Modelling a quantity that decreases by the same percentage in each unit of time with the exponential decay model. Includes converting a decay rate into a decay factor, evaluating and graphing a depreciation model, classifying a model as growth or decay from its base, and reading why a decay curve flattens without ever reaching zero.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 8 — Exponents and Exponential Functions
Exponential Decay Functions
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.7 Exponential Decay Functions §8.7, pp. 482-488 — the lesson these objectives are drawn from
Warm-up
Lesson 8.6 grew a quantity by a fixed percentage. Losing one works the same way, with a single sign changed.
Discussion prompt
A car worth 16 000 loses 12 per cent of its value in a year. What is it worth afterwards, and what did you multiply by?
Hint: Losing twelve per cent leaves eighty-eight per cent.
Answer:
\[ 16\,000 - 0.12(16\,000) = 16\,000(0.88) = 14\,080 \]
Multiplying by 0.88 does the subtraction in one step, because keeping eighty-eight per cent is exactly what losing twelve per cent means. That single number is the decay factor this lesson is built on.
Concept
A quantity is decreasing exponentially if it decreases by the same percentage r in each unit of time. Such decay is modelled by y equals C times the quantity one minus r, raised to t.
exponential decay — A decrease in which a quantity loses the same percentage in each unit of time, modelled by y equals C times one minus r, all raised to the power t, with r between nought and one.
The expression one minus r is called the decay factor.
Figure (svg): The exponential decay model with each part labelled
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.7 Exponential Decay Functions §8.7, pp. 482-482
Section
Section 1
Concept
In the model, C is the initial amount before any decay occurs, r is the decay rate written as a decimal, and t is the time. The bracket one minus r is the decay factor, and it lies between nought and one.
\[ y = C(1 - r)^t, \quad 0 < r < 1 \]
That restriction on r is what keeps the factor below one.
Figure (svg): The exponential decay model with each part labelled
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.7 Exponential Decay Functions §8.7, pp. 482-482 — the exponential decay model and its definitions
Picture it
Twelve per cent lost, eighty-eight kept.
Figure (svg): A percentage rate converted into a decay factor
The rate and the factor add to one, which is a useful check: a rate of 0.12 and a factor of 0.88 must sum to a whole.
Worked example
This is Example 1 from the textbook.
\[ \text{A car costing } 16\,000 \text{ dollars depreciates at } 12\% \text{ a year. Write a decay model.} \]
Name the initial amount
Why: The price before any depreciation.
\[ C = 16000 \]
Name the decay rate
Why: Twelve per cent as a decimal.
\[ r = 0.12 \]
Substitute into the model
Why: One minus twelve hundredths.
\[ 16 \, 000(1 - 0.12) ^{t} \]
Subtract inside the bracket
Why: That gives the decay factor.
\[ y = 16 \, 000(0.88) ^{t} \]
Figure (svg): A percentage rate converted into a decay factor
\[ y = 16\,000(0.88)^t \]
Verify: test the model after one year
Why: One year gives sixteen thousand times 0.88, which is fourteen thousand and eighty — a loss of one thousand nine hundred and twenty, and twelve per cent of sixteen thousand is exactly that. The first step reproduces the described loss.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.7 Exponential Decay Functions §8.7, pp. 482-482
Matching
Divide by a hundred, then subtract from one.
Match the pairs
Why: Every factor here is below one, because something is being kept but not all of it. Each rate and its factor sum to one, which is the fastest check on the conversion.
Worked example
Guided Practice 1. A different price and rate.
\[ \text{A car costing } 24\,000 \text{ dollars depreciates at } 10\% \text{ a year. Write a decay model.} \]
Name the initial amount
Why: Twenty-four thousand.
\[ C = 24000 \]
Name the decay rate
Why: Ten per cent as a decimal.
\[ r = 0.10 \]
Subtract from one
Why: Keeping ninety per cent.
\[ 1 - 0.10 = 0.90 \]
Write the model
Why: The factor raised to the power t.
\[ y = 24 \, 000(0.9) ^{t} \]
Figure (svg): The solution to Worked example a second depreciation model shown as a ladder of expressions, one row per algebraic move
\[ y = 24\,000(0.9)^t \]
Verify: check the factor against the rate
Why: The rate 0.10 and the factor 0.90 sum to one, which they must, since what is lost and what is kept make up the whole. A factor above one or below nought would signal an arithmetic slip immediately.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.7 Exponential Decay Functions §8.7, pp. 482-482
Trap
\[ y = 16\,000(1 + 0.12)^t = 16\,000(1.12)^t \]
Substitute into the model from the previous lesson
Why: The model looked the same, so the plus sign came along with it.
A factor of 1.12 makes the car gain twelve per cent a year, so after eight years the model would give more than forty thousand for a used car.
\[ y = 16\,000(1 - 0.12)^t = 16\,000(0.88)^t \]
Subtract the rate, because the quantity is losing value
Why: Depreciate, decrease and lose all call for the minus sign.
The factor's size is the check: below one for decay and above one for growth, always.
Faded example
Initial amount, then the factor.
Fill in the blanks
C = 16\,000, \; r = 0.12 \;\Longrightarrow\; y = 16\,000(1 - 0.12)^t = 16\,000(0.88)^t
Why: The subtraction happens once, when the model is written, and the resulting factor is then used for every period. Writing one plus twelve hundredths would model a car that appreciates.
Elimination
A 24 000 dollar car depreciating at 10 per cent a year.
Eliminate the wrong options
Which is the decay model?
Survives elimination: A
Why: Losing ten per cent means keeping ninety, so the factor is 0.9. Option C is the commonest slip: the rate itself was used as the factor, which confuses what is lost with what remains.
Socratic
The restriction says r is between nought and one.
Discussion prompt
Explain why the decay factor must be between nought and one. Then say what a factor of exactly one, or of less than nought, would mean.
Hint: Ask what fraction of the amount survives each period.
Answer:
The factor is the fraction of the amount that survives each period. Since something is lost, the fraction is less than one; since a quantity cannot lose more than all of itself, the fraction is more than nought. Those two facts are exactly the restriction that r lies between nought and one.
A factor of exactly one would mean nothing is lost, so the quantity would stay at C forever — a horizontal line rather than a curve. A negative factor would mean the value flips sign each period, which no depreciating car does, and it is why the model insists on r below one.
Section
Section 2
Concept
Once the model is written, finding a value at any time is a substitution and a calculator step. The factor is applied once for each period.
Rounding to the nearest dollar is usual for money.
Figure (svg): The value of a sixteen thousand dollar car depreciating at twelve per cent a year
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.7 Exponential Decay Functions §8.7, pp. 483-483 — Example 2, Use an Exponential Decay Model
Picture it
From sixteen thousand down.
Figure (svg): The value of a sixteen thousand dollar car depreciating at twelve per cent a year
The two marked points are Example 2's answer and Example 3's graphical estimate. The curve carries every year between them.
Worked example
This is Example 2 from the textbook.
\[ \text{Use } y = 16\,000(0.88)^t \text{ to find the car's value after } 8 \text{ years.} \]
Write the model
Why: The one built in Example 1.
\[ y = 16 \, 000(0.88) ^{t} \]
Substitute eight for t
Why: Eight years of ownership.
\[ 16 \, 000(0.88) ^{8} \]
Evaluate the power
Why: A calculator gives about 0.3596.
\[ 0.3596 \]
Multiply
Why: Sixteen thousand times that.
\[ \text{about } 5754 \]
Figure (svg): The value of a sixteen thousand dollar car depreciating at twelve per cent a year
\[ 16\,000(0.88)^8 \approx 5754 \]
Verify: check the size against the halving point
Why: 0.88 to the sixth is about 0.46, so the value roughly halves in six years; eight years should therefore leave rather more than a third, and 5754 is about thirty-six per cent of sixteen thousand. The answer sits where the halving estimate says it should.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.7 Exponential Decay Functions §8.7, pp. 483-483
Faded example
Substitute, then power, then multiply.
Fill in the blanks
y = 16\,000(0.88)^8} \approx 5754
Why: The exponent counts the periods and the factor does the rest. Multiplying by 0.88 eight separate times gives the same result, which is worth doing once to see why the power is there.
Worked example
Guided Practice 2. The 24 000 dollar model.
\[ \text{Use } y = 24\,000(0.9)^t \text{ to find the value after } 6 \text{ years.} \]
Substitute six for t
Why: Six years of ownership.
\[ 24 \, 000(0.9) ^{6} \]
Evaluate the power
Why: 0.9 to the sixth is about 0.5314.
\[ 0.5314 \]
Multiply
Why: Twenty-four thousand times that.
\[ \text{about } 12755 \]
Round for money
Why: To the nearest dollar.
\[ \text{about } \$ 12 \, 755 \]
Figure (svg): The solution to Worked example the friend's car after six years shown as a ladder of expressions, one row per algebraic move
\[ 24\,000(0.9)^6 \approx 12\,755 \]
Verify: compare with a rough halving
Why: A ten per cent yearly loss roughly halves a value in about seven years, so after six years a little over half should remain. Twelve thousand seven hundred is just over half of twenty-four thousand, which fits.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.7 Exponential Decay Functions §8.7, pp. 483-483
Trap
A car worth 16 000 loses 12 per cent a year for 8 years.
Multiply twelve per cent by eight to get ninety-six per cent lost
Why: Eight years of twelve per cent looks like ninety-six per cent.
\[ 16\,000 - 0.96(16\,000) = 640 \]
That treats each year's loss as twelve per cent of the original price. After the first year the car is worth less, so the second year's loss is smaller — and the true answer is nearly nine times this one.
\[ 16\,000(0.88)^8 \approx 5754 \]
Apply the factor once per year to the current value
Why: The exponent does exactly that.
Adding percentage losses would eventually take a quantity below nought, which is a sign the method is wrong.
Prediction
The percentage lost each year is constant.
Predict first
What happens to the number of dollars lost each year?
Correct: It shrinks, since the same percentage of a smaller value is less.
\[ 1920, \; 1690, \; 1487, \; 1309, \; \ldots \]
Why: The first year costs one thousand nine hundred and twenty dollars, twelve per cent of sixteen thousand. The second costs twelve per cent of fourteen thousand and eighty, which is about one thousand six hundred and ninety. A constant rate applied to a shrinking amount gives shrinking losses, which is precisely why the graph flattens out.
Translation
Write the model, then evaluate.
Match the pairs
Why: The last three come straight from Example 3's table of values. Notice that doubling the time does not halve the value — four years leaves about sixty per cent and eight years about thirty-six, which is sixty per cent of sixty per cent.
Socratic
Eight years of twelve per cent is not ninety-six per cent.
Discussion prompt
Explain why percentage losses cannot be added across periods. Then say what adding them would predict for nine years, and why that is impossible.
Hint: Ask what each year's percentage is taken of.
Answer:
Each year's twelve per cent is taken of that year's value, not of the original price. Once the car has depreciated once, the second year's twelve per cent is twelve per cent of a smaller number, so the dollar losses are not equal and cannot simply be totalled.
Adding them for nine years would give a hundred and eight per cent lost, which would make the car worth less than nothing. Multiplying by 0.88 nine times gives about four thousand seven hundred dollars instead, and no number of multiplications by 0.88 ever reaches nought.
Section
Section 3
Concept
To graph a decay model, make a table of values, plot the points and draw a smooth curve through them. The graph can then be read for estimates at times not in the table.
Example 3 uses t equal to 0, 2, 4, 6 and 8.
Figure (svg): A table of the car's value every two years
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.7 Exponential Decay Functions §8.7, pp. 483-483 — Example 3, Graph an Exponential Decay Model, and its table of values
Picture it
Five points, evenly spaced in time.
Figure (svg): A table of the car's value every two years
The third row is not in the textbook's table but is worth adding yourself. Seeing the losses shrink explains the curve's shape better than the curve does.
Worked example
This is Example 3, part a.
\[ \text{Graph } y = 16\,000(0.88)^t \text{ for } t \text{ from } 0 \text{ to } 8. \]
Take t equal to nought and two
Why: The start and two years on.
\[ 16000, \; 12390 \]
Take t equal to four and six
Why: Two more points.
\[ 9595, \; 7430 \]
Take t equal to eight
Why: The last point.
\[ 5754 \]
Plot and join smoothly
Why: A falling curve that flattens.
Figure (svg): The value of a sixteen thousand dollar car depreciating at twelve per cent a year
\[ (0, 16\,000), \; (2, 12\,390), \; (4, 9595), \; (6, 7430), \; (8, 5754) \]
Verify: check the gaps between consecutive values
Why: The drops are about 3610, 2795, 2165 and 1676 — each roughly seventy-seven per cent of the one before, which is 0.88 squared. Regularly shrinking gaps are the signature of exponential decay in a table.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.7 Exponential Decay Functions §8.7, pp. 483-483
Sorting
Look at how the values change.
Sort into buckets
Sort each sequence of values by the kind of change it shows.
Dividing each value by the one before is the test. A constant quotient means exponential and a constant difference means linear, and the two are easy to confuse when only three values are shown.
Worked example
Example 3, part b, then check it with the model.
\[ \text{Use the graph to estimate the car's value after } 5 \text{ years, then check.} \]
Find five on the horizontal axis
Why: Between the four-year and six-year points.
\[ t = 5 \]
Read up to the curve
Why: Between 9595 and 7430.
\[ \text{about } 8400 \]
Check with the model
Why: Sixteen thousand times 0.88 to the fifth.
\[ 16 \, 000(0.88) ^{5} \]
Evaluate
Why: About eight thousand four hundred and forty.
\[ 8444 \]
Figure (svg): The value of a sixteen thousand dollar car depreciating at twelve per cent a year
\[ 16\,000(0.88)^5 \approx 8444 \]
Verify: ask whether the halfway guess would have worked
Why: Halfway between 9595 and 7430 is 8513, which is close but too high — the curve bends, so it lies below the straight line joining the two points. The graph reading of 8400 was actually the better estimate.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.7 Exponential Decay Functions §8.7, pp. 483-483
Error analysis
The student estimated the value at five years from the four-year and six-year table entries.
Annotate
On: \( \begin{aligned} \text{halfway between } 9595 \text{ and } 7430 &= \frac{9595 + 7430}{2} \\ &= 8512.5 \\ \text{so } y(5) &= 8513 \end{aligned} \)
Averaging neighbours is fine for a linear model, where it is exact, and always slightly wrong for an exponential one. It is worth knowing which direction the error goes: for a decay curve the average is always too high.
Faded example
Multiply by the factor each step.
Fill in the blanks
16\,000 \;\to\; 14\,080 \;\to\; 12390 \;\to\; 10\,894 \quad \text0.88 ___ \text___
Why: Each entry is 0.88 of the one before, which is the same as computing 16 000 times 0.88 to the power of the year. The step-by-step and the formula give identical tables, which is a useful thing to confirm once.
Hypothesis
Joining two points of a decay curve with a straight line.
Predict first
Compared with the curve, where does the straight line between two of its points lie?
Correct: Above the curve, so averaging overestimates.
\[ \frac{9595 + 7430}{2} = 8513 \quad \text{against} \quad 16\,000(0.88)^5 \approx 8444 \]
Why: A decay curve falls steeply first and then flattens, so between two points it dips below the straight line joining them. Averaging the endpoint values therefore always gives a value slightly too high, as the student's 8513 against the true 8444 shows. The gap widens as the two points move further apart.
Socratic
The percentage lost never changes.
Discussion prompt
Explain why a decay graph is steep at first and flat later. Then say what its steepness has to do with its height.
Hint: Ask how many dollars twelve per cent is at each stage.
Answer:
Twelve per cent of sixteen thousand is one thousand nine hundred and twenty, but twelve per cent of six thousand is only seven hundred and twenty. The same percentage of a smaller value is a smaller loss, so the curve drops fast while the value is large and slowly once it is small.
The steepness is proportional to the height at every point, which is the defining property of exponential functions and the reason the same curve shape appears in growth, decay, cooling, radioactive half-lives and drug concentrations. It is the same observation as in Lesson 8.6, now pointing downwards.
Section
Section 4
Concept
For a model of the form y equals C times b to the t, the base b is the growth or decay factor. The model is exponential growth if b is greater than one and exponential decay if b lies between nought and one.
\[ y = Cb^t: \quad b > 1 \text{ growth}, \quad 0 < b < 1 \text{ decay} \]
Growth replaces b by one plus r, and decay by one minus r.
Figure (svg): A number line showing which values of b give growth and which give decay
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.7 Exponential Decay Functions §8.7, pp. 484-484 — the comparison of growth and decay models preceding Example 4
Picture it
Left of one falls, right of one rises.
Figure (svg): A number line showing which values of b give growth and which give decay
This single line replaces the whole classification. No conversion is needed — only a comparison with one.
Worked example
This is Example 4 from the textbook.
\[ \text{Classify } y = 30(1.2)^t \text{ and } y = 30\left(\tfrac{3}{5}\right)^t \text{ as growth or decay.} \]
Look at the first base
Why: 1.2 is greater than one.
Name its factor
Why: The growth factor is the base.
\[ 1.2 \]
Look at the second base
Why: Three fifths is 0.6, which is below one.
Name its factor
Why: The decay factor is the base.
\[ \frac{3}{5} \]
Figure (svg): A number line showing which values of b give growth and which give decay
\[ 1.2 > 1 \Rightarrow \text{growth}; \quad 0 < \tfrac{3}{5} < 1 \Rightarrow \text{decay} \]
Verify: evaluate each after one period
Why: The first gives thirty-six, up from thirty, and the second gives eighteen, down from thirty. The classification and the arithmetic agree, and one substitution is all it takes to confirm it.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.7 Exponential Decay Functions §8.7, pp. 484-484
Sorting
Compare each base with one.
Sort into buckets
Sort each model by what it describes.
These are Guided Practice 5 to 8 together with the lesson's two running examples. Only the base matters — the number in front changes the starting height and never the direction.
Worked example
Three fifths as a percentage loss.
\[ \text{What decay rate does a factor of } \tfrac{3}{5} \text{ represent?} \]
Write the factor as a decimal
Why: Three divided by five.
\[ 0.6 \]
Set it equal to one minus r
Why: The definition of the decay factor.
\[ 1 - r = 0.6 \]
Solve for r
Why: Subtract 0.6 from one.
\[ r = 0.4 \]
Write it as a percentage
Why: Multiply by a hundred.
Figure (svg): The solution to Worked example recover the rate from the factor shown as a ladder of expressions, one row per algebraic move
\[ 1 - r = 0.6 \Rightarrow r = 0.4 = 40\% \]
Verify: check the two numbers sum to one
Why: Keeping sixty per cent and losing forty per cent account for the whole, and thirty times 0.6 is eighteen, which is thirty less forty per cent of thirty. Both readings describe the same step.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.7 Exponential Decay Functions §8.7, pp. 484-484
Trap
\[ y = 30\left(\tfrac{3}{5}\right)^t \]
Report the decay rate as three fifths
Why: The three fifths is the number in the model, so it looks like the rate.
Three fifths is the decay factor — the fraction kept. The rate is what is lost, which is two fifths, or forty per cent.
\[ 1 - r = \tfrac{3}{5} \Rightarrow r = \tfrac{2}{5} = 40\% \]
Subtract the factor from one to recover the rate
Why: The factor sits in the model and the rate does not.
The same distinction as growth factor against growth rate in Lesson 8.6, with the subtraction pointing the other way.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Growth | Decay | |
|---|---|---|
| The model | y = C(1 + r)^t | y = C(1 - r)^t |
| The base | greater than 1 | between 0 and 1 |
| The curve | rises | falls, flattening towards zero |
One sign in the first row produces every other difference. That is worth noticing, because it means nothing genuinely new was learnt in this lesson beyond a minus.
Faded example
Compare with one, then read off the base.
Fill in the blanks
In y = 5(0.2)^t the base is below one, so the model shows decay, and its decay factor is 0.2.
Why: A base of 0.2 keeps a fifth of the quantity each period, which is a decay rate of eighty per cent — a very fast decline. The five in front only sets the starting value.
Socratic
A base below one decays whatever number sits in front of it.
Discussion prompt
Explain why C plays no part in deciding growth or decay. Then say what would happen if C were negative.
Hint: Ask which number gets raised to a power.
Answer:
Only the base is raised to the power of t, so only the base is applied repeatedly. C is a single multiplication that fixes where the curve starts on the vertical axis, and multiplying every value by the same positive number cannot turn a falling sequence into a rising one.
A negative C would flip the whole graph below the axis, so a decay model would rise towards nought from underneath rather than fall towards it from above. The textbook's models all take C positive, since they describe amounts of things, but the base would still be what decides whether the curve moves towards nought or away from it.
Section
Section 5
Concept
Multiplying a positive number by a factor below one always leaves something positive, so a decay model never reaches nought. It gets arbitrarily close, and the graph flattens towards the horizontal axis.
The car is scrapped; the model carries on.
Figure (svg): A decay curve flattening towards the horizontal axis without touching it
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.7 Exponential Decay Functions §8.7, pp. 482-487 — the decay model's behaviour over long times
Picture it
Close to nought, never nought.
Figure (svg): A decay curve flattening towards the horizontal axis without touching it
Eighty-eight per cent of any positive number is positive, however small the number. That single fact is the whole of this idea.
Worked example
A question the model cannot answer.
\[ \text{Using } y = 16\,000(0.88)^t, \text{ find when the car's value reaches } 0. \]
Try twenty years
Why: Sixteen thousand times 0.88 to the twentieth.
\[ \text{about } 1235 \]
Try forty years
Why: Twice as long again.
\[ \text{about } 95 \]
Try a hundred years
Why: Far beyond any car's life.
\[ \text{about } 0.04 \]
Draw the conclusion
Why: Small, but never nought.
\[ \text{never reaches } 0 \]
Figure (svg): A decay curve flattening towards the horizontal axis without touching it
\[ 16\,000(0.88)^t > 0 \text{ for every } t \]
Verify: say what happens to a real car instead
Why: A real car is sold for scrap at some small positive value, or stops running altogether, so the model stops describing it well after a decade or so. The mathematics and the situation diverge, and the mathematics is not the one that is wrong — it is simply being asked about something outside its range.
Prediction
A base between nought and one.
Predict first
What does the model predict as t grows without bound?
Correct: Values that get arbitrarily close to nought but stay positive.
\[ t = 20: \; 1235 \qquad t = 40: \; 95 \qquad t = 100: \; 0.04 \]
Why: Multiplying a positive number by 0.88 gives another positive number, however many times it is done, so nought is never reached. The values do fall below any bound you name, which is why the graph looks as though it settles on the axis while never actually touching it.
Worked example
A question the model answers well.
\[ \text{Estimate when } y = 16\,000(0.88)^t \text{ reaches } 8000. \]
Try five years
Why: About eight thousand four hundred.
\[ 8444 \]
Try six years
Why: About seven thousand four hundred.
\[ 7430 \]
Narrow between them
Why: Eight thousand lies in that gap.
\[ \text{between } 5\text{ and } 6 \]
Refine
Why: Closer to five and a half.
\[ \text{about } 5.4\text{ years} \]
Figure (svg): The value of a sixteen thousand dollar car depreciating at twelve per cent a year
\[ 16\,000(0.88)^{5.4} \approx 8000 \]
Verify: check that halving takes the same time again
Why: From eight thousand, another five and a half years should give about four thousand, and 16000 times 0.88 to the eleventh is about 4030. Equal times for equal halvings is characteristic of exponential decay, and it is the idea behind half-life.
Trap
Losing 12 per cent a year, the car will be worth nothing after about eight and a half years.
Divide a hundred per cent by twelve per cent
Why: Eight and a bit twelve-per-cents make a whole.
That adds the percentages again, which the graph flatly contradicts: after eight years the car is still worth more than five and a half thousand dollars.
\[ 16\,000(0.88)^t > 0 \text{ for every } t \]
Recognise that a positive number times a positive factor stays positive
Why: No number of multiplications by 0.88 ever gives nought.
The right question is when the value falls below some threshold, not when it reaches nought.
Faded example
The signature of exponential decay.
Fill in the blanks
The car falls to half its value in about 5.4 years, and to a quarter after about 10.8 years in total.
Why: Each halving takes the same time, because halving depends on the ratio rather than on the starting amount. That constant is called a half-life, and the same idea describes radioactive decay and drug clearance.
Elimination
A depreciation model over a car's life.
Eliminate the wrong options
Which question is the model well suited to?
Survives elimination: A
Why: Three years is well inside the period over which cars depreciate at a fairly steady percentage, so the model is on solid ground there. Its weakness is at the extremes, which is where every model's weakness lies.
Socratic
The mathematics carries on indefinitely.
Discussion prompt
Say where a depreciation model stops matching a real car, and why. Then say what should be reported alongside a prediction from one.
Hint: Ask what happens to very old cars.
Answer:
Depreciation is fastest in the first years and steadier afterwards, and eventually a car reaches a floor set by its scrap or parts value rather than by any percentage. Some cars become collectible and rise in value, which no decay model can produce at all.
Any prediction should be reported with the assumption that the twelve per cent rate continues and with the period over which that is plausible. A value at three years is an estimate; a value at fifty years is an arithmetic exercise, and saying which is which is part of using the model honestly.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Growth | Decay | |
|---|---|---|
| 12% a year means | factor 1.12 | factor 0.88 |
| The bracket is | 1 + r | 1 - r |
| Over a long time | grows past any bound | approaches zero without reaching it |
The first row is where most mistakes are made and the last row is where most misreadings are. Between them sits a single sign.
Pattern
Whether the problem asks for a model, a value or a classification, these five moves cover it.
Step five is quick and catches the two commonest errors at once: a factor above one means the rate was added, and a factor equal to the rate means the subtraction was skipped.
OpenStax Intermediate Algebra 2e, §10.2 Evaluate and Graph Exponential Functions §10.2
Check
Subtract the rate from one.
Check your understanding
A 20 000 dollar machine depreciates at 15 per cent a year. What is the model?
Answer: A
Why: Losing fifteen per cent means keeping eighty-five, so the decay factor is 0.85. Testing one year gives seventeen thousand, which is twenty thousand less fifteen per cent of it.
Check
Only the base decides.
Check your understanding
Which model shows exponential decay?
Answer: A
Why: A base of 0.2 lies between nought and one, so the quantity keeps a fifth of itself each period. The small coefficient in option B is a starting value, not a base, and 1.1 makes it growth.
Check
Apply the factor once per year.
Check your understanding
Using y = 16 000(0.88)^t, roughly what is the car worth after 4 years?
Answer: A
Why: 0.88 to the fourth is about 0.5997, and sixteen thousand times that is about 9595 — a shade under sixty per cent of the original price.
Real world
This is the lesson's running situation. You buy a car for 16 000 dollars and expect it to depreciate at 12 per cent a year.
Discussion prompt
Write the model, find its value after eight years, and estimate when it falls below 4000. Then say how far into the future you would be willing to quote a figure.
Hint: Try t equal to ten and t equal to twelve.
Answer:
\[ y = 16\,000(0.88)^t \;\Longrightarrow\; y(8) = 16\,000(0.88)^8 \approx 5754 \]
For four thousand, eleven years gives about four thousand and thirty and twelve years about three thousand five hundred and fifty, so the value drops below four thousand a little after eleven years.
Quoting a figure three or four years out is reasonable, since depreciation over that span is fairly steady. Beyond ten years the real value is dominated by condition, mileage and model rather than by any percentage, and the model's answer becomes an assumption rather than an estimate. Saying so is part of reporting it.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
A quantity decays with factor three fifths. What is its decay rate?
Correct: Two fifths, or 40 per cent.
\[ 1 - r = \tfrac{3}{5} \Rightarrow r = \tfrac{2}{5} = 40\% \]
Why: The factor is what remains and the rate is what is lost, and the two make a whole. A factor of three fifths keeps sixty per cent, so forty per cent is lost each period. Checking with a number settles it: thirty times three fifths is eighteen, and eighteen is thirty less forty per cent of thirty. The same relationship ran through Lesson 8.6 with an addition instead — there the factor was one plus the rate, here it is one minus it, and in both cases the factor is the number that actually appears in the model.
Explain it
They keep writing the decay rate as the base of the model.
Discussion prompt
In no more than four sentences, explain the difference between the decay rate and the decay factor. Then give them the one-line check.
Hint: Lost against kept.
Answer:
A usable answer: the rate is the fraction you lose and the factor is the fraction you keep, so they add to one. Losing twelve per cent leaves eighty-eight, which is why 0.88 and not 0.12 goes into the model.
The check is to work out one period by hand. If your factor turns sixteen thousand into fourteen thousand and eighty you have it right, and if it turns sixteen thousand into one thousand nine hundred and twenty you have used the rate by mistake.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The conversion is fixed by remembering that the rate and the factor sum to one, then testing a single period. Evaluating is fixed by applying the factor once per period rather than adding percentages. Classifying is fixed by comparing the base with one and ignoring everything else. The never-reaching-nought point is fixed by noticing that a positive number times a positive factor stays positive. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page write the growth and decay models side by side with the differing sign circled in both, and label the initial amount, rate, factor and time on each. Underneath, convert three decay rates into factors, showing the division by a hundred and the subtraction from one as separate steps, and beside each write the check that the rate and factor sum to one. In the middle, build the table for sixteen thousand at twelve per cent for eight years in two-year steps, add a third row giving the dollars lost in each interval, and graph the points with a smooth curve. Mark the five-year estimate on the graph and write the exact value beside it. In the lower half, draw a number line from nought to two, shade the decay half and the growth half, and place the six bases from this lesson's classification exercise on it. Finally, in the margin, write one sentence saying why the curve never touches the horizontal axis.
Your third table row should shrink steadily. If those losses come out equal you have subtracted a fixed amount rather than a fixed percentage, which is the linear model and not this one.
Recap
Five things, and the first is one sign away from the last lesson.
| If the question says | Your first move is |
|---|---|
| Depreciates at 12% a year | Decay factor 0.88 |
| Base is between 0 and 1 | Exponential decay |
| Base is greater than 1 | Exponential growth |
| Find the value after 8 years | Substitute 8 for t and evaluate |
| When does it reach zero? | It never does; ask for a threshold instead |
That completes Chapter 8. Chapter 9 turns to radicals and connections to geometry, beginning with square roots and the way they undo the squaring that has appeared throughout this chapter.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 8 Exponents and Exponential Functions — Lesson 8.7 Exponential Decay Functions §8.7, pp. 482-488 — everything on these slides traces back here
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