Comparing Functions — I can compare two functions represented in different ways and justify which changes faster over an interval. - Grade 9 Math

SECTION 1

Historical Context & Motivation

Throughout history, mathematicians and scientists have needed to compare how different quantities change. Whether predicting the growth of a population, the speed of a falling object, or the interest earned in a bank account, people have always asked the same question: which quantity is changing faster? The idea of comparing rates of change is at the heart of algebra, and it connects directly to how we use functions to model the real world.

~300 BCE • Euclid's Proportional Reasoning

Greek mathematician Euclid formalized the idea of ratios and proportions, giving early tools for comparing how quantities relate to one another.

1637 • Descartes Introduces Coordinate Graphing

René Descartes published the Cartesian coordinate system, allowing functions to be visualized as curves on a graph — making visual comparison of rates possible.

1700s • Euler Formalizes the Function Concept

Leonhard Euler introduced modern function notation f(x), enabling mathematicians to express, manipulate, and compare functions using algebra.

1800s–Today • Functions in Science and Economics

Functions became the universal language for modeling change — from population growth to stock prices — and comparing them became essential for making real-world decisions.

Today, the ability to compare functions is a foundational skill in algebra. In real life, functions rarely come to you in the same format. One might be given as an equation, another as a table of values, and a third as a graph. The central question this lesson addresses is: How do you compare two functions that are represented differently, and how do you determine which one changes faster over a specific interval?

SECTION 2

Core Principles & Definitions

Before you can compare functions, you need to understand the key ideas that make comparison possible. Every function has an output that depends on an input, and the way that output changes is what we measure. The tool we use to measure that change is called the average rate of change. It tells you how much the output of a function changes, on average, per unit change in the input over a given interval.

Function Representations

A function can be expressed as an equation (e.g., f(x) = 3x + 2), a table of input-output pairs, a graph on the coordinate plane, or a verbal description. All four forms carry the same information.

Average Rate of Change

The average rate of change over an interval [a, b] equals the change in output divided by the change in input: (f(b) − f(a)) ÷ (b − a). This is identical to the slope of the line connecting two points on the graph.

Interval Matters

A function may change faster than another over one interval but slower over a different interval. Always specify the interval when comparing rates of change.

Linear vs. Non-Linear

Linear functions have a constant rate of change (the slope never varies). Non-linear functions — such as quadratics or exponentials — have rates of change that vary from interval to interval.

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KEY TAKEAWAY

Think of comparing functions like comparing two road trips. One car might be represented by a speedometer reading (an equation), while the other car's journey is recorded in a logbook of distances at certain times (a table). Even though the information looks different, you can still figure out which car covered more ground per hour over any given stretch. The average rate of change is like the average speed — it lets you compare any two functions on equal footing, no matter how they're presented.

SECTION 3

Visual Explanation — Comparing on a Graph

The most intuitive way to compare functions is to look at their graphs. When two functions are plotted on the same coordinate plane, the one whose graph is steeper over an interval is changing faster. The diagram below shows a linear function f(x) = 2x + 1 and a quadratic function g(x) = x² plotted together. Notice how the quadratic starts out growing more slowly than the linear function but eventually overtakes it.

On the interval [1, 3], the cyan line (f) has a constant slope of 2, while the pink curve (g) has an average rate of change of 4. Over this interval, the quadratic is changing faster, as shown by its steeper dashed secant line.

In the diagram above, the shaded amber region highlights the interval from x = 1 to x = 3. The dashed lines connecting the two endpoints on each function are called secant lines, and the slope of each secant line equals the average rate of change of that function over the interval. Even though f(x) = 2x + 1 starts out ahead (at x = 0, f gives 1 while g gives 0), the quadratic g(x) = x² catches up because its rate of change accelerates. This is the visual key to comparing functions: the steeper the secant line, the faster the function is changing.

SECTION 4

Mathematical Framework — Average Rate of Change

The mathematical tool that lets you compare any two functions — regardless of how they are represented — is the average rate of change formula. This formula works the same whether you pull values from an equation, a table, or a graph.

AVERAGE RATE OF CHANGE

Average Rate of Change = [f(b) − f(a)] ÷ (b − a)

Where a and b are the x-values at the endpoints of the interval, f(a) is the output at x = a, and f(b) is the output at x = b. The result tells you how many units the output changes for each 1-unit increase in x, on average.

If you recognize this formula, it's because it is identical to the slope formula you learned in earlier algebra: (y₂ − y₁) ÷ (x₂ − x₁). The average rate of change between two points is literally the slope of the line through those two points.

SLOPE FORMULA CONNECTION

m = (y₂ − y₁) ÷ (x₂ − x₁) = [f(b) − f(a)] ÷ (b − a)

For a linear function, the average rate of change is the same over every interval (it equals the constant slope m). For non-linear functions, the rate changes depending on which interval you choose.

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Extracting Values from Different Representations

From an equation, plug in x = a and x = b to compute f(a) and f(b). From a table, look up the rows where x = a and x = b. From a graph, read the y-coordinates at those x-values. From a verbal description, translate the words into numbers (e.g., "earns $50 per week" means the rate of change is 50). Once you have two output values and two input values, the formula works the same way every time.

SECTION 5

Detailed Breakdown — Side-by-Side Representations

One of the trickiest parts of comparing functions is that the two functions might be given in completely different formats. Below is a detailed example showing how to extract comparable information from a table and an equation. Consider Function A given as a table, and Function B given as an equation: h(x) = 3x + 2.

x	Function A: p(x)	Function B: h(x) = 3x + 2
0	1	2
1	3	5
2	9	8
3	27	11
4	81	14

Function A (p(x)) is exponential (3ˣ), while Function B (h(x)) is linear. Their values are compared side by side.

This side-by-side view shows how to extract rates from a table (left, violet) versus an equation (right, cyan). On the interval [1, 3], the exponential function A changes at an average rate of 12, while the linear function B changes at a constant rate of 3. Function A is changing faster.

The diagram above illustrates the core strategy. You take each function, regardless of its form, and compute the average rate of change over the same interval. The function with the larger absolute value of rate of change is increasing (or decreasing) faster. Notice that the exponential function p(x) = 3^x has a rate of change that grows as x increases, while the linear function h(x) = 3x + 2 always has the same rate of 3. This is why exponential functions eventually outpace linear functions.

SECTION 6

Worked Example

Let's work through a complete problem where two functions are given in different representations. Function A is described verbally, and Function B is given as a graph.

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Problem Statement

Function A: A plant grows 4 cm in its first week and then doubles its total height every week after that. Function B is shown as a graph passing through the points (1, 6), (2, 10), (3, 14), and (4, 18). Which function is growing faster on the interval from week 2 to week 4?

Comparing a Verbal Function and a Graph

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Step 1 — Translate Function A into values

Function A starts at 4 cm after week 1 and doubles each week. So at week 1: 4 cm. Week 2: 4 × 2 = 8 cm. Week 3: 8 × 2 = 16 cm. Week 4: 16 × 2 = 32 cm. We can write this as A(x) = 4 × 2^(x−1) or simply list the values: A(2) = 8 and A(4) = 32.

A(2) = 8, A(4) = 32
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Step 2 — Read Function B's values from the graph

From the points on the graph, we read B(2) = 10 and B(4) = 18. The graph is a straight line, so Function B is linear.

B(2) = 10, B(4) = 18
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Step 3 — Compute average rate of change for Function A

Using the formula: Rate_A = [A(4) − A(2)] ÷ (4 − 2) = (32 − 8) ÷ 2 = 24 ÷ 2 = 12 cm per week.

Average rate of change for A = 12 cm/week
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Step 4 — Compute average rate of change for Function B

Rate_B = [B(4) − B(2)] ÷ (4 − 2) = (18 − 10) ÷ 2 = 8 ÷ 2 = 4 cm per week.

Average rate of change for B = 4 cm/week
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Step 5 — Compare and justify

Since 12 > 4, Function A is growing faster than Function B on the interval [2, 4]. This makes sense because Function A is exponential (doubling each week), so its rate of change increases over time, while Function B is linear with a constant rate of 4 cm per week.

Function A grows faster on [2, 4] — 12 cm/week vs. 4 cm/week

SECTION 7

Strengths, Limitations, and Representation Trade-Offs

Each way of representing a function has strengths and weaknesses when you are trying to compare rates of change. Understanding these trade-offs helps you choose the best strategy for a given problem.

Representation	Strengths for Comparison	Limitations
Equation	You can compute exact output values for any x. Easy to evaluate the average rate of change formula algebraically.	Requires algebraic skill; harder to see the overall shape or behavior at a glance.
Table	Values are already computed — just read and substitute. Patterns in outputs are easy to spot.	Only gives values at listed x-values. If the interval endpoints aren't in the table, you can't compute the rate directly.
Graph	Visually shows which function is steeper. You can quickly see where one overtakes the other.	Reading exact values can be imprecise, especially between grid lines. Estimation errors can affect calculations.
Verbal Description	Describes real-world context, making the comparison meaningful. Can highlight the type of growth (constant, doubling, etc.).	Must be translated into numbers first. Ambiguous wording can lead to misinterpretation.

Trade-offs of each function representation when comparing rates of change

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KEY TAKEAWAY

Think of each representation as a different "language" that describes the same function. An equation is like giving exact GPS coordinates; a table is like listing your favorite stops along a road; a graph is like a satellite photo of the route; and a verbal description is like spoken directions from a friend. They all describe the same trip, but some are more useful than others depending on what you need. The average rate of change formula is your universal translator — it works regardless of the "language" the function comes in.

SECTION 8

Connection to Advanced Topics

The concept of comparing how fast functions change is a stepping stone to some of the most powerful ideas in mathematics. In this lesson you computed average rates of change over intervals. In more advanced courses, you'll learn to find the instantaneous rate of change — the rate at a single point — which is the central idea of calculus.

Concept in This Lesson	Advanced Version
Average rate of change over an interval [a, b]	Derivative — the instantaneous rate of change at a single point (Calculus)
Comparing slopes of secant lines	Comparing slopes of tangent lines at specific points (Differential Calculus)
"Which function changes faster?" between two x-values	Optimization — finding where a function changes fastest or reaches a maximum (Calculus & beyond)
Linear vs. exponential growth comparison	Asymptotic analysis — classifying growth rates of algorithms (Computer Science)

How today's concepts connect to more advanced mathematics

You don't need to worry about these advanced ideas right now, but know that the skill you're building — comparing functions and interpreting their rates of change — is a foundation you'll use in every math and science course from here on out. The ability to look at a table, an equation, and a graph and make sense of all three is genuinely powerful.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A linear function and an exponential function both pass through the point (0, 5). Without any calculations, explain why there must exist some interval over which the exponential function has a greater average rate of change than the linear function.

Show answer

An exponential function's rate of change increases as x increases, while a linear function's rate of change stays the same forever. Even if the linear function grows faster at first, the exponential function's ever-increasing rate will eventually surpass the constant rate of the linear function. Therefore, for sufficiently large x-values, there must be an interval where the exponential function changes faster. This is a fundamental property: exponential growth always overtakes linear growth eventually.

Question 2: BASIC CALCULATION

Function f is given by f(x) = 5x − 3. Function g is given by the table: g(1) = 4, g(2) = 9, g(3) = 16, g(4) = 25. Find the average rate of change of each function on the interval [1, 4]. Which function changes faster?

Show answer

For f: f(1) = 5(1) − 3 = 2 and f(4) = 5(4) − 3 = 17. Average rate of change = (17 − 2) ÷ (4 − 1) = 15 ÷ 3 = 5. For g: g(1) = 4 and g(4) = 25. Average rate of change = (25 − 4) ÷ (4 − 1) = 21 ÷ 3 = 7. Since 7 > 5, function g changes faster on the interval [1, 4]. Looking at the table values, g appears to follow the pattern g(x) = (x + 1)², a quadratic function, which is why its rate of change exceeds the constant rate of the linear function f over this interval.

Question 3: INTERMEDIATE

Function A is described as: "A car starts 10 miles from home and drives at a constant speed of 60 miles per hour." Function B is given by B(t) = 5t² + 10, where t is in hours and B is in miles. Compare the average rates of change of both functions on the interval [0, 2] and then on [2, 4]. Does the same function change faster on both intervals?

Show answer

Function A translates to A(t) = 60t + 10. On [0, 2]: Rate of A = (60(2) + 10 − (60(0) + 10)) ÷ 2 = (130 − 10) ÷ 2 = 60 mph. Rate of B = (5(4) + 10 − (5(0) + 10)) ÷ 2 = (30 − 10) ÷ 2 = 10 mph. Function A is faster on [0, 2]. On [2, 4]: Rate of A = (250 − 130) ÷ 2 = 60 mph (still constant). Rate of B = (5(16) + 10 − (5(4) + 10)) ÷ 2 = (90 − 30) ÷ 2 = 30 mph. Function A is still faster, but B's rate has tripled from 10 to 30. The quadratic B is accelerating and will eventually overtake A at larger t-values, but it hasn't done so yet in the interval [0, 4].

Question 4: APPLIED

Two social media accounts are growing in followers. Account X's followers are modeled by f(w) = 200 + 50w, where w is the number of weeks. Account Y's weekly follower counts are recorded in a table: w = 0 → 180, w = 2 → 220, w = 4 → 310, w = 6 → 480, w = 8 → 820. Over which 2-week interval does Account Y first grow faster than Account X? Justify your answer with calculations.

Show answer

Account X has a constant rate of 50 followers per week (the slope of the linear equation). Now compute Account Y's rate for each 2-week interval: [0, 2]: (220 − 180) ÷ 2 = 20 per week. [2, 4]: (310 − 220) ÷ 2 = 45 per week. [4, 6]: (480 − 310) ÷ 2 = 85 per week. [6, 8]: (820 − 480) ÷ 2 = 170 per week. Account Y first exceeds X's rate of 50 per week on the interval [4, 6], where Y grows at 85 followers per week compared to X's 50. This pattern suggests Y is growing exponentially, and from week 4 onward, Y is consistently outpacing X.

Question 5: CRITICAL THINKING

Two functions, f and g, are both defined on the interval [0, 6]. You are told that f(0) = g(0) = 0 and f(6) = g(6) = 36. Does this guarantee that f and g have the same average rate of change on the interval [0, 6]? Does it guarantee they change at the same rate on every sub-interval within [0, 6]? Explain, and give an example of two different functions that satisfy these conditions but behave differently within the interval.

Show answer

Yes, both functions must have the same average rate of change on [0, 6] because (36 − 0) ÷ (6 − 0) = 6 for both. However, this does NOT mean they change at the same rate on every sub-interval. For example, let f(x) = 6x (linear) and g(x) = x² (quadratic). Both satisfy f(0) = g(0) = 0 and f(6) = g(6) = 36. But on [0, 3]: f's rate = (18 − 0) ÷ 3 = 6, while g's rate = (9 − 0) ÷ 3 = 3. On [3, 6]: f's rate = (36 − 18) ÷ 3 = 6, while g's rate = (36 − 9) ÷ 3 = 9. The functions have identical overall averages but very different behavior within the interval. This shows that matching endpoints doesn't mean matching behavior — the path matters, not just the destination.

SUMMARY

Lesson Summary

Summary

Functions can be represented as equations, tables, graphs, or verbal descriptions, and you can compare any two functions even when they come in different formats. The key is the average rate of change formula: (f(b) − f(a)) ÷ (b − a). This tells you how quickly a function's output changes per unit of input over any interval [a, b]. The function with the greater average rate of change is increasing faster on that interval.

Remember that the interval matters — a function that changes faster on one interval may be slower on another, especially when comparing linear functions (constant rate) to non-linear functions like quadratics or exponentials (changing rates). To justify which function changes faster, always state the interval, calculate both rates, compare the numerical values, and explain why the result makes sense given the types of functions involved.