Chapter 9 · Functions

Compare Functions in Context and Justify Growth

Compare two linear functions in real situations to see who starts ahead and who gains faster. Justify which one grows faster using rates of change, units and the domain.

16 min read

Two friends save money. Two tanks fill with water. Two hikers climb a trail. In each story, two quantities change, and you can ask two different questions.

Who is ahead? Compare the two quantities at the same moment.
Who is gaining faster? Compare how quickly the two quantities change.

These questions can have different answers. Someone who starts behind can gain faster and finish ahead. In this lesson, you will answer both questions for stories like these and back up each answer with numbers and units.

Each story describes a function: a rule that gives every permitted input exactly one output. The input is the quantity you start from, such as the number of weeks. The output is the quantity that depends on it, such as the money saved. Different inputs may share an output, and the rule is still a function.

The domain is the set of permitted inputs. A story often limits it. For example, a savings plan might run only for the whole weeks $0$ through $8$ . Always state the domain, because your answers hold only for inputs inside it.

Every function in this lesson is linear on its domain: its output changes at a constant rate, so equal changes in the input always give equal changes in the output. You can write it as $y = mx + b$ , where $x$ is the input and $y$ is the output. Two numbers describe it.

The initial value $b$ is the output when the input is $0$ , as long as $0$ is in the domain.
The rate of change $m$ is the change in the output for each $1$ -unit increase in the input.

A proportional relationship is the special case $y = kx$ . Its output is $0$ when its input is $0$ , so its graph passes through the origin $(0, 0)$ . Dividing both sides by $x$ , which is allowed when $x \neq 0$ , shows that $y/x$ equals the same number $k$ for every nonzero input. A function with a nonzero initial value can be linear, but it is not proportional.

You already know how to read the initial value and the rate of change from an equation, a table or a description. Now you will use them to compare two functions in the same story.

Worked example

Two savings plans

Maya and Leo save for $8$ weeks. Let $w$ be the number of weeks, a whole number from $0$ to $8$ , and let $s$ be the savings in dollars. Maya's savings follow $s = 12w + 40$ . Leo starts with $25$ dollars and adds $15$ dollars every week. Who starts with more? Who saves more each week? Who has more after $8$ weeks?

Write Leo's rule from his description. He adds the same 15 dollars every week, so his savings change at a constant rate and his function is linear. The rate of change is 15 dollars per week, and the initial value is the 25 dollars he starts with.
$s = 15w + 25$
Compare the initial values. Week 0 is in the domain, and each rule holds for every week in the domain, so you may substitute 0 for w. Since 40 is greater than 25, Maya starts ahead by 15 dollars.
$\text{Maya: } 12(0) + 40 = 40 \qquad \text{Leo: } 15(0) + 25 = 25$
Compare the rates of change. In each rule, the number that multiplies w is the dollars added per week: 12 for Maya and 15 for Leo. Both rates are in dollars per week, so you can compare them directly. Leo saves 3 more dollars each week, so his savings grow faster.
$15 > 12$
Compare the savings after 8 weeks. Week 8 is in the domain, so substitute 8 for w in each rule. Multiply first, then add. Since 145 is greater than 136, Leo finishes ahead by 9 dollars.
$\text{Maya: } 12(8) + 40 = 96 + 40 = 136 \qquad \text{Leo: } 15(8) + 25 = 120 + 25 = 145$
Check that the answers fit together. Leo begins 15 dollars behind but gains 3 dollars on Maya every week. Over 8 weeks he gains 24 dollars, which covers the 15-dollar gap and leaves him 9 dollars ahead.
$3 \times 8 = 24 \qquad 24 - 15 = 9$

Answer

Maya starts with more: $40$ dollars compared with Leo's $25$ . Leo saves more each week: $15$ dollars compared with Maya's $12$ . After $8$ weeks, Leo has more: $145$ dollars compared with Maya's $136$ .

Savings for every week in the domain. The gap changes by 3 dollars each week.

Week	Maya's savings (dollars)	Leo's savings (dollars)	Who is ahead, and by how many dollars
$0$	$40$	$25$	Maya, by $15$
$1$	$52$	$40$	Maya, by $12$
$2$	$64$	$55$	Maya, by $9$
$3$	$76$	$70$	Maya, by $6$
$4$	$88$	$85$	Maya, by $3$
$5$	$100$	$100$	Tie
$6$	$112$	$115$	Leo, by $3$
$7$	$124$	$130$	Leo, by $6$
$8$	$136$	$145$	Leo, by $9$

Key concept

Linear functions: faster growth means a greater rate of change

Take two linear functions with the same input unit and the same output unit, each with a positive rate of change on a stated domain. The one with the greater rate of change grows faster: for each $1$ -unit increase in the input, its output rises by more than the other's. Which one is ahead is a separate question. You answer it by comparing outputs at the same input.

To find a rate of change from two input-output pairs $(x_1, y_1)$ and $(x_2, y_2)$ of a linear function, divide the change in output by the change in input. The inputs must differ, so you never divide by $0$ . Two different outputs for one input would not be a function: on a graph, those points lie on a vertical line, whose slope (rate of change) is undefined.

\text{rate of change} = \frac{y_2 - y_1}{x_2 - x_1}, \quad x_2 \neq x_1

To justify a claim is to give the evidence and the reasoning that make it true. A complete justification that one function grows faster has four parts.

Say what the input and the output measure, with their units, and state the domain.
Give each rate of change with its unit, and say where it comes from: the number that multiplies the input in an equation, a phrase such as "each minute", or a calculation from two table rows.
Compare the two rates.
Say where your claim holds.

The unit of a rate of change is an output unit per input unit, such as liters per minute.

Take care with tables. A table shows only the input-output pairs listed in it, so by itself it supports claims only about those values. To say anything about other inputs in the domain, you need a rule, such as a statement that the amount changes at a constant rate.

Tank B: liters of water at four times

Time (minutes)	Water in Tank B (liters)
$2$	$42$
$5$	$60$
$8$	$78$
$12$	$102$

Worked example

Justify which tank fills faster

Two tanks fill at constant rates for $12$ minutes. Let $t$ be the time in minutes, with $0 \le t \le 12$ , and let $V$ be the amount of water in a tank, in liters. Tank A follows $V = 9t$ . The table above shows readings for Tank B. In which tank does the amount of water grow faster? Justify your answer.

Set up the comparison. Both functions have time in minutes as the input and water in liters as the output, on the same domain from 0 to 12 minutes. The problem says both tanks fill at constant rates, so both functions are linear, and Tank B's rate applies to every time in the domain, not only to the four times in the table.
Read Tank A's rate from its equation. The number that multiplies t is 9, so Tank A gains 9 liters each minute. Tank A is proportional: it holds 0 liters at time 0, and dividing both sides of the rule by t, which is allowed when t is not 0, always gives 9.
$\frac{V}{t} = \frac{9t}{t} = 9 \quad (t \neq 0)$
Find Tank B's rate from two rows of the table: 2 minutes and 5 minutes. The inputs differ, so the change in input is 3, not 0. Divide the change in liters by the change in minutes.
$\frac{60 - 42}{5 - 2} = \frac{18}{3} = 6$
Check the other rows. With a constant rate, every pair of rows must give the same result, and they do. Tank B gains 6 liters each minute.
$\frac{78 - 60}{8 - 5} = \frac{18}{3} = 6 \qquad \frac{102 - 78}{12 - 8} = \frac{24}{4} = 6$
Compare the rates. Both are positive and both are in liters per minute, so the greater rate means faster growth. Tank A gains 3 more liters than Tank B every minute.
$9 > 6$

Answer

The amount of water grows faster in Tank A. For $0 \le t \le 12$ , Tank A gains $9$ liters each minute, read from $V = 9t$ , and Tank B gains $6$ liters each minute, found from its table. Both rates are in liters per minute, and $9 > 6$ .

A shortcut that fails. Suppose you find Tank B's rate by dividing one output by its input: $42 \div 2 = 21$ liters per minute. Then Tank B seems to grow faster than Tank A, because $21 > 9$ . Its true rate is $6$ .

For a nonzero input, output divided by input gives the rate of change only when the linear function is proportional, $y = kx$ . Tank B is linear but not proportional. Its ratios are not constant: $42 \div 2 = 21$ , $60 \div 5 = 12$ and $102 \div 12 = 8.5$ .

Tank B did not start empty. Going back $2$ minutes from $42$ liters at $6$ liters per minute gives $42 - 2 \times 6 = 30$ liters at $t = 0$ . Those $30$ liters are in every reading, but they are not growth. Divide the change in output by the change in input instead.

Ben's elevation at four times

Time (minutes)	Ben's elevation (meters)
$5$	$360$
$15$	$440$
$25$	$520$
$40$	$640$

Worked example

Who climbs faster, and who is higher?

Ana and Ben climb a trail for $40$ minutes, each at a constant rate. Let $t$ be the time in minutes, with $0 \le t \le 40$ , and let $E$ be the elevation (height above sea level) in meters. Ana starts at $420$ meters and climbs $5$ meters each minute. The table above shows Ben's elevations. Who gains elevation faster? Who is higher after $30$ minutes, and who is higher after $40$ minutes? Justify each answer.

Find Ben's rate from the rows for 5 and 15 minutes. The inputs differ, so the change in input is 10, not 0. The other rows agree with the constant rate stated in the problem: from 15 to 25 minutes, 80 divided by 10 is 8, and from 25 to 40 minutes, 120 divided by 15 is 8.
$\frac{440 - 360}{15 - 5} = \frac{80}{10} = 8$
Compare the rates. Ana's rate comes from the words: 5 meters each minute. Ben's rate is 8 meters each minute. Both rates are positive and in meters per minute, so the greater rate means faster growth. Ben gains elevation faster throughout the domain from 0 to 40 minutes.
$8 > 5$
Write Ana's rule from her starting elevation and her rate, then find her elevation at 30 and at 40 minutes. Both times are in the domain, so you may substitute them for t.
$E = 5t + 420 \qquad 5(30) + 420 = 150 + 420 = 570 \qquad 5(40) + 420 = 200 + 420 = 620$
Find Ben's elevation at 30 minutes. That time is not in the table, but the constant rate is a rule that covers the whole domain. Start from 520 meters at 25 minutes and add 8 meters for each of the next 5 minutes. At 40 minutes, the table shows 640 meters.
$520 + 8(30 - 25) = 520 + 8(5) = 520 + 40 = 560$
Compare the elevations at each time. At 30 minutes, 570 is greater than 560, so Ana is 10 meters higher. At 40 minutes, 640 is greater than 620, so Ben is 20 meters higher.
$570 > 560 \qquad 640 > 620$
Check that the answers fit together. Ben gains 8 minus 5, or 3, more meters than Ana each minute. From 30 to 40 minutes that is 30 meters, enough to erase Ana's 10-meter lead and put Ben 20 meters ahead.
$3 \times 10 = 30 \qquad 30 - 10 = 20$

Answer

Ben gains elevation faster: $8$ meters each minute compared with Ana's $5$ , for $0 \le t \le 40$ . After $30$ minutes, Ana is higher, at $570$ meters compared with Ben's $560$ . After $40$ minutes, Ben is higher, at $640$ meters compared with Ana's $620$ .

Here is the idea to keep. When a story gives you two linear functions, ask two separate questions. Who is ahead? Compare the outputs at the same input. At input $0$ , when it is in the domain, these are the initial values. Who grows faster? Compare the rates of change, in the same units, on the stated domain. When both rates are positive, the greater rate means faster growth.

For a linear function, read the rate from its equation or from a phrase such as "each minute", or compute change in output divided by change in input from two table rows with different inputs. Dividing one output by its nonzero input gives the rate only for a proportional relationship $y = kx$ . A strong justification names both rates, their units, where they come from, and the domain where your claim holds.

Up next

Evaluate a linear function and complete its table

15 min read

Continue