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B2.2 · Identify exponential growth and decay and contextual restrictions

Learn to identify exponential growth and decay and contextual restrictions through clear examples and targeted practice.

Ontario Grade 11 Mathematics

Exponential Functions

How repeated percent change and context shape an exponential model

A savings balance might increase by the same percentage each month. A quantity of medicine in the body might decrease by the same percentage each hour. Both situations can be described by exponential models. To identify the type of change, look for a repeated multiplier rather than a repeated addition. Then use the context to decide which input values make sense. This lesson focuses on recognizing growth and decay and describing those contextual restrictions.

What you will learn

1. Prerequisite bridge: repeated addition and repeated multiplication

A sequence is a list of values in a set order. In a linearly changing sequence, the same amount is added or subtracted each step. For example, CAD 5, 8, 11, 14 increases by 33 each time.
In an exponentially changing sequence, each value is multiplied by the same positive number to get the next value. For example, CAD 5, 10, 20, 40 is multiplied by 22 each time. This repeated multiplication is the key feature to notice.
The multiplier is the number used in each repeated multiplication. In the second sequence, the multiplier is 22. If the multiplier is greater than 11, repeated multiplication makes positive values grow. If the multiplier is between 00 and 11, repeated multiplication makes positive values shrink.
next value=current value×multiplier\text{next value}=\text{current value}\times\text{multiplier}

2. Growth and decay in words, tables, and equations

Exponential growth means a quantity is multiplied by the same factor over equal time steps, and the factor is greater than 11. The values increase when the starting value is positive. A balance that rises by 4% each year has a yearly multiplier of 1.041.04, because the new amount is the old amount plus 4% of the old amount.
Exponential decay means a quantity is multiplied by the same factor over equal time steps, and the factor is greater than 00 but less than 11. The values decrease when the starting value is positive. If a quantity falls by 20% each hour, then 80% remains each hour. Its hourly multiplier is 0.800.80.
A common equation for an exponential situation is y=a(b)ty=a(b)^t. Here, aa is the starting value at t=0t=0, bb is the repeated multiplier, and tt counts equal time steps. The value of yy is the amount after tt steps. Growth is identified by b>1b>1; decay is identified by 0<b<10<b<1, assuming a positive starting value.
A table can make the multiplier visible. In a growth table, divide each value by the previous value. In a decay table, do the same. A constant quotient greater than 11 indicates growth; a constant quotient between 00 and 11 indicates decay. For instance, values CAD 80, 64, 51.2 have the same multiplier, 0.80.8.
y=a(b)ty=a(b)^t

3. Reading a graph and applying contextual restrictions

An exponential graph shows how the output changes as the input changes. For a positive starting value, a growth model rises as time increases, while a decay model falls. The graph helps show the trend, but the equation or table helps confirm that the change is by a constant multiplier.
A contextual restriction is a limit on which input or output values make sense in the situation. The equation may accept values that the real situation does not. For example, if tt measures years since a deposit was made, negative values of tt are usually not part of the situation being described. The relevant inputs may be whole years, or they may include any non-negative time, depending on how the situation is measured.
The output also has meaning and units. A model for the number of living plants should not be interpreted as a negative number of plants. A model for money should be understood in the stated currency and time period. State the restriction using the situation, not just the graph or equation.
Do not assume that a model describes a situation forever. A growth model for savings may be useful only while its stated interest rate and conditions apply. A decay model for a medicine amount may be intended only for the time interval given. Use restrictions supplied by the problem, and explain any direct restriction that follows from the meaning of the variables.

4. A reliable identification routine

Start by naming the quantity and the equal time step. Check whether the amount changes by the same number each step or is multiplied by the same factor. If a percent change is stated, convert it to the fraction that remains or to the multiplier for the increase.
Next, classify the multiplier. A multiplier above 11 signals growth. A multiplier between 00 and 11 signals decay. Finally, interpret the variables and state restrictions that come from the setting, such as non-negative time or a stated observation period.
This routine prevents a common mix-up: a quantity can increase by a fixed number and still not be exponential. Exponential change depends on the current amount because the same percentage, and therefore the same multiplier, is applied each step.

Recognizing the multiplier

SituationMultiplierIdentification
Amount rises by 4% each step1.041.04Growth
80% remains each step0.800.80Decay
1212 is added each stepNo constant multiplier is givenNot identified as exponential from this information

Worked example

Classifying a repeated decrease

A lab sample contains 250250 milligrams of a substance. At the end of each hour, 80% of the amount from the previous hour remains. Identify the type of change, write a model, and state a suitable restriction on time.
  1. Find the hourly multiplier
    The phrase “80% remains” gives the fraction kept each hour directly. The amount is multiplied by 0.800.80 at each one-hour step.
    b=0.80b=0.80
  2. Classify the change
    The multiplier is positive and less than 11, so the amount decreases by the same factor each hour. This is exponential decay.
    0<0.80<10<0.80<1
  3. Write the model
    The starting amount is 250250 milligrams at time t=0t=0. Use tt for elapsed hours and multiply the starting amount by the hourly multiplier raised to the number of hours.
    A(t)=250(0.80)tA(t)=250(0.80)^t
  4. State the contextual restriction
    Elapsed time cannot be negative. If the model is being used only for the first six hours, the relevant time values are from 00 through 66. The problem must specify that interval; without it, the direct restriction is t≥0t\geq 0.
    t≥0t\geq 0
Answer: The sample shows exponential decay. A model is A(t)=250(0.80)tA(t)=250(0.80)^t, where A(t)A(t) is the amount in milligrams and tt is elapsed time in hours. The contextual restriction is t≥0t\geq 0, unless the situation gives a shorter observation interval.
Check: At t=0t=0, the model gives 250250 milligrams. At t=1t=1, it gives 200200 milligrams, which is 80% of 250250. This matches the description.

Common mistakes and how to avoid them

Calling any increase exponential growth.
Correction: Check how the increase happens. A constant amount added each step is not the repeated-multiplier pattern used to identify exponential growth.
Using the percent decrease as the multiplier.
Correction: For a decrease, find the percent that remains. If 20% is lost, then 80% remains, so the multiplier is 0.800.80.
Assuming every input allowed by the equation is allowed by the situation.
Correction: Read the variable definitions and setting. State restrictions such as non-negative elapsed time or a given observation interval.
Treating a multiplier below 11 as a negative number.
Correction: A decay multiplier is positive and less than 11. For example, 0.80.8 is positive and produces a decrease when repeatedly applied to a positive amount.

Lesson summary

Check your understanding

Question 1

A quantity changes from 4040 to 5050 to 62.562.5 over equal time steps. Which statement best identifies the pattern?
  1. Exponential growth with multiplier 1.251.25
  2. Exponential decay with multiplier 0.80.8
  3. A constant increase of 1010 each step
  4. The values do not have a constant multiplier
Show answer and explanation
Exponential growth with multiplier 1.251.25
Both successive quotients are 1.251.25, so each value is multiplied by the same factor greater than 11. This is exponential growth.

Question 2

A value decreases by 35% during each equal time step. What is the multiplier, and is the change growth or decay?
  1. The multiplier is 0.350.35; it is decay.
  2. The multiplier is 0.650.65; it is decay.
  3. The multiplier is 1.351.35; it is growth.
  4. The multiplier is 0.650.65; it is growth.
Show answer and explanation
The multiplier is 0.650.65; it is decay.
If 35% is lost, then 65% remains. The multiplier is 0.650.65, which is between 00 and 11, so the change is decay.

Question 3

A model uses tt for years after a tree is planted. Which restriction follows directly from the meaning of tt?
  1. t<0t<0
  2. t≤0t\leq 0
  3. t≥0t\geq 0
  4. There is no restriction on tt
Show answer and explanation
t≥0t\geq 0
Years after planting cannot be negative in this context. The planting time is t=0t=0, and later times are positive.

Key terms

Exponential change
A pattern in which each value is found by multiplying the previous value by the same factor over equal steps.
Multiplier
The factor used to multiply one value to get the next value.
Contextual restriction
A limit on input or output values that follows from what the variables represent in a situation.
Starting value
The amount at the beginning of the model, when the step count is zero.

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCF3M), expectation B2.2. It is a study resource, not an official curriculum publication.

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