Lesson 3: Modeling with Linear Functions

Property

A linear model describes a variable that increases or decreases at a constant rate. It has the form

Examples

April's income is 200 dollars per week plus 9% commission on sales $S$. Her income $I$ is $I = 200 + 0.09S$.

Property

Variables that increase or decrease at a constant rate can be described by linear equations. To model this, treat two related data pairs as points $(x_1, y_1)$ and $(x_2, y_2)$. First, compute the slope (rate of change), then substitute the slope and either point into the point-slope formula to find the governing equation.

Examples

• A taxi ride costs 10 dollars for 2 miles and 16 dollars for 4 miles. Let cost be $C$ and distance be $d$. The points are $(2, 10)$ and $(4, 16)$. The slope (cost per mile) is $m = \frac{16-10}{4-2} = 3$. The equation is $C - 10 = 3(d - 2)$.
• A tree was 8 feet tall in 2015 and 14 feet tall in 2018. Let height be $H$ and the year be $t$ (with $t=0$ in 2015). The points are $(0, 8)$ and $(3, 14)$. The slope is $m = \frac{14-8}{3-0} = 2$ feet per year. The equation is $H = 2t + 8$.
• A phone plan costs 40 dollars for 5 GB of data and 50 dollars for 10 GB. Let cost be $C$ and data be $D$. The points are $(5, 40)$ and $(10, 50)$. The slope is $m = \frac{50-40}{10-5} = 2$ dollars per GB. The equation is $C - 40 = 2(D - 5)$.

Explanation

Real-world scenarios with a steady rate of change can be modeled using a linear equation. This allows you to make predictions by finding the line's equation from just two data points, like cost over time or distance versus speed.