Introduction

We often use linear functions in real-world situations to describe relationships between quantities.

For example, suppose a writer is tracking their progress while typing a draft. They observe that

  • after 2 minutes, they have typed 300 words, and

  • after 5 minutes, they have typed 750 words.

We can build a linear function to model this situation.

Let x represent the number of minutes, and g(x) represent the number of words typed after x minutes. Then, the linear function modeling this situation is

g(x) = mx + b,

for some slope m and value b, which we need to find.

From the two observations, we know two points on the line:

\begin{align*} (x_1, y_1) &= (2, 300) \\[5pt] (x_2, y_2) &= (5, 750) \end{align*}

Using the given two points, we find the slope m using the slope formula:

\begin{align*} m &= \frac{y_2 - y_1}{x_2 - x_1} \\[5pt] &= \frac{750 - 300}{5 - 2} \\[5pt] &= \frac{450}{3} \\[5pt] &= 150 \end{align*}

This means that the writer is typing at a rate of 150 words per minute. So, the linear function is \begin{align*} g(x) &= mx + b \\[5pt] &= 150x + b. \end{align*}

To find the value of b, we substitute one of the points, say (2, 300), into this equation:

\begin{align*} 300 &= 150 \times 2 + b \\[5pt] 300 &= 300 + b \\[5pt] b &= 0 \end{align*}

This means the starting number of words is 0 - the writer began typing from scratch. Therefore, the linear equation that defines g(x) is

g(x) = 150x.

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Example: Constructing Linear Functions Using Slope-Intercept Form

A linear function g(x) gives the remaining mobile data, in gigabytes, on a monthly plan after x days of continuous usage. There are 48 GB remaining after 2 days, and 30 GB remaining after 5 days. Which equation defines g(x)?

EXPLANATION

The linear function modeling this situation is

g(x) = mx + b.

We know two points on the line:

\begin{align*} (x_1, y_1) &= (2, 48) \\[5pt] (x_2, y_2) &= (5, 30) \end{align*}

Using the given two points, we find the slope m{:}

\begin{align*} m &= \dfrac{y_2 - y_1}{x_2 - x_1} \\[5pt] &= \dfrac{30 - 48}{5 - 2} \\[5pt] &= \dfrac{-18}{3} \\[5pt] &= -6 \end{align*}

So, the linear function is \begin{align*} g(x) &= mx + b \\[5pt] &= -6x + b. \end{align*}

To find the value of b, we substitute one of the points, say (2, 48), into this equation:

\begin{align*} 48 &= -6 \times 2 + b \\[5pt] 48 &= -12 + b \\[5pt] b &= 60 \end{align*}

Therefore, the linear equation that defines g(x) is

g(x) = -6x + 60.

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Practice: Constructing Linear Functions Using Slope-Intercept Form

A linear function $g(x)$ gives the total cost, in dollars, of producing $x$ custom tables. The total cost is $\$520$ to produce $4$ tables, and $\$820$ to produce $10$ tables. Which equation defines $g(x)?$

a
 $g(x) = 80x + 160$
b
 $g(x) = 60x + 200$
c
 $g(x) = 50x - 50$
d
 $g(x) = 50x + 320$
e
 $g(x) = 50x$
Practice: Constructing Linear Functions Using Slope-Intercept Form

A linear function $g(x)$ gives the remaining battery life, in minutes, after $x$ hours of video streaming. The battery lasts $180$ minutes after $2$ hours, and $60$ minutes after $6$ hours. Which equation defines $g(x)?$

a
 $g(x) = -20x + 200$
b
 $g(x) = -30x$
c
 $g(x) = -30x + 240$
d
 $g(x) = -30x + 180$
e
 $g(x) = -40x + 300$
Example: Constructing Functions From Tabular Data
x r(x)
6 22
9 31
12 40

Some values of the linear function r are shown in the table above. Which of the following defines r(x)?

EXPLANATION

Since r is a linear function, we know it must be of the form r(x) = mx + b where

  • m is the slope, and

  • b is the y -intercept.

First, we find the slope using two points. For x = 6 and x = 9, we have
\begin{align} m &= \frac{r(x_2) - r(x_1)}{x_2 - x_1} \\[5pt] &= \frac{31 - 22}{9 - 6} \\[5pt] &= \frac{9}{3} \\[5pt] &= 3. \end{align}

So, the linear function is \begin{align*} r(x) &= mx + b \\[5pt] &= 3x + b. \end{align*}

Next, we find the y -intercept using one point. For x = 6, we have
\begin{align*} r(6) &= 3(6) + b \\[5pt] 22 &= 18 + b \\[5pt] b &= 4. \end{align*}

Therefore, the linear equation that defines r(x) is r(x) = 3x + 4.

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Practice: Constructing Functions From Tabular Data
$x$ $y$
$0$ $3$
$2$ $9$
$4$ $15$

The table shows three values of $x$ and their corresponding values of $y.$ There is a linear relationship between $x$ and $y.$ Which of the following equations represents this relationship?

a
 $y = 3x + 9$
b
 $y = -3x + 3$
c
 $y = -3x - 3$
d
 $y = 3x + 3$
e
 $y = -3x + 9$
Practice: Constructing Functions From Tabular Data
$x$ $y$
$0$ $20$
$1$ $15$
$2$ $10$

The table shows three values of $x$ and their corresponding values of $y.$ There is a linear relationship between $x$ and $y.$ Which of the following equations represents this relationship?

a
 $y = 5x + 15$
b
 $y = -5x + 15$
c
 $y = 5x + 20$
d
 $y = 5x -15$
e
 $y = -5x + 20$
Practice: Constructing Functions From Tabular Data
$x$ $g(x)$
$2$$7$
$4$$13$
$6$$19$

Some values of the linear function $g$ are shown in the table above. Which of the following defines $g?$

a
 $g(x) = 3x - 1$
b
 $g(x) = 4x - 1$
c
 $g(x) = 3x + 1$
d
 $g(x) = 2x + 1$
e
 $g(x) = 3x$
Linear Functions With Alternative Domains

Sometimes, when modeling real-life situations with linear functions, the independent variable does not start at zero.

For example, suppose a puzzle enthusiast is working on a large jigsaw puzzle. Their progress follows this pattern:

  • On the first day, they place 50 pieces.

  • They add 30 more pieces on each of the following days.

With this, we can construct a linear function P for the total number of puzzle pieces placed after d days, where d \geq 1.

First, let's break down the pattern to understand it:

  • On day d=1, they place P=50 pieces.

  • On day d=2, they place P=50+30=80 pieces in total.

  • On day d=3, they place P=50+30+30=110 pieces in total.

  • \cdots

And so on.

Notice that the amount added each day after the first is always 30. So the total number of pieces placed includes

  • the initial 50 pieces placed on the first day, and

  • 30 pieces for each additional day beyond the first. Since the first 50 pieces are placed on day 1, the number of additional pieces placed by day d equals 30(d - 1).

Therefore, the total number of pieces P placed after d days, where d \geq 1, is

\begin{align*} P(d) &= 50 + 30(d - 1) \\[5pt] &= 50 + 30d - 30 \\[5pt] &= 30d + 20. \end{align*}

Let's see another example.

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Example: Constructing Functions With Alternative Domains

A theme park charges \[math]65 per person for the first 30 people in a group and \[/math]50 for each additional person. Which function C gives the total cost, in dollars, for a group of g people, where g \geq 30?

EXPLANATION

The total cost consists of two parts.

  • For the first 30 people, the park charges

65 \times 30 = 1,950\,\,\textrm{dollars}.

  • For the additional persons beyond 30, the park charges

50(g - 30)\,\,\textrm{dollars}.

Therefore, the total cost C, in dollars, for a group of g people, where g\geq 30, is

\begin{align*} C(g) &= 1,950 + 50(g - 30) \\[5pt] &= 1,950 + 50g - 1,500 \\[5pt] &= 50g + 450. \end{align*}

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Practice: Constructing Functions With Alternative Domains

The cost of renting a large party tent for up to $7$ days is $\$200$ for the first day and $\$85$ for each additional day. Which of the following equations gives the cost $C,$ in dollars, of renting the tent for $d$ days, where $d$ is a positive integer and $1 \leq d \leq 7?$

a
 $C(d) = 200d + 85$
b
 $C(d) = 85d + 200$
c
 $C(d) = 85d - 85$
d
 $C(d) = 85d + 115$
e
 $C(d) = 200d + 115$
Practice: Constructing Functions With Alternative Domains

For groups of $55$ or more people, a ski resort charges $\$100$ per person for the first $55$ people and $\$80$ for each additional person. Which function $C$ gives the total charge, in dollars, for a group with $p$ people, where $p \geq 55?$

a
 $C(p) = 100p + 1,100$
b
 $C(p) = 100p + 2,200$
c
 $C(p) = 80p + 1,100$
d
 $C(p) = 80p + 4,400$
e
 $C(p) = 80p + 5,500$
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