B3.3 Solve real-world problems using exponential graphs and equations

Ontario Grade 11 Mathematics

Exponential Functions

MCR3U · Strand B: Exponential Functions · Expectation B3.3

Every time a savings account earns compound interest, a bacterial colony doubles in size, or a used car loses value over the years, an exponential function is at work. These situations share one key feature: the quantity changes by a constant multiplying factor over equal time intervals, not by a constant amount. In this lesson you will move from recognising that pattern to writing an equation, reading a graph, and answering practical questions — all using tools already in your Grade 11 toolkit.

What you will learn

Identify the key features of an exponential model (initial value, growth or decay factor, and base) from a real-world context.
Write and evaluate an exponential equation of the form $y = a \cdot b^x$ to model growth and decay situations.
Read and interpret an exponential graph to answer questions about a real-world situation.
Solve real-world problems by substituting values into an exponential equation and reasoning about the result.

Prerequisite Bridge: What You Already Know

In Grade 10 you studied linear functions, where the graph is a straight line and the quantity changes by a constant amount each step. For example, earning CAD 5 every hour is linear because you add the same value repeatedly.

Exponential functions are different. Instead of adding, you multiply by the same factor each time. If a rabbit population triples every year, you multiply by 3 each year. That multiplying factor is called the base of the exponential function.

The general form of an exponential equation used in this course is

y = a \cdot b^x

, where

a

is the initial value (the amount when

x = 0

),

b

is the growth or decay factor, and

x

is the number of time periods. You need

a > 0

and

b > 0

,

b \neq 1

. If

b > 1

, the function models growth; if

0 < b < 1

, it models decay.

y = a \cdot b^x

Linear: constant addition each step. Exponential: constant multiplication each step.
General model: $y = a \cdot b^x$
$a$ = initial value, $b$ = growth or decay factor, $x$ = number of periods.
Growth: $b > 1$ . Decay: $0 < b < 1$ .

Recognising Exponential Situations and Building the Equation

The first step in any real-world problem is deciding whether the situation is exponential. Ask yourself: does the quantity get multiplied by the same number over each equal time interval? If yes, you have an exponential model.

To find

a

, look for the starting amount — the value at time zero. To find

b

, look for the multiplying factor. If a population grows by 20% each year, the factor is

1 + 0.20 = 1.20

because you keep the original amount and add 20% of it. If a car loses 15% of its value each year, the factor is

1 - 0.15 = 0.85

because 85% of the value remains after each year.

Once you have

a

and

b

, substitute them into

y = a \cdot b^x

and define what

x

represents (for example,

x

= number of years since the start). Always state the units of

x

and

y

clearly — this prevents the most common errors on tests.

A quick reasonableness check: if the situation is growth, the value of

y

should increase as

x

increases. If it is decay,

y

should decrease and approach zero but never actually reach it.

b = 1 \pm \frac{r}{100}

Exponential clue: quantity multiplies by the same factor each period.
Growth rate r% gives factor $b = 1 + \frac{r}{100}$ .
Decay rate r% gives factor $b = 1 - \frac{r}{100}$ .
Always define what $x$ and $y$ represent, including units.
Check: growth means $y$ increases; decay means $y$ decreases toward zero.

Reading and Interpreting Exponential Graphs

A graph can answer questions quickly without algebra. The horizontal axis usually represents time (

x

), and the vertical axis represents the quantity (

y

). On a growth graph the curve rises steeply to the right; on a decay graph the curve falls and flattens.

The y-intercept of the graph is the initial value

a

. You can read it directly by finding where the curve crosses the vertical axis. To find the value at a specific time, move along the horizontal axis to that

x

-value, go straight up to the curve, and then read across to the vertical axis.

To find when a quantity reaches a particular value, start on the vertical axis at that value, move horizontally to the curve, and then drop straight down to read the time on the horizontal axis. This graph-reading skill is especially useful when the time value is not a whole number and an exact match by substitution is hard to find directly. If the answer falls between two grid lines, estimate by interpolating (reading between the lines).

Also pay attention to the shape: exponential curves are always curved (not straight), always positive (never touch or cross the horizontal axis), and the rate of change itself keeps changing — that is what makes them different from linear functions.

y-intercept on the graph = initial value $a$ .
To find $y$ at a given $x$ : read up from the axis to the curve, then across.
To find $x$ for a given $y$ : read across from the value to the curve, then down.
Exponential curves are always positive and always curving.
Use the graph to estimate values that are hard to pin down exactly by substitution alone.

Solving Problems Algebraically

When you know

a

,

b

, and

x

, finding

y

is straightforward: substitute and evaluate using order of operations. Always apply the exponent before multiplying by

a

.

A slightly harder task is finding

x

when you know

y

. In this course, you can solve these problems in two ways. First, if the answer is a whole number, you can substitute trial values of

x

(try

x = 1, 2, 3, \ldots

) until the equation balances — this is called solving by systematic trial. Second, you can read the answer from a graph as described in the previous section. Both methods are valid at this course level.

Some problems ask you to compare two exponential situations or to find a break-even point. Set the two expressions equal and use trial or graph-reading to find where they match.

Remember to interpret your answer in context. If

x

represents years and you get

x = 3

, state '3 years after the start'. If

y

represents population and you get

y = 4800

, state 'the population is 4 800 individuals'. Numbers without context do not fully answer a real-world problem.

y = a \cdot b^x

To find $y$ : substitute $x$ into $y = a \cdot b^x$ and evaluate.
Apply the exponent first, then multiply by $a$ .
To find $x$ : use systematic trial (whole-number answers) or read from a graph.
Always interpret your numerical answer in the context of the problem.

Putting It All Together: Strategy for Real-World Problems

A reliable four-step strategy will handle most MCR3U exponential problems. Step 1 — Identify: read the problem carefully and confirm the situation is exponential. Write down the initial value

a

and the factor

b

. Step 2 — Model: write the equation

y = a \cdot b^x

with

x

and

y

clearly defined. Step 3 — Solve: substitute known values to find the unknown, using evaluation, systematic trial, or graph-reading as needed. Step 4 — Interpret: write a sentence that answers the original question using correct units.

Be especially careful with the time variable. If a problem says 'after 6 months' and your model uses years, convert first: 6 months = 0.5 years. Mismatched units are among the most common sources of error.

For problems involving a percentage increase or decrease, double-check your factor. A 30% increase gives

b = 1.30

, not

0.30

. A 30% decrease gives

b = 0.70

, not

1.70

. Mixing these up completely reverses the model.

Four steps: Identify → Model → Solve → Interpret.
Convert all time values to the same unit before substituting.
Percentage increase: $b = 1 + \frac{r}{100}$ . Percentage decrease: $b = 1 - \frac{r}{100}$ .
Finish every solution with a sentence that answers the question in context.

Bacterial Colony Growth at 3-Hour Intervals (Example 1 Check)

Time (hours)	Number of Periods ( $x$ )	Bacteria Count ( $y$ )
0	0	500
3	1	1 000
6	2	2 000
9	3	4 000
12	4	8 000

Laptop Value by Year — Systematic Trial (Example 2)

Year ( $x$ )	Calculation	Value (CAD)	Below CAD 400?
0	$1200 \cdot (0.75)^0$	1 200.00	No
1	$1200 \cdot (0.75)^1$	900.00	No
2	$1200 \cdot (0.75)^2$	675.00	No
3	$1200 \cdot (0.75)^3$	506.25	No
4	$1200 \cdot (0.75)^4$	379.69	Yes

Worked example

Example 1 — Bacterial Growth (Finding a Future Value)

A biology student starts an experiment with 500 bacteria. The colony doubles every 3 hours. How many bacteria will there be after 12 hours?

Identify the initial value and the growth factor
The starting amount is 500, so $a = 500$ . The colony doubles each period, so the growth factor is $b = 2$ . Each time period is 3 hours.
$a = 500, \quad b = 2$
Determine the number of time periods
The question asks about 12 hours. Since each period is 3 hours, the number of periods is found by dividing the total time by the length of one period.
$x = \frac{12}{3} = 4$
Write the exponential equation
Substituting $a = 500$ and $b = 2$ into the model $y = a \cdot b^x$ gives the equation below.
$y = 500 \cdot 2^x$
Substitute and evaluate
Replace $x$ with 4. Apply the exponent first, since $2^4 = 16$ , then multiply by the initial value.
$y = 500 \cdot 2^4 = 500 \cdot 16 = 8000$
Interpret the answer
The value $y = 8000$ represents the number of bacteria. State the answer in context.

Answer: After 12 hours there will be 8 000 bacteria in the colony.

Check: Trace each 3-hour period: 500, then 1000, then 2000, then 4000, then 8000. Four doublings from 500 gives 8 000, which matches the calculated answer.

Worked example

Example 2 — Depreciation (Finding the Time by Systematic Trial)

A new laptop is purchased for CAD 1 200. Its value decreases by 25% each year. In which year does the laptop's value first fall below CAD 400?

Identify the initial value and the decay factor
The purchase price is CAD 1 200, so $a = 1200$ . The value decreases by 25% each year, meaning 75% of the value remains each year.
$a = 1200, \quad b = 1 - 0.25 = 0.75$
Write the exponential equation
Let $x$ represent the number of years since purchase and $y$ represent the value in dollars.
$y = 1200 \cdot (0.75)^x$
Use systematic trial to test whole-number years
Substitute whole-number values of $x$ one at a time and evaluate until the value drops below CAD 400. Testing $x = 1$ gives 900, $x = 2$ gives 675, $x = 3$ gives 506.25, and $x = 4$ gives approximately 379.69.
$y(4) = 1200 · (0.75)^4 \approx 379.69$
Identify the first year the condition is met
At $x = 3$ the value is approximately CAD 506.25, which is still above CAD 400. At $x = 4$ the value is approximately CAD 379.69, which is below CAD 400. So the laptop's value first drops below CAD 400 during year 4.
Interpret the answer
State the answer clearly in the context of the problem, including units.

Answer: The laptop's value first falls below CAD 400 during the 4th year after purchase (approximately CAD 379.69).

Check: Recompute directly:

1200 \times 0.421875 = 506.25

for

x=3

, which is at least 400, and

1200 \times 0.31640625 = 379.69

for

x=4

, which is below 400. The 4th year is confirmed.

Common mistakes and how to avoid them

Using the percentage rate itself as the base — for example, writing

b = 0.25

for a 25% decay instead of

b = 0.75

.

Correction: The base is the fraction that remains or is added. For decay at rate r%, the base is

b = 1 - \frac{r}{100}

. For growth, it is

b = 1 + \frac{r}{100}

.

Multiplying the exponent and the base before applying the power — for example, evaluating

500 \cdot 2^4

as

(500 \cdot 2)^4

.

Correction: Apply the exponent to the base first, so

2^4 = 16

, then multiply by

a

to get

500 \times 16 = 8000

. Follow order of operations: exponents before multiplication.

Forgetting to convert the time to the correct number of periods — for example, using

x = 12

when 12 hours means 4 periods of 3 hours each.

Correction: Divide the total time by the length of one period to get the correct value of

x

. Define

x

clearly at the start of every solution.

Stopping at a numerical answer without interpreting it in context.

Correction: Always write a sentence that states what the number means, including its units and what it refers to, such as population or value in dollars.

Confusing growth and decay graphs — expecting a decay curve to cross the horizontal axis.

Correction: An exponential decay curve gets closer and closer to the horizontal axis but never touches it, because multiplying a positive number by a positive factor can never produce zero.

Lesson summary

An exponential function has the form $y = a \cdot b^x$ , where $a$ is the initial value and $b$ is the constant growth or decay factor.
Growth situations have $b > 1$ ; decay situations have $0 < b < 1$ . The factor $b$ is found from the percentage rate using $b = 1 + \frac{r}{100}$ or $b = 1 - \frac{r}{100}$ .
To find a future or past value, substitute the known time $x$ into the equation and evaluate — applying the exponent before multiplying.
To find the time when a quantity reaches a target, use systematic trial (testing whole-number values of $x$ ) or read the answer from a graph.
Graphs of exponential functions are always curved and always positive; the y-intercept equals the initial value $a$ .
Every solution must end with an interpreted sentence: state what the number means, with correct units and context.

Check your understanding

Question 1

A savings account holds CAD 2 000 and grows by 10% each year. Which equation models the account balance

y

after

x

years?

$y = 2000 \cdot (0.10)^x$
$y = 2000 \cdot (1.10)^x$
$y = 2000 \cdot (0.90)^x$
$y = 2000 + 10x$

Show answer and explanation

y = 2000 \cdot (1.10)^x

A 10% annual growth means the factor is

b = 1 + 0.10 = 1.10

, and the initial value is

a = 2000

. The model is

y = 2000 \cdot (1.10)^x

. The first option uses the rate itself as the base, the third option uses a decay factor in the wrong direction, and the fourth option is a linear model instead of an exponential one.

Question 2

A radioactive sample starts at 800 g and loses half its mass every 5 years. How many grams remain after 15 years?

400 g
200 g
100 g
50 g

Show answer and explanation

100 g

The number of 5-year periods in 15 years is

x = 15 \div 5 = 3

. The decay factor is

b = 0.5

, so

y = 800 \cdot (0.5)^3 = 800 \cdot 0.125 = 100

grams. After three halvings the mass goes 800, then 400, then 200, then 100.

Question 3

On an exponential decay graph, what does the y-intercept represent?

The decay factor $b$
The time when the quantity reaches zero
The initial value $a$ at time $x = 0$
The rate of decrease per period

Show answer and explanation

The initial value

a

at time

x = 0

The y-intercept is the point where

x = 0

. Substituting

x = 0

into

y = a \cdot b^0 = a \cdot 1 = a

shows the y-intercept always equals the initial value

a

.

Question 4

A car worth CAD 20 000 depreciates at 20% per year. Using systematic trial, in which year does its value first drop below CAD 10 000?

Year 2
Year 3
Year 4
Year 5

Show answer and explanation

Year 4

The model is

y = 20000 \cdot (0.80)^x

. Testing values gives

x=2

: 12 800,

x=3

: 10 240, and

x=4

: 8 192. At

x=3

the value is still above 10 000, but at

x=4

it drops to 8 192, which is below 10 000. So the value first drops below CAD 10 000 in Year 4.

Key terms

Exponential function: A function of the form $y = a \cdot b^x$ where the variable $x$ appears as an exponent and $b$ is a positive constant not equal to 1.
Initial value ( $a$ ): The value of $y$ when $x = 0$ ; the starting amount in a real-world model. It equals the y-intercept of the graph.
Growth factor: The base $b$ in an exponential model when $b > 1$ ; represents the constant multiplier that increases the quantity each period.
Decay factor: The base $b$ in an exponential model when $0 < b < 1$ ; represents the constant multiplier that decreases the quantity each period.
Time period: The fixed interval of time over which the quantity is multiplied by the factor $b$ once. Examples include one year, one hour, or one half-life.
Systematic trial: A method of solving an equation by testing whole-number values of the variable one at a time until the equation is satisfied or a target condition is met.
Depreciation: The decrease in the monetary value of an asset over time, often modelled as exponential decay when expressed as a fixed percentage loss per period.
y-intercept: The point where a graph crosses the vertical axis, occurring at $x = 0$ . For $y = a \cdot b^x$ , the y-intercept is always $(0, a)$ .

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCR3U), expectation B3.3. It is a study resource, not an official curriculum publication.

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B3.3 · Solve real-world problems using exponential graphs and equations

Ontario Grade 11 Mathematics

MCR3U · Strand B: Exponential Functions · Expectation B3.3

What you will learn

Prerequisite Bridge: What You Already Know

Recognising Exponential Situations and Building the Equation

Reading and Interpreting Exponential Graphs

Solving Problems Algebraically

Putting It All Together: Strategy for Real-World Problems

Bacterial Colony Growth at 3-Hour Intervals (Example 1 Check)

Laptop Value by Year — Systematic Trial (Example 2)

Example 1 — Bacterial Growth (Finding a Future Value)

Example 2 — Depreciation (Finding the Time by Systematic Trial)

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through MCR3U

About this lesson