Learn on PengienVision, Mathematics, Grade 8Chapter 3: Use Functions to Model Relationships

Lesson 4: Construct Functions to Model Linear Relationships

In this Grade 8 lesson from enVision Mathematics Chapter 3, students learn how to construct linear functions in the form y = mx + b to model real-world relationships. The lesson covers writing equations by identifying slope and y-intercept from graphs, tables, and two given data points, then using those equations to make predictions. It is part of a broader unit on using functions to model relationships and builds students' ability to interpret rate of change and initial value in context.

Section 1

Writing an Equation from a Graph

Property

To write an equation from a graph in slope-intercept form $y = mx + b$ :

Identify the y-intercept $b$ where the line crosses the y-axis
Find the slope $m = \frac{\text{rise}}{\text{run}}$ using two clear points on the line
Substitute $m$ and $b$ into $y = mx + b$

Examples

Section 2

Writing a Function from a Table of Values

Property

To write a linear function in the form $y = mx + b$ from a table of values, first find the slope ( $m$ ) and then determine the y-intercept ( $b$ ).

Find the slope ( $m$ ): Use any two points $(x_1, y_1)$ and $(x_2, y_2)$ from the table.

m = \frac{{\text{change in } y}}{{\text{change in } x}} = \frac{{y_2 - y_1}}{{x_2 - x_1}}

Find the y-intercept ( $b$ ): Identify the value of $y$ when $x=0$ . If $x=0$ is not in the table, use the slope $m$ and any point $(x, y)$ from the table to solve for $b$ in the equation $y = mx + b$ .

Section 3

Writing an Equation for a Linear Function

Property

To write an equation for a linear function, you need to find the slope $(m)$ and the $y$ -intercept $(b)$ .

Identify two points on the line.
Use the two points to calculate the slope: $m = \frac{y_2 - y_1}{x_2 - x_1}$ .
Use the slope and one point $(x_1, y_1)$ in the point-slope form, $y - y_1 = m(x - x_1)$ , and solve for $y$ . Alternatively, substitute $m$ and a point into $y = mx+b$ and solve for $b$ .

Examples

A line has a slope of 4 and passes through $(2, 5)$ . Using $y = mx + b$ , we get $5 = 4(2) + b$ , so $5 = 8 + b$ , and $b = -3$ . The equation is $y = 4x - 3$ .
A line passes through $(1, 2)$ and $(4, 11)$ . The slope is $m = \frac{11-2}{4-1} = \frac{9}{3} = 3$ . Using the point $(1,2)$ , we have $y-2 = 3(x-1)$ , which simplifies to $y = 3x - 1$ .
A gym charges a 50 dollars sign-up fee and 25 dollars per month. The cost function is $C(x) = 25x + 50$ , where $x$ is the number of months.

Explanation

To define a specific line, you need to know its direction (slope) and one point it passes through. Once you have these two pieces of information, you can create a unique formula for that line.

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Writing an Equation from a Graph

Property

To write an equation from a graph in slope-intercept form $y = mx + b$ :

Identify the y-intercept $b$ where the line crosses the y-axis
Find the slope $m = \frac{\text{rise}}{\text{run}}$ using two clear points on the line
Substitute $m$ and $b$ into $y = mx + b$

Examples

Section 2

Writing a Function from a Table of Values

Property

To write a linear function in the form $y = mx + b$ from a table of values, first find the slope ( $m$ ) and then determine the y-intercept ( $b$ ).

Find the slope ( $m$ ): Use any two points $(x_1, y_1)$ and $(x_2, y_2)$ from the table.

m = \frac{{\text{change in } y}}{{\text{change in } x}} = \frac{{y_2 - y_1}}{{x_2 - x_1}}

Find the y-intercept ( $b$ ): Identify the value of $y$ when $x=0$ . If $x=0$ is not in the table, use the slope $m$ and any point $(x, y)$ from the table to solve for $b$ in the equation $y = mx + b$ .

Section 3

Writing an Equation for a Linear Function

Property

To write an equation for a linear function, you need to find the slope $(m)$ and the $y$ -intercept $(b)$ .

Identify two points on the line.
Use the two points to calculate the slope: $m = \frac{y_2 - y_1}{x_2 - x_1}$ .
Use the slope and one point $(x_1, y_1)$ in the point-slope form, $y - y_1 = m(x - x_1)$ , and solve for $y$ . Alternatively, substitute $m$ and a point into $y = mx+b$ and solve for $b$ .

Examples

A line has a slope of 4 and passes through $(2, 5)$ . Using $y = mx + b$ , we get $5 = 4(2) + b$ , so $5 = 8 + b$ , and $b = -3$ . The equation is $y = 4x - 3$ .
A line passes through $(1, 2)$ and $(4, 11)$ . The slope is $m = \frac{11-2}{4-1} = \frac{9}{3} = 3$ . Using the point $(1,2)$ , we have $y-2 = 3(x-1)$ , which simplifies to $y = 3x - 1$ .
A gym charges a 50 dollars sign-up fee and 25 dollars per month. The cost function is $C(x) = 25x + 50$ , where $x$ is the number of months.

Explanation

To define a specific line, you need to know its direction (slope) and one point it passes through. Once you have these two pieces of information, you can create a unique formula for that line.

Overview

Lesson overview

Review the lesson summary and core properties without leaving the current page.

Overview

Section 1

Writing an Equation from a Graph

Property

To write an equation from a graph in slope-intercept form $y = mx + b$ :

Identify the y-intercept $b$ where the line crosses the y-axis
Find the slope $m = \frac{\text{rise}}{\text{run}}$ using two clear points on the line
Substitute $m$ and $b$ into $y = mx + b$

Examples

Section 2

Writing a Function from a Table of Values

Property

To write a linear function in the form $y = mx + b$ from a table of values, first find the slope ( $m$ ) and then determine the y-intercept ( $b$ ).

Find the slope ( $m$ ): Use any two points $(x_1, y_1)$ and $(x_2, y_2)$ from the table.

m = \frac{{\text{change in } y}}{{\text{change in } x}} = \frac{{y_2 - y_1}}{{x_2 - x_1}}

Find the y-intercept ( $b$ ): Identify the value of $y$ when $x=0$ . If $x=0$ is not in the table, use the slope $m$ and any point $(x, y)$ from the table to solve for $b$ in the equation $y = mx + b$ .

Section 3

Writing an Equation for a Linear Function

Property

To write an equation for a linear function, you need to find the slope $(m)$ and the $y$ -intercept $(b)$ .

Identify two points on the line.
Use the two points to calculate the slope: $m = \frac{y_2 - y_1}{x_2 - x_1}$ .
Use the slope and one point $(x_1, y_1)$ in the point-slope form, $y - y_1 = m(x - x_1)$ , and solve for $y$ . Alternatively, substitute $m$ and a point into $y = mx+b$ and solve for $b$ .

Examples

A line has a slope of 4 and passes through $(2, 5)$ . Using $y = mx + b$ , we get $5 = 4(2) + b$ , so $5 = 8 + b$ , and $b = -3$ . The equation is $y = 4x - 3$ .
A line passes through $(1, 2)$ and $(4, 11)$ . The slope is $m = \frac{11-2}{4-1} = \frac{9}{3} = 3$ . Using the point $(1,2)$ , we have $y-2 = 3(x-1)$ , which simplifies to $y = 3x - 1$ .
A gym charges a 50 dollars sign-up fee and 25 dollars per month. The cost function is $C(x) = 25x + 50$ , where $x$ is the number of months.

Explanation

To define a specific line, you need to know its direction (slope) and one point it passes through. Once you have these two pieces of information, you can create a unique formula for that line.