Students can use the **Spectrum Math Grade 8 Answer Key** **Chapter 4 Lesson 4.3 Functions and Linear Relationships**Â as a quick guide to resolve any of their doubts.

## Spectrum Math Grade 8 Chapter 4 Lesson 4.3 Functions and Linear Relationships Answers Key

Data in tables can be used to create equations. If the table of values represents a function, a linear relationship in the form of y = mx + b exists.

Step 1: Find the rate of change by calculating the slope, or rate of change, between the two variables.

\(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

\(\frac{9-1}{99-27}\) = \(\frac{8}{72}\) = \(\frac{1}{9}\)

Step 2: Substitute known values of x and y with the slope into the formula y = mx + b.

9 = (\(\frac{1}{9}\))(99) + b

9 = (\(\frac{1}{9}\))(99) + b

9 – 11 = 11 + b – 11

-2 = b

Step 3: Use the found values in the linear function to complete the table.

y = (\(\frac{1}{9}\))(72) – 2

y = 8 – 2 = 6

y = (\(\frac{1}{9}\))(54) – 2

y = 6 – 2 = 4

**Find the relationship for each function table and then complete the table.**

Question 1.

a.

Function: _____

Answer:

1, 2, y = \(\frac{1}{12}\)x

Explanation:

The table of values represents a function.

A linear relationship is in the form of y = mx + b exists.

y = mx + b

Step 1: Find the rate of change by calculating the slope, or rate of change, between the two variables.

\(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

m = \(\frac{10-7}{120-84}\) = \(\frac{3}{36}\) = \(\frac{1}{12}\)

Step 2: Substitute known values of x and y with the slope into the formula y = mx + b.

y = mx + b

10 = (\(\frac{1}{12}\))(120) + b

10 = (\(\frac{120}{12}\)) + b

10 – 10 =Â b = 0

Step 3: Use the found values in the linear function to complete the table.

y = mx + b.

y = \(\frac{1}{12}\)12 + 0.

for x = 12 :: y = 1

y = \(\frac{1}{12}\)24 + 0.

for x = 24 :: y = 2

So, the relationship of the function table is y = \(\frac{1}{12}\)x

The value of y = 1, 2

b.

Function: _____

Answer:

5, 8, y = x + 3

Explanation:

The table of values represents a function.

A linear relationship is in the form of y = mx + b exists.

y = mx + b

Step 1: Find the rate of change by calculating the slope, or rate of change, between the two variables.

\(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

m = \(\frac{14-7}{11- 4}\) = \(\frac{7}{7}\) = 1

Step 2: Substitute known values of x and y with the slope into the formula y = mx + b.

y = mx + b

7 = 1 (4) + b

7- 4 =Â b

b = 3

Step 3: Use the found values in the linear function to complete the table.

y = mx + b.

y = (1)2 + 3.

for x = 2 :: y = 5

y = (1)5 + 3.

for x = 5 :: y = 8

So, the relationship of function table is y = x + 3

The value of y = 5, 8

Question 2.

a.

Function: _____

Answer:

44, 60, y = 10x + 10

Explanation:

The table of values represents a function.

A linear relationship is in the form of y = mx + b exists.

y = mx + b

Step 1: Find the rate of change by calculating the slope, or rate of change, between the two variables.

\(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

m = \(\frac{36 – 12}{6 – 3}\) = \(\frac{24}{3}\) = 8

Step 2: Substitute known values of x and y with the slope into the formula y = mx + b.

y = mx + b

12 = 8 (3) + b

12 – 24 =Â b

b = -12

Step 3: Use the found values in the linear function to complete the table.

y = mx + b.

y = (8)7Â – 12 = 56 – 12 = 44

for x = 7 :: y = 44

y = (8)9 – 12 = 72 – 12 = 60

for x = 9 :: y = 60

So, the relationship of the function table is y = 10x + 10

The value of y = 44, 60

b.

Function: _____

Answer:

30, 110, y = 10 x + 10

Explanation:

The table of values represents a function.

A linear relationship is in the form of y = mx + b exists.

y = mx + b

Step 1: Find the rate of change by calculating the slope, or rate of change, between the two variables.

\(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

m = \(\frac{60 – 50}{5 – 4}\) = \(\frac{10}{1}\) = 10

Step 2: Substitute known values of x and y with the slope into the formula y = mx + b.

y = mx + b

50 = 10 (4) + b

50 – 40 =Â b

b = 10

Step 3: Use the found values in the linear function to complete the table.

y = mx + b.

y = (10)2Â + 10 = 20 + 10 = 30

for x = 2 :: y = 80

y = (10)10 + 10 = 100 + 10 = 110

for x = 10 :: y = 100

So, the relationship of the function table is y = 10 x + 10

The value of y = 30, 110

Question 3.

a.

Function: _____

Answer:

40, 75, y = 7x + 5

Explanation:

The table of values represents a function.

A linear relationship is in the form of y = mx + b exists.

y = mx + b

Step 1: Find the rate of change by calculating the slope, or rate of change, between the two variables.

\(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

m = \(\frac{26 – 19}{3 – 2}\) = \(\frac{7}{1}\) = 7

Step 2: Substitute known values of x and y with the slope into the formula y = mx + b.

y = mx + b

19 = 7 (2) + b

19 – 14 =Â b

b = 5

Step 3: Use the found values in the linear function to complete the table.

y = mx + b.

y = (7)5Â + 5 = 35 + 5 = 40

for x = 5 :: y = 40

y = (7)10 + 5 = 70 + 5 = 75

for x = 10 :: y = 75

So, the relationship of the function table isÂ y = 7x + 5

The value of y = 40, 75

b.

Function: _____

Answer:

0, 10, y = \(\frac{1}{4}\)x – 2

Explanation:

The table of values represents a function.

A linear relationship is in the form of y = mx + b exists.

y = mx + b

Step 1: Find the rate of change by calculating the slope, or rate of change, between the two variables.

\(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

m = \(\frac{4 – 1}{24 – 12}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)

Step 2: Substitute known values of x and y with the slope into the formula y = mx + b.

y = mx + b

1 = \(\frac{1}{4}\) (12) + b

1 – 3 =Â b

b = -2

Step 3: Use the found values in the linear function to complete the table.

y = mx + b.

y = (\(\frac{1}{4}\))8Â – 2 = 2 – 2 = 0

for x = 8 :: y = 40

y = (\(\frac{1}{4}\))48 – 2 = 12 – 2 = 10

for x = 48 :: y = 10

So, the relationship of the function table is y = \(\frac{1}{4}\)x – 2

The value of y = 0, 10

**Find the relationship for each function table and then complete the table.**

Question 1.

a.

Function: _____

Answer:

32, 67, y = 7x – 3

Explanation:

The table of values represents a function.

A linear relationship is in the form of y = mx + b exists.

y = mx + b

Step 1: Find the rate of change by calculating the slope, or rate of change, between the two variables.

\(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

m = \(\frac{18 – 11}{3 – 2}\) = \(\frac{7}{1}\) = 7

Step 2: Substitute known values of x and y with the slope into the formula y = mx + b.

y = mx + b

11 = 7 (2) + b

11 – 14 =Â b

b = -3

Step 3: Use the found values in the linear function to complete the table.

y = mx + b.

y = (7)5Â – 3 = 35 – 3 = 32

for x = 5 :: y = 32

y = (7)10 – 3 = 70 – 3 = 67

for x = 10 :: y = 67

y = 7x – 3

So, the relationship of the function table is y = 7x – 3

The value of y = 32, 67

b.

Function: _____

Answer:

4, 14, y = \(\frac{1}{4}\)x + 2

Explanation:

The table of values represents a function.

A linear relationship is in the form of y = mx + b exists.

y = mx + b

Step 1: Find the rate of change by calculating the slope, or rate of change, between the two variables.

\(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

m = \(\frac{8 – 5}{24 – 12}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)

Step 2: Substitute known values of x and y with the slope into the formula y = mx + b.

y = mx + b

5 = \(\frac{1}{4}\) (12) + b

5 – 3 =Â b

b = 2

Step 3: Use the found values in the linear function to complete the table.

y = mx + b.

y = (\(\frac{1}{4}\))8Â + 2 = 2 + 2 = 4

for x = 8 :: y = 4

y = (\(\frac{1}{4}\))48 + 2 = 12 + 2 = 14

for x = 48 :: y = 14

So, the relationship of the function is y = \(\frac{1}{4}\)x + 2

The value of y = 4, 14

Question 2.

a.

Function: _____

Answer:

6, 13, y = \(\frac{1}{5}\)x + 3

Explanation:

The table of values represents a function.

A linear relationship is in the form of y = mx + b exists.

y = mx + b

Step 1: Find the rate of change by calculating the slope, or rate of change, between the two variables.

\(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

m = \(\frac{11 – 7}{40 – 20}\) = \(\frac{4}{20}\) = \(\frac{1}{5}\)

Step 2: Substitute known values of x and y with the slope into the formula y = mx + b.

y = mx + b

7 = \(\frac{1}{5}\) (20) + b

7 – 4 =Â b

b = 3

Step 3: Use the found values in the linear function to complete the table.

y = mx + b.

y = (\(\frac{1}{5}\))15Â + 3 = 3 + 3 = 6

for x = 15 :: y = 6

y = (\(\frac{1}{5}\))50 +3 = 10 + 3 = 13

for x = 50 :: y = 13

So, the relationship of function is y = \(\frac{1}{5}\)x + 3

The value of y = 6, 13

b.

Function: _____

Answer:

7, 18, y = 11x – 4

Explanation:

The table of values represents a function.

A linear relationship is in the form of y = mx + b exists.

y = mx + b

Step 1: Find the rate of change by calculating the slope, or rate of change, between the two variables.

\(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

m = \(\frac{106 – 40}{10 – 4}\) = \(\frac{66}{6}\) = 11

Step 2: Substitute known values of x and y with the slope into the formula y = mx + b.

y = mx + b

40 = 11 (4) + b

40 – 44 =Â b

b = -4

Step 3: Use the found values in the linear function to complete the table.

y = mx + b.

y = (11)1Â – 4 = 11 – 4 = 7

for x = 1 :: y = 7

y = (11)2 – 4 = 22 – 4 = 18

for x = 10 :: y = 18

So, the relationship of the function is y = 11x – 4

The value of y = 7, 18

Question 3.

a.

Function: _____

Answer:

22, 42, y = 4x – 6

Explanation:

The table of values represents a function.

A linear relationship is in the form of y = mx + b exists.

y = mx + b

Step 1: Find the rate of change by calculating the slope, or rate of change, between the two variables.

\(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

m = \(\frac{18 – 2}{6 – 2}\) = \(\frac{16}{4}\) = 4

Step 2: Substitute known values of x and y with the slope into the formula y = mx + b.

y = mx + b

2 = 4 (2) + b

2 – 8 =Â b

b = -6

Step 3: Use the found values in the linear function to complete the table.

y = mx + b.

y = (4)7Â – 6 = 28 – 6 = 22

for x = 7 :: y = 22

y = (4)12 – 6 = 48 – 6 = 42

for x = 12 :: y = 42

So, the relationship of the function is y = 4x – 6

The value of y = 22, 42

b.

Function: _____

Answer:

6 , 13, y = \(\frac{1}{7}\)x + 4

Explanation:

The table of values represents a function.

A linear relationship is in the form of y = mx + b exists.

y = mx + b

Step 1: Find the rate of change by calculating the slope, or rate of change, between the two variables.

\(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

m = \(\frac{12 – 8}{56 – 28}\) = \(\frac{4}{28}\) = \(\frac{1}{7}\)

Step 2: Substitute known values of x and y with the slope into the formula y = mx + b.

y = mx + b

8= \(\frac{1}{7}\) (28) + b

8= \(\frac{28}{7}\) + b

8 – 4 =Â b

b = 4

Step 3: Use the found values in the linear function to complete the table.

y = mx + b.

y = (\(\frac{1}{7}\))14Â + 4 = 2 + 4 = 6

for x = 14 :: y = 6

y = (\(\frac{1}{7}\))63 + 4 = 9 + 4 = 13

for x = 63 :: y = 13

So, the relationship of the functionÂ is y = \(\frac{1}{7}\)x + 4

The value of y = 6, 13

Question 4.

a.

Function: _____

Answer:

12, 18, y = x + 6

Explanation:

The table of values represents a function.

A linear relationship is in the form of y = mx + b exists.

y = mx + b

Step 1: Find the rate of change by calculating the slope, or rate of change, between the two variables.

\(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

m = \(\frac{15 – 10}{9 – 4}\) = \(\frac{5}{5}\) = 1

Step 2: Substitute known values of x and y with the slope into the formula y = mx + b.

y = mx + b

10 = 1 (4) + b

10 – 4 =Â b

b = 6

Step 3: Use the found values in the linear function to complete the table.

y = mx + b.

y = (1)6Â + 6 = 6 + 6 = 12

for x = 6 :: y = 12

y = (1)12 + 6 = 12 + 6 = 18

for x = 12 :: y = 18

So, the relationship of the function isÂ y = x + 6

The value of y = 12, 18

b.

Function: _____

Answer:

2, 4, y = \(\frac{1}{3}\)x + 1

Explanation:

The table of values represents a function.

A linear relationship is in the form of y = mx + b exists.

y = mx + b

Step 1: Find the rate of change by calculating the slope, or rate of change, between the two variables.

\(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

m = \(\frac{3 – 1}{6 – 0}\) = \(\frac{2}{6}\) = \(\frac{1}{3}\)

Step 2: Substitute known values of x and y with the slope into the formula y = mx + b.

y = mx + b

3 = \(\frac{1}{3}\) (6) + b

3 – 2 =Â b

b = 1

Step 3: Use the found values in the linear function to complete the table.

y = mx + b.

y = (\(\frac{1}{3}\))3Â + 1 = 1 + 1 = 2

for x = 3 :: y = 2

y = (\(\frac{1}{3}\))9 + 1 = 3 + 1 = 4

for x = 9 :: y = 4

So, The relationship of the function is y = \(\frac{1}{3}\)x + 1

The value of y = 2, 4