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Module 5 Appendix
Suggested Answers
Lesson 1
SC 1. This question can be solved in many ways. Sample solutions show the graphing and table of values methods.
Method 1: Graphing
Step 1: Graph the ordered pairs.
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Step 2: Check to see if a single straight line can pass through the points.
In this case, a line will pass through every point. Therefore, the relation is linear.
Method 2: Table of Values
Step 1: Arrange the ordered pairs in ascending order of x-values.
x y
-2 8
2 2
6 -4
8 -7
Step 2: Calculate the changes in each variable.
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Step 3: Evaluate the slope at each level.
− =− =−64
64
32
Since the slope is constant, the relation is linear.
SC 2.
a. Linear. The graph is a straight line.b. Linear. For every 4 units increase in the y-column, there is a corresponding increase of 1 unit in the x-column.c. Plot the points. Check whether a single straight line will pass through the points.
Since a single straight line does not pass through the points, the relation is non-linear.
d. Linear. For every 1 unit increase in the y-column, there is a corresponding increase of 1 unit in the x-column.
x y = x
0 0
1 1
2 2
3 3
4 4
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e. The change in the number of bacteria is not constant as each hour passes. Therefore, the relation is non-linear.
Time(hours)
Number of Bacteria
0 1
1 2
2 4
3 8
Lesson 2
SC 1.a. The x-intercept is (3, 0). The y-intercept is (0, 3).b. The x-intercept is (–1, 0). The y-intercept is (0, 2).c. The x-intercept is (0, 0). The y-intercept is (0, 0).
SC 2.
SC 3.
a. C The slope is the change in the dependent variable compared to the change in the independent variable.
Therefore, the slope is change in costchange in minutes
or cost per minute.
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b. B The y-intercept has coordinates (0, 12), which is the cost for 0 minutes of cellphone use. The context of the question indicates that there is a $12 monthly charge regardless of minutes used. Therefore, the y-intercept represents the recurring monthly charge.
SC 4.
a. The domain and range for the function described is as follows.
Words • The domain includes the integers from 0 to 10.• The range is in multiples or 100 from 2000 to 3000.
Number line
• domain:
• range:
Set Notation • domain: { , }x x x0 10≤ ≤ ∈ I
• range: { , }2000 100 0 10+ ≤ ≤ ∈y y y I
List • The domain is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.• The range is {2000, 2100, 2200, 2300, 2400, 2500, 2600, 2700, 2800, 2900, 3000}.
b. It does not make sense to have negative numbers when dealing with cost and time. In addition, the graph should also be limited to ten days for the total length of the stay in Brazil.
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Lesson 3
SC 2.
a. This shows a graph of a line passing through the points (0, -1), (3, -3), (6, -5), (9, -7).
b.
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c.
SC 3. Your completed table should look like the following.
Graph 1: Unrestricted Graph
Graph 2: Restricted Graph
(based on a specific situation)
If there is a difference between the unrestricted and restricted graph, explain
why.
Independent Variable x month The independent variable in Graph 2 represents something specific—in this case, time in months.
Dependent Variable y cost ($) The dependent variable represents something specific—in this case, cost in dollars.
Domain all real numbers 0 12£ £xGraph 2 only shows from the time of purchase (month 0) to the end the first year (month 12) of the contract.
Range all real numbers 300 900£ £y Graph 2 shows the cost from the date of purchase until the end of the first year of the contract.
Slope 50 50 There is no difference.
y-intercept (0, 300) (0, 300)There is no difference in value; however, in Graph 2, this represents the cost at month 0—the purchase price of the phone.
SC 4.
a. This is a table of values showing expenses and minimum profit for the first $25 spent.
Expenses Sales Profit$5 $25 $20
$10 $50 $40
$15 $75 $60
$20 $100 $80
$25 $125 $100
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b. The independent variable is the amount of money spent on supplies (expenses). The dependent variable is the profit made.
c.
d. The y-intercept is 0.
e. The y-intercept represents fixed costs.
f. You can calculate the slope by using the graph. Use two points that weren't plotted, such as (6, 24) and (21, 84).
my y
x x
m
m
m
=−−
= −−
=
=
2 1
2 1
84 2421 6
60154
g. The slope represents the profit per dollar spent on supplies.
h. The domain is 0 25£ £x . The range is 0 100£ £y .
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Lesson 4
SC 1.
a. y x− =− −( )3 34
8
b. Apply distributive property to the slope and brackets.
y x
y x
y x
y x
− =− −( )
− =− +( )
− =− +
− + =− + +
3 34
8
3 34
34
8
3 34
6
3 3 34
6 33
34
9y x=− +
SC 2.
a. y x
y x
y x
y y x y
x y
= +
( )= +
= +− = + −
= − +
43
5
3 3 43
5
3 4 15
3 3 4 15 3
0 4 3 155
4 3 15 0x y− + =
b. y x
y x
y x
y x
− =− +( )
−( )= − +( )
− =− +( )− =− −
7 12
3
2 7 2 12
3
2 14 3
2 14 3
22 14 3 3 3
2 11 0
y x x x
x y
− + + =− − + ++ − =
SC 3.
a. (0.25, 4) and (1.75, 10)
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b. my y
x x=
−−
= −−
=
=
2 1
2 1
10 41 75 0 256
1 54
. .
.
c. y x− = −( )10 4 1 75.
d. y x
y x
y x
y x
− = −( )− = − ( )
− + = − += +
10 4 1 75
10 4 4 1 75
10 10 4 7 10
4 3
.
.
e. The y-intercept has coordinates (0, 3). This means that the person started timing his walk at a point 3 km away from the school.
f. y x
x y
= +− + =
4 3
4 3 0
Lesson 5
SC 1.
a. The slope of y x=− + −23
23
2 is .
b. From the equation, you can identify the y-intercept as (0, 2).
c. From the information in SC 1.b., the y-intercept is (0, 2). Plot this point.
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From the information in SC 1.a., the slope of the line is - 23
. Recall that slope is riserun
. Therefore, the rise is
-2 and the run is 3. You can apply this slope to the y-intercept by marking a second point that is two units
down and three units right of (0, 2).
You only need two points to draw a linear graph. Once you have two points, you can simply connect the two points with a straight line and extend your graph in either direction.
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d. Go to the multimedia DVD and watch the short video “Graphing a Linear Function,” which shows how you can graph the linear function y x=− +2
32 on a graphing calculator.
SC 2. From the equation y - 4 = 3(x -2), you know that one point on the graph occurs at (2, 4). Recall that (x1, y
1)
represents a point on the graph. Be careful with the signs, as the equation subtracts the points—the subtraction sign means they are positive x and y values. So, start with the point (2, 4) on your graph.
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The slope of the graph is 3 or 31 . Recall that slope is rise
run. Therefore, the rise is 3 and the run is 1. You can apply this
slope to the point (2, 4) by marking a second point that is 3 units above and 1 unit right of (2, 4).
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Now that you have two points on your graph, join them with a line to obtain the graph of y - 4 = 3(x - 2).
SC 3. Recall that in order to graph any line, you need to know either the location of two points on the graph or the location of a single point and the slope of the line.
The following shows two methods, one for each set of information provided.
Method 1: Slope and Point
From the previous discussion, the slope is given by -AB
and the y-intercept is -CB
.
In the equation 3x + 4y -12 = 0, A = 3, B = 4, and C = -12.
Therefore, the slope is − −=AB
34
and the y-intercept is −− −( )
= = =CB
12
4124
3.
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You can also obtain the slope and y-intercept by rearranging the original equation:
3 4 12 0
3 4 12 3 12 0 3 12
4 3 12
44
34
124
34
x y
x y x x
y x
y x
y
+ − =+ − − + = − +
=− +
=−
+
=− xx+3
Plot the y-intercept and apply the slope.
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Now join the points with a line to complete the graph.
Method 2: Two Intercepts
You can also graph a linear function if you know two points on the line. Two points that can easily be found when you are given an equation are the x-intercept and the y-intercept. In Part B of the Math Lab, you were asked to find the intercepts of an equation. Compare your method to the one outlined here.
To determine the x-intercept, remember that anywhere on the x-axis, y has a value of 0. Therefore, you can substitute y = 0 and solve for x to find the x-intercept.
3 4 12 0
3 4 0 12 0
3 12 0
3 12 12 0 12
3 12
33
123
4
x y
x
x
x
x
x
x
+ − =
+ ( )− =
− =− + = +
=
=
=
The x-intercept is 4 and is described by the point (4, 0).
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To determine the y-intercept, remember that anywhere on the y-axis, x has a value of 0. Therefore, you can substitute x = 0 and solve for y to find the y-intercept.
3 4 12 0
3 0 4 12 0
4 12 0
4 12 12 0 12
4 12
44
124
3
x y
y
y
y
y
y
y
+ − =
( )+ − =
− =− + = +
=
=
=
The y-intercept is 3 and is described by the point (0, 3).
Plot the x-intercept and the y-intercept.
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Now you have two points on the graph. Join these two points to graph the equation 3x + 4y - 12 = 0.
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SC 4.
a. The slope of y x=− − −12
12
2 is . From the equation, you can identify the y-intercept as -2.
Choose a point on the line to verify that the graph is correct. For example, choose (-4, 0).
y x=− 12
2
0 12
4 2
0 2 2
0 0
−
=− −( )−
= −=
Since the left side is equal to the right side, the point satisfies the equation. Therefore, the graph is correct.
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b. Theslopeof y x+ = +( )3 623
23
is . Fromtheequationyoucanidentify(-6,-3)asapointonthegraph.
Chooseapointonthelinetoverifythatthegraphiscorrect.Forexample,choose(0,1)
y x+ = +( )
+ = +( )
= ( )
=
3 23
6
1 3 23
0 6
4 23
6
4 4
Sincetheleftsideisequaltotherightside,thepointsatisfiestheequation.Therefore,thegraphiscorrect.
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c. Changex+2y-6=0tothey-interceptform.
x y
y x
y x
y x
+ − ==− +
=− +
=− +
2 6 0
2 6
22 2
62
12
3
Chooseapointonthelinetoverifythatthegraphiscorrect.Forexample,choose(6,0)
x y+ − =
( )+ ( )− =
− ==
2 6 0
6 2 0 6 0
6 6 0
0 0
Sincetheleftsideisequaltotherightside,thepointsatisfiestheequation.Therefore,thegraphiscorrect.
SC 5.
a. Cb. Ac. Fd. Be. Ef. D
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