Change in Linear and Exponential Functions
Why this matters
Imagine you and your friend Priya are offered two different summer job plans for a new cafe in Boston.
- Plan AYou get a $50 signing bonus and earn exactly $15 for every hour you work.
- Plan BPriya gets a $1 signing bonus but earns $2 for her first hour, $4 for her second, $8 for her third, and so on, with her hourly wage doubling each hour.
At first, your plan seems way better. After one hour, you've made $65 while Priya has only made $3. But who is better off after a full 8-hour shift? Whose earnings grow steadily, and whose earnings explode?
This is the core of what we're exploring today: the fundamental difference between steady, additive change and rapid, multiplicative change. We'll connect these ideas to the two most important function types you'll see: linear and exponential.
Concept overview
flowchart TD
A[Start: Given a data pattern] --> B{Is the change between outputs constant?};
B -->|Additive change<br>(y2 - y1 = constant)| C[Linear Function];
B -->|Multiplicative change<br>(y2 / y1 = constant)| D[Exponential Function];
C --> E[Form: f(x) = mx + b];
C --> F[Analogy: Arithmetic Sequence];
C --> G[Graph: Straight Line];
D --> H[Form: f(x) = ab^x];
D --> I[Analogy: Geometric Sequence];
D --> J[Graph: J-Curve];
Core explanation
At its heart, the difference between linear and exponential functions is about addition versus multiplication. That’s the big idea. Let's break down how this simple difference creates two completely distinct worlds of growth.
Linear Functions: The Steady Climb
Think about Plan A from our cafe example: a $50 bonus plus $15 per hour.
- After 0 hours (you just signed up): $50
- After 1 hour: $50 + $15 = $65
- After 2 hours: $65 + $15 = $80
- After 3 hours: $80 + $15 = $95
Notice the pattern? For every equal step in time (one hour), your total pay goes up by an equal amount ($15). This is constant additive change.
The function for your pay, let's call it L(x), would be:
L(x) = 15x + 50
This is the classic linear form f(x) = mx + b.
bis the initial value or y-intercept. It's your starting point. Here, it's the $50 bonus.mis the slope or rate of change. It's the constant amount you add for each step. Here, it's the $15 you earn per hour.
Connection to Arithmetic Sequences
This is almost identical to an arithmetic sequence, where you start with a term and add a common difference (d) repeatedly.
a_n = a_0 + dn
See the parallel? The linear function f(x) = b + mx is the continuous version of the discrete arithmetic sequence. A sequence gives you values at integer steps (1st hour, 2nd hour), while the function can tell you your pay at any time, like after 2.5 hours. This is a key domain difference: sequences are for discrete steps, functions can be for continuous intervals.
What if you don't know the starting value?
Sometimes, you're given a rate of change and a point that isn't the start. For example: "After 4 hours, Carlos had earned $110." We can use the point-slope form:
f(x) = y_1 + m(x - x_1)
Here, (x_1, y_1) is your known point (4, 110) and m is the slope (15).
L(x) = 110 + 15(x - 4)
If you distribute and simplify, you'll get the exact same 15x + 50 function. It's just a different way to build the same line.
Exponential Functions: The Rocket Launch
Now let's look at Priya's plan: a $1 bonus, with her earnings doubling each hour.
- After 1 hour: $1 (bonus) + $2 = $3
- After 2 hours: $1 (bonus) + $2 + $4 = $7
- After 3 hours: $1 (bonus) + $2 + $4 + $8 = $15
Let's look at the rate itself, not the total. The amount she earns in that hour is what's changing. Let's model the function for her total earnings differently, focusing on the multiplicative pattern. A better way to model this is to think of an initial value that gets repeatedly multiplied.
Let's reframe Priya's deal to fit our standard exponential model f(x) = ab^x. Let's say she starts with an initial value of $3 and it doubles every hour.
- After 0 hours: $3
- After 1 hour: $3 * 2 = $6
- After 2 hours: $6 * 2 = $12
- After 3 hours: $12 * 2 = $24
Here, for every equal step in time (one hour), the output value is multiplied by an equal amount (2). This is constant multiplicative change.
The function for this pattern, let's call it E(x), would be:
E(x) = 3 * 2^x
This is the standard exponential form f(x) = ab^x.
ais the initial value. It's the value whenx=0. Here, it's $3.bis the base or common ratio. It's the constant you multiply by for each step. Here, it's 2 (for doubling).
Connection to Geometric Sequences
This perfectly mirrors a geometric sequence, where you start with a term and multiply by a common ratio (r) repeatedly.
g_n = g_0 * r^n
Just like before, the exponential function f(x) = ab^x is the continuous cousin of the discrete geometric sequence. The function can tell you the value at x = 1.8, while the sequence is locked into integer steps.
What if you don't know the starting value?
Similarly, we can build an exponential function from any point, not just the y-intercept. For example: "After 2 hours, the value was $12." We can use a point-ratio form:
f(x) = y_1 * b^(x - x_1)
Here, (x_1, y_1) is our known point (2, 12) and b is the common ratio (2).
E(x) = 12 * 2^(x - 2)
With exponent rules, this simplifies to 12 * 2^x * 2^(-2), which is 12 * 2^x * (1/4), giving you the exact same 3 * 2^x function.
The Big Takeaway
The most important skill is to look at a table of data and determine the pattern.
- To check for a linear functionAre the differences in the y-values constant? (Do you add/subtract the same number each time?)
- To check for an exponential functionAre the ratios of the y-values constant? (Do you multiply/divide by the same number each time?)
Both function types can be defined by just two distinct points. For a line, two points give you a slope. For an exponential curve, two points let you solve for the base and initial value. The key is the kind of change that happens between those points.
Worked examples
What's the Pattern?
A biologist is tracking two different bacteria populations, A and B, in a lab. The data is recorded in the table below. Determine if each population's growth is linear or exponential, and write a function for each.
| Hours (x) | Population A | Population B |
|---|---|---|
| 0 | 50 | 1000 |
| 1 | 200 | 1250 |
| 2 | 350 | 1500 |
| 3 | 500 | 1750 |
Solution Walkthrough:
Population A:
- 1Test for linear change (additive)Let's find the difference in population between each hour.
- From x=0 to x=1: 200 - 50 = 150
- From x=1 to x=2: 350 - 200 = 150
- From x=2 to x=3: 500 - 350 = 150 The difference is a constant +150. This is linear growth.
- 2Write the functionWe need the form
f(x) = mx + b.- The constant difference is our slope,
m = 150. - The initial value is the population at
x=0, which is 50. So,b = 50. - The function for Population A is
A(x) = 150x + 50.
- The constant difference is our slope,
Population B:
- 1Test for linear change (additive)
- From x=0 to x=1: 1250 - 1000 = 250
- From x=1 to x=2: 1500 - 1250 = 250
- From x=2 to x=3: 1750 - 1500 = 250 Wait, this is also a constant difference! So Population B is also growing linearly.
- 2Write the functionWe use
f(x) = mx + b.- The slope is
m = 250. - The initial value at
x=0isb = 1000. - The function for Population B is
B(x) = 250x + 1000.
- The slope is
This is a critical point: Don't assume a problem will always give you one of each. You have to test the data. Let's try another one for Population B where it's exponential.
Revised Example 1 (Population B only):
| Hours (x) | Population B (revised) |
|---|---|
| 0 | 100 |
| 1 | 200 |
| 2 | 400 |
| 3 | 800 |
Solution Walkthrough (Revised B):
- 1Test for linear change200 - 100 = 100. But 400 - 200 = 200. The differences are not constant. It's not linear.
- 2Test for exponential change (multiplicative)Let's find the ratio of the population between each hour.
- From x=0 to x=1: 200 / 100 = 2
- From x=1 to x=2: 400 / 200 = 2
- From x=2 to x=3: 800 / 400 = 2 The ratio is a constant 2. This is exponential growth.
- 3Write the functionWe need the form
f(x) = ab^x.- The common ratio is our base,
b = 2. - The initial value at
x=0isa = 100. - The function for the revised Population B is
B(x) = 100 * 2^x.
- The common ratio is our base,
Finding a Function from Two Points
A town's property value is growing exponentially. In 2020, a house was valued at $300,000. In 2022, the same house was valued at $330,750. Write an exponential function V(t) that models the house's value, where t is the number of years since 2020.
Solution Walkthrough:
- 1Identify your pointsWe have two points
(t, V(t)).- Let
t=0be the year 2020. So our first point is(0, 300000). - The year 2022 is 2 years later, so
t=2. Our second point is(2, 330750).
- Let
- 2Use the points to find the common ratio (b)We can use the point-ratio form
f(x) = y_1 * b^(x - x_1)or set up a system of equations. Let's use the standard formV(t) = ab^t.- From point 1:
300000 = a * b^0. Sinceb^0 = 1, this tells usa = 300000. This is our initial value! - From point 2:
330750 = a * b^2.
- From point 1:
- 3Solve for bNow we can substitute the value of
awe found into the second equation.330750 = 300000 * b^2- Divide both sides by 300,000:
330750 / 300000 = b^21.1025 = b^2 - Take the square root of both sides:
b = sqrt(1.1025) = 1.05(We only take the positive root since this is a growth model).
- Divide both sides by 300,000:
- 4Write the final functionWe have our initial value
a = 300000and our baseb = 1.05.V(t) = 300000 * (1.05)^t
Try it yourself
- 1Scenario AnalysisMaya is considering two payment plans for a graphic design project.
- Plan 1A flat fee of $200, plus $50 for each revision round.
- Plan 2A fee of $10 for the first design, which doubles for each subsequent revision round ($20 for the 1st revision, $40 for the 2nd, and so on). Identify which plan represents linear growth and which represents exponential growth. Write a function for each, where
xis the number of revision rounds. Which plan is better for the client if they expect to need 4 rounds of revisions?
- Plan 1
- 2From Two PointsA function passes through the points
(2, 24)and(5, 192).- Part AIf the function is linear, what is its equation? (Hint: First find the slope
m). - Part BIf the function is exponential, what is its equation? (Hint: Set up
r^(x2-x1) = y2/y1to find the common ratiorfirst).
- Part A
Practice — 8 questions
In simple terms, this topic is about the two main ways things change: by adding the same amount each time (linear), or by multiplying by the same amount each time (exponential).
- 2.2.A: Construct functions of the real numbers that are comparable to arithmetic and geometric sequences.
- 2.2.B: Describe similarities and differences between linear and exponential functions.
- 2.2.A.1
- Linear functions of the form f (x) = b + mx are similar to arithmetic sequences of the form a_n = a_0 + dn, as both can be expressed as an initial value (b or a_0) plus repeated addition of a constant rate of change, the slope (m or d).
- 2.2.A.2
- Similar to arithmetic sequences of the form a_n = a_k + d(n − k), which are based on a known difference, d, and a kth term, linear functions can be expressed in the form f(x) = y_i + m(x − x_i) based on a known slope, m, and a point, (x_i, y_i).
- 2.2.A.3
- Exponential functions of the form f (x) = ab^x are similar to geometric sequences of the form g_n = g_0 r^n, as both can be expressed as an initial value (a or g_0) times repeated multiplication by a constant proportion (b or r).
- 2.2.A.4
- Similar to geometric sequences of the form g_n = g_k r^{(n-k)}, which are based on a known ratio, r, and a kth term, exponential functions can be expressed in the form f (x) = y_i r^{(x−x_i)} based on a known ratio, r, and a point, (x_i, y_i).
- 2.2.A.5
- Sequences and their corresponding functions may have different domains.
- 2.2.B.1
- Over equal-length input-value intervals, if the output values of a function change at constant rate, then the function is linear; if the output values of a function change proportionally, then the function is exponential.
- 2.2.B.2
- Linear functions of the form f (x) = b + mx and exponential functions of the form f(x) = ab^x can both be expressed analytically in terms of an initial value and a constant involved with change. However, linear functions are based on addition, while exponential functions are based on multiplication.
- 2.2.B.3
- Arithmetic sequences, linear functions, geometric sequences, and exponential functions all have the property that they can be determined by two distinct sequence or function values.
flowchart TD
A[Start: Given a data pattern] --> B{Is the change between outputs constant?};
B -->|Additive change<br>(y2 - y1 = constant)| C[Linear Function];
B -->|Multiplicative change<br>(y2 / y1 = constant)| D[Exponential Function];
C --> E[Form: f(x) = mx + b];
C --> F[Analogy: Arithmetic Sequence];
C --> G[Graph: Straight Line];
D --> H[Form: f(x) = ab^x];
D --> I[Analogy: Geometric Sequence];
D --> J[Graph: J-Curve];
Read what Saavi narrates
Hello everyone, it's Saavi. Let's talk about something you see every day: how things grow and change.
Imagine you and your friend Priya are offered two different summer job plans for a new cafe in Boston. Your plan: you get a fifty-dollar signing bonus and earn exactly fifteen dollars for every hour you work. Priya's plan: she gets a tiny one-dollar bonus, but her hourly wage doubles every hour. So, two dollars for the first hour, four for the second, eight for the third, and so on.
At first, your plan seems way better, right? But whose earnings grow steadily, and whose earnings... explode?
This is the core of our lesson today. We're comparing two fundamental patterns of growth. The first is steady, additive change, like your job offer. This is what we call a linear function. The second is rapid, multiplicative change, like Priya's offer. That's an exponential function. Understanding the difference is one of the most important skills in all of math.
Let's look at an example. Imagine a biologist is tracking a bacteria population. At hour zero, there are 100 bacteria. At hour one, there are 200. At hour two, 400. At hour three, 800. Is this growth linear or exponential?
First, let's check for linear change. We ask: are we *adding* the same amount each time? From 100 to 200, we added 100. But from 200 to 400, we added 200. The amount we add isn't constant. So, it's not linear.
Now, let's check for exponential change. We ask: are we *multiplying* by the same amount each time? To get from 100 to 200, we multiply by two. To get from 200 to 400, we multiply by two. And from 400 to 800... we multiply by two again. Yes! We have a constant ratio.
So, the growth is exponential. The function would be `f(x) = ab^x`. The 'a' is our initial value, which is 100. The 'b' is our common ratio, which is 2. So the function is `f(x) = 100 times 2 to the power of x`.
A common mistake I see all the time is mixing this up. You might look at that table and see the first jump is "plus 100" and the second is "plus 200" and get stuck. You have to remember to check for multiplication too. Always test for a constant difference first. If that doesn't work, then test for a constant ratio by dividing the second y-value by the first.
You're building a powerful toolkit here. Recognizing these two patterns—additive and multiplicative—will help you model everything from money to populations to physics. You can do this. Keep practicing.
You'll look at a table like (1, 5), (2, 10) and say the "change" is 5. But the next point is (3, 20), and the change is 10. You conclude it's neither linear nor exponential, but you missed the multiplicative pattern (10/5 = 2, 20/10 = 2).
Always test for a common difference first. If that fails, *then* test for a common ratio by dividing consecutive output values (`y_2 / y_1`).
`a` is the starting amount (at x=0), and `b` is the growth factor. Saying a function is `50 * 1.02^x` is completely different from `1.02 * 50^x`. The first is a slow 2% growth on $50; the second is an impossibly fast explosion.
Remember "a" is the "anchor" or initial value. "b" is the "base" that has the exponent attached to it, representing the repeated multiplication.
Given `f(x) = 3 * 2^x`, a student might calculate `f(4)` as `(3 * 2)^4 = 6^4 = 1296`. The correct way is `3 * (2^4) = 3 * 16 = 48`.
Remember PEMDAS (Parentheses, Exponents, Multiplication/Division, Addition/Subtraction). The exponent `x` applies *only* to the base `b`, not to the product `ab`.
If you're given the points (2, 20) and (3, 40), it's tempting to say the initial value is 20. But the initial value is the `y`-value when `x=0`. Here, you'd have to work backward from (2, 20) with a ratio of 2 to find that `f(1)=10` and `f(0)=5`. The initial value is 5.
The initial value `a` (for exponential) or `b` (for linear) is *always* the `y`-coordinate of the point where `x=0`. If you aren't given that point, you must solve for it.
Given points (2, 12) and (4, 48), some students divide the x's and y's: `48/12 = 4` and `4/2 = 2` and get stuck. Others divide the wrong way: `12/48 = 1/4`.
The ratio `r` is `y_2 / y_1`. The exponent on `r` is the difference in x-values, `x_2 - x_1`. So, `r^(4-2) = 48/12`, which means `r^2 = 4`, and `r=2`.