiving. He begins his drive at 3am. He drives
210 miles in 3 hours. He then stops for a half hour break and drives another 90 miles in the
next 2 hours. His trip ends at 10 am after driving a total of 377.5 miles.
• Create a graph to illustrate this problem.
• What is Joe’s average rate of change between 3 am and 6 am?
• If Joe’s rate of change for the first three hours had remained constant throughout the entire
trip, what time would he have arrived at his sister’s house?
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
13. A final exam contains multiple choice questions that are worth 2 points and free response
questions that are worth 3 points. The test is worth 60 points. The equation that represents x
multiple choice questions and y free response questions is:
2x +3y = 60.
• Graph the equation on the grid. Let x = the number of multiple choice questions and
y = the number of free response questions.
• If there are 15 multiple choice questions on the test, how many free response questions are
on the test? Justify your answer.
• Identify one other possibility for the number of each question on the test: Justify your
answer.
There could be _______ multiple choice and _______ free response.
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
Graphing Equations Chapter Test – Answer Key
1. Which line on the graph has a slope of 2/3? (1 Point)
2. Which equation is represented on the graph? (1 Point)
A. Line A
B. Line B
C. Line C
D. Line D
A. y = 4x – 6
B. y = -4x – 6
C. y = 4x +6
D. y = 6x +4
A
**Note: You can eliminate C & D
immediately because they have negative
slopes.
Then you only need to count the slope for A
& B.
A
.
**This is another problem where you can
eliminate answers that don’t make sense. You
know the y – intercept is -6, so C & D can be
eliminated!
The line is rising so the slope is positive.
Therefore, B can also be eliminated!
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
3. Find the slope of the line that has a y-intercept of 3 and contains the point (6,-3). (2 points)
Patti is tracking the price of gas over a 10 week period. See the graph below to answer questions 4
– 6.
4. What is the y-intercept in this problem? What does it mean in the context of this problem?
(2 points)
The y-intercept is 2.40. In this problem the y-intercept indicates that when Patti first began tracking
gas prices (week 0), the price was $2.40.
5. What is the rate of change between week 3 and week 6? (2 Points)
Ordered pairs: (3, 3.20) (6, 2.20)
y2 – y1 = 2.20 – 3.20 = -1
x2 – x1 6 – 3 3
Slope = rise from one point to the next
run
Slope = 1 = -1
-1
The rate of change between week 3 and 6 is -1/3. This means
that the price of gas dropped about $0.33 each week.
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
6. What is the rate of change between week 0 and Week 3? Explain what the rate of change means
in the context of this problem. (3 Points)
Ordered pairs: (0, 2.40) (3, 3.20)
y2 – y1 = 3.20 – 2.40= .8 = .27
x2 – x1 3 – 0 3
7. What is the x intercept for the equation: 3x + 5y = 27? (2 points)
X intercept: Let y = 0
3x +5(0) = 27 3x = 27 x = 9 The x intercept is 9.
3x = 27 3 3
8. Judy bought 10 shares of a stock on February 1, 2008 for $350. On October 1, 2008 she sold the
stock for $210. Find Judy’s average monthly rate of change on the stock. (2 points)
Ordered pairs: (2, 350) (10, 210) y2 – y1 = 210 – 350 = -140 = -17.5
x2 – x1 10-2 8
Feb is month 2 October is month 10
9. Graph the following equation on the grid. Identify the slope and the y intercept. (3 points)
2x +5y = -15
10. Which equation is equivalent to: 4x +2y = -12 (1 point)
A. y = -2x – 6 C. y = -4x - 12
B. y = 2x – 6 D. y = -2x + 6
Slope = -2/5
Y intercept = -3
The rate of change between week 0 and 3 is .27. This means
that the price of gas rose about $0.27 each week between
weeks 0 and 3.
Judy’s average monthly rate of change is -17.5.
She lost 17.5 dollars per month while holding
that stock.
I am going to rewrite it in slope
intercept form:
2x -2x +5y = -15 – 2x
5y = -2x – 15
5y = -2x – 15
5 5 5
y = -2/5x -3
4x – 4x +2y = -12 – 4x
2y = -4x – 12
2y = -4x – 12
2 2 2
y = -2x -6
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
11. Identify the graph that represents the function: y = |x+3| (1 point)
A. B. C. D.
You can eliminate answer B because it represents a negative absolute value, since it is upside
down. We also know that the graph shifts up if there is a constant after the absolute value.
Therefore, we’ve narrowed it down to A or D. Since y = 0, which x value would make sense, -3 or
3? -3 + 3 = 0, so the correct answer is A.
You can also make a table of value and see which graph has the same points as the table of
values.
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
12. Joe is traveling to his sister’s house for Thanksgiving. He begins his drive at 3am. He drives
210 miles in 3 hours. He then stops for a half hour break and drives another 90 miles in the
next 2 hours. His trip ends at 10 am after driving a total of 377.5 miles.
(4 points)
• Create a graph to illustrate this problem. (2/4 points)
• What is Joe’s average rate of change between 3 am and 6 am?
• If Joe’s rate of change for the first three hours had remained constant throughout the entire
trip, what time would he have arrived at his sister’s house?
Rate of change = slope Two points: 3am (3, 0) 6am (6, 210)
Slope = y2 – y1 = 210-0 = 210 = 70
X2 – x1 6 – 3 3
The rate of change between 3am and 6 am is 70. Therefore, Joe maintained a constant rate
of 70 miles per hour during this three hour time period.
If Joe had been able to maintain a constant rate of 70 miles per hour throughout his entire trip he would have arrived
at his sister’s house at 8:24 am.
D= rt Distance = 377.5 miles Rate = 70 miles per hour Time = ?
377.5 = 70t 377.5 = 70t
70 70
T = 5.4 (5 hours and 24 minutes) ** .4 = 4/10 = 2/5. 2/5 of 60 minutes = 2/5 •60 = 24
3 am plus 5 hours and 24 minutes = 8:24 am.
Copyright© 2009 Algebra-class.com
Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations
12. A final exam contains multiple choice questions that are worth 2 points and free response
questions that are worth 3 points. The test is worth 60 points. The equation that represents x
multiple choice questions and y free response questions is:
2x +3y = 60. (4 points)
• Graph the equation on the grid. Let x = the number of multiple choice questions and
y = the number of free response questions.
• If there are 15 multiple choice questions on the test, how many free response questions are
on the test? Justify your answer.
•
• Identify one other possibility for the number of each question on the test: Justify your
answer.
There could be _______ multiple choice and _______ free response.
In order to graph the equation, we must find
the x and y intercepts:
2x +3y = 60
X Intercept Y Intercept
2x +3(0) = 60 2(0) +3y = 60
2x = 60 3y = 60
2 2 3 3
x = 30 y = 20
X intercept = 30 Y intercept = 20
Since we know there are 15 multiple choice questions, we can substitute 15 for x and solve for y.
2(15) + 3y = 60
30+3y = 60
30 – 30 +3y = 60 – 30
3y = 30
3y/3 = 30/3
Y= 10
If there are 15 multiple choice questions, then there are 10 free
response questions.
Justify: 2(15) + 3(10) = 60
30+30 = 60
There could be 21 multiple choice and 6 free response.
2(21) + 3(6) = 60
42 + 18 = 60
60 = 60
This test is worth 28 points