B2.3 Solve applications using exponential graphs and equations

Ontario Grade 11 Mathematics

Exponential Functions

Build a model, read its graph, and use it to answer a real question

A quantity changes exponentially when it is multiplied by the same factor over equal time intervals. For example, a savings balance might increase by the same percentage each year, or a substance might lose the same percentage each hour. In this lesson, you will connect these situations to tables, graphs, and equations. You will also practise finding when a quantity reaches a target without using methods from later courses.

What you will learn

Recognize when a situation can be represented by an exponential model.
Use a table, graph, and equation to describe and solve an exponential application.
Explain what the starting value and repeated growth or decay factor mean in context.
Check that an answer makes sense for the situation.

1. Prerequisite bridge: repeated multiplication

An exponential situation is built from repeated multiplication. This differs from a linear situation, where the same amount is added or subtracted each time. Suppose a quantity starts at

5

and doubles each hour. The values are

5

,

10

,

20

, and

40

. Each new value is found by multiplying the previous value by

2

.

The starting value is the amount at time zero. The repeated factor tells how the amount changes over one time interval. A factor greater than

1

represents growth. A positive factor less than

1

represents decay, or a decrease by the same proportion each interval.

The variable in an equation names a quantity that can change. In an application, state what it measures and its units. For instance, time might be measured in hours and an amount in grams. This helps you choose sensible values and interpret your answer.

y=a(b)^t

Exponential change uses a constant multiplier over equal intervals.
Growth uses a factor greater than $1$ ; decay uses a factor between $0$ and $1$ .
Always identify the starting value, time interval, and units.

2. From a situation to a table, graph, and equation

In the model

y=a(b)^t

,

t

is the number of time intervals,

a

is the starting amount, and

b

is the repeated factor. The output

y

is the amount after

t

intervals. The model applies when the same factor is used for each interval.

For a growth rate of

r

per interval, write the rate as a decimal and add it to

1

to get the factor. For example, growth of 8% gives a factor of

1.08

. For a decrease of 8%, subtract the decimal rate from

1

, giving a factor of

0.92

. The factor is not the percentage itself.

A table makes repeated change visible. A graph places time on the horizontal axis and amount on the vertical axis. For growth, the plotted points rise and become farther apart as time passes. For decay, they fall and get closer to the horizontal axis. In both cases, the graph is curved rather than a straight line.

A graphing tool can display the model and help estimate an output or locate a time when the amount reaches a target. Use the labels and scale carefully. A graph gives an estimate when the crossing falls between marked time values. If time must be a whole number of intervals, use whole-number inputs and explain that choice.

y=a(1+r)^t

For growth rate $r$ , use $b=1+r$ when $r$ is written as a decimal.
For decrease rate $r$ , use $b=1-r$ .
A table, graph, and equation describe the same relationship in different ways.

3. Solving an application and checking the result

Begin by deciding whether the situation describes repeated percentage change or another repeated multiplier. Identify the starting amount and the length of one interval. Then write an equation with variables that match the question. State whether your answer is an exact value from the model or an estimate read from a graph.

To find an amount at a given time, substitute the time into the equation. To find when an amount reaches a target, graph the model and the target value together, or use a table of values to find the interval where the amount passes the target. With a graph, the intersection is the point where the two plotted relationships have the same output. Read its time coordinate, then report a suitable estimate.

Technology can help draw an exponential graph, but it does not decide whether the model suits the situation. Check that the graph uses the correct starting value and factor. Check the axes, units, and time scale. Finally, test the answer against the story: a decay model should not predict an increase, and a time should not be negative if the situation begins at time zero.

Some applications count only whole intervals, such as the number of completed years. In those cases, report the first whole-number interval that meets the target, if that is what the question asks. If the context allows part of an interval, a graph estimate may be appropriate. Make the interpretation clear rather than reporting a bare number.

t\geq 0

Match the equation to the story before calculating.
For a target question, find where the model reaches the target and interpret the time coordinate.
Check units, scale, and whether the answer fits the situation.

First values for the cooling sample

Time, $t$ (hours)	Amount, $A$ (grams)
$0$	$160$
$1$	$120$
$2$	$90$
$3$	$67.5$
$4$	$50.625$
$5$	$37.97$

Worked example

A cooling sample

A sample contains

160

grams of a substance. Every hour, the amount is reduced by 25%. Estimate when the sample first contains

50

grams, assuming time can be measured to the nearest tenth of an hour.

Identify the repeated factor
The sample keeps 75% of its amount each hour, because $100\%-25\%=75\%$ . As a decimal, this is $0.75$ . The starting amount is $160$ grams.
$a=160,\quad b=0.75$
Write the model
Let $t$ be time in hours and let $A$ be the amount in grams. Repeatedly multiplying by the hourly factor gives the exponential model.
$A=160(0.75)^t$
Use a table to narrow the time
Evaluate the model at whole-hour times. At $4$ hours the amount is about $50.6$ grams, which is still above the target. At $5$ hours it is about $38.0$ grams, which is below the target. Therefore, the target is reached between $4$ and $5$ hours.
$A(4)=50.625,\quad A(5)=37.96875$
Estimate from the graph
Graph $A=160(0.75)^t$ together with the horizontal target line $A=50$ . Their intersection occurs at about $t=4.1$ . The graph estimate to the nearest tenth is therefore $4.1$ hours.
$160(0.75)^t=50$

Answer: The sample contains about

50

grams after

4.1

hours.

Check: The amount at

4

hours is slightly above

50

grams, and at

5

hours it is below

50

grams. An estimate just after

4

hours is reasonable.

Common mistakes and how to avoid them

Using

0.25

as the hourly factor because the sample decreases by 25%.

Correction: The sample keeps 75% each hour, so the factor is

0.75

. The decrease rate and the repeated factor are not the same.

Treating the starting amount as the amount after the first interval.

Correction: The starting value is at time zero. In this example, the amount at

t=0

is

160

grams.

Reporting a graph estimate without units or context.

Correction: State what the time measures and connect it to the target. Here, the sample reaches about

50

grams after about

4.1

hours.

Assuming a curved graph must be linear because it passes through listed table values.

Correction: The table entries come from repeated multiplication. The points follow an exponential curve, not a straight-line pattern.

Lesson summary

Exponential applications involve multiplying by the same factor over equal intervals.
Use $y=a(b)^t$ , where $a$ is the starting amount and $b$ is the repeated factor.
Convert percentage growth or decrease to a multiplier before building the model.
Use tables and graphs to estimate when a quantity reaches a target, then check the estimate in context.

Check your understanding

Question 1

A plant population begins at

240

and grows by 10% each week. Which equation models the population

P

after

w

weeks?

$P=240(1.10)^w$
$P=240(0.10)^w$
$P=240(0.90)^w$
$P=240+10w$

Show answer and explanation

P=240(1.10)^w

A 10% increase means the population is multiplied by

1.10

each week. The starting value is

240

, so the model is

P=240(1.10)^w

.

Question 2

A quantity follows

Q=90(0.8)^t

. What does the factor

0.8

mean?

The quantity increases by 80% each interval.
The quantity keeps 80% of its amount each interval.
The quantity decreases by $80$ units each interval.
The starting amount is $0.8$ units.

Show answer and explanation

The quantity keeps 80% of its amount each interval.

Multiplying by

0.8

means each new amount is 80% of the previous amount. This is a decrease of 20% per interval.

Key terms

Exponential model: An equation that represents repeated multiplication by the same factor over equal intervals.
Starting value: The amount at time zero in a model.
Repeated factor: The number by which the amount is multiplied during each interval.
Decay: A repeated decrease by the same proportion or factor.
Intersection: A point where two graphs meet and have the same output.

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCF3M), expectation B2.3. It is a study resource, not an official curriculum publication.

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B2.3 · Solve applications using exponential graphs and equations