F. Financial Literacy · Financial Decisions: Interest, Borrowing, Appreciation, Depreciation and Budgets

Percent Appreciation and Depreciation: Curved Graphs

F1.2 A1About 70 minutes

Learning goals

I can model appreciation and depreciation by a constant percent with V = P(1 + r)ᵗ or V = P(1 − r)ᵗ.
I can tell from a table whether a value changes by a constant amount (linear) or a constant percent (curved).
I can read a curved appreciation or depreciation graph to answer questions, such as when something is worth half its price.
I can compare a straight-line model with a percent model and explain which fits a situation better.

Words to know

percent (declining-balance) depreciation
Losing the same percent of the current value each year. The dollar loss gets smaller every year.

percent appreciation
Gaining the same percent of the current value each year. The dollar gain gets bigger every year.

growth or decay factor
The number you multiply by each year: 1 + r for appreciation (for example 1.04), 1 − r for depreciation (for example 0.85).

constant ratio
When each value divided by the one before gives the same number. It is the sign of a percent (exponential) model.

first differences
The changes between consecutive values in a table. Constant first differences are the sign of a linear model.

Hook: why cars lose the most at the start

In Lesson 8 you saw that a new car lost $4 500 in its first year but much less later on. A straight line cannot show that, because a straight line loses the same amount every year. A better model says a car loses about the same percent of its value each year, for example 15%. The graph of that model is a curve.

This is the same idea as compound interest (Lesson 3), but used for the value of things. It connects to the non-linear relations you study in the Algebra strand, and you will see these "exponential" relations again in later high-school math courses.

The percent model

Depreciation: V = P(1 − r)ᵗ Appreciation: V = P(1 + r)ᵗ

P = original value
r = percent change per year, as a decimal
t = number of years

A $32 000 car that loses 15% a year keeps 85% of its value each year: V = 32 000(0.85)ᵗ.

t (years)	0	1	2	3	4	5
V ($)	$32 000.00	$27 200.00	$23 120.00	$19 652.00	$16 704.20	$14 198.57
Dollars lost that year	–	$4 800.00	$4 080.00	$3 468.00	$2 947.80	$2 505.63

Each value is 0.85 times the one before. The dollar loss shrinks every year, even though the percent stays at 15%.

A percent-depreciation curve. It falls steeply at first and then flattens out. It gets closer and closer to $0 but never reaches it.

Worked example

Example 1: Answer questions from the model

Use V = 32 000(0.85)ᵗ.

(a) Find the value after 3 years. (b) How much value is lost in the 4th year? (c) During which year does the car become worth less than half its price?

(a) V = 32 000 × 0.85³ = 32 000 × 0.614125 = $19 652.00.
(b) After 4 years: 32 000 × 0.85⁴ = $16 704.20. Lost in year 4: 19 652.00 − 16 704.20 = $2 947.80.
(c) Half the price is $16 000. After 4 years the car is worth $16 704.20 (more than half). After 5 years it is worth $14 198.57 (less than half).
So it drops below half during the 5th year. The graph shows this at about t = 4.3.

(a) $19 652.00; (b) $2 947.80; (c) during the 5th year (at about 4.3 years).

Linear or percent? Look at the table

If the differences (subtract each value from the next) are the same, the model is linear (straight-line).
If the ratios (divide each value by the one before) are the same, the model is a percent model (curved).

For 1000, 900, 800, 700 the differences are all −100: linear. For 1000, 900, 810, 729 the ratios are all 0.9: percent (10% depreciation each year).

Two models for the same car. The percent model drops faster at first, but the straight-line model reaches $0 after 10 years while the percent model still shows some value. The models cross between years 6 and 7.

Worked example

Example 2: Percent appreciation of a home

A home is worth $500 000. Suppose it appreciates 4% per year (an example rate; real home prices can also fall). Find its value after 5 and 10 years, and estimate when it doubles.

Model: V = 500 000(1.04)ᵗ.
After 5 years: 500 000 × 1.04⁵ ≈ 500 000 × 1.216653 ≈ $608 326.45.
After 10 years: 500 000 × 1.04¹⁰ ≈ 500 000 × 1.480244 ≈ $740 122.14.
Doubling: by the Rule of 72, 72 ÷ 4 = 18 years. Check: 500 000 × 1.04¹⁸ ≈ $1 012 908, just over $1 000 000.

About $608 326 after 5 years, $740 122 after 10 years, and it roughly doubles in about 18 years.

Percent appreciation curves upward: each year's gain is bigger than the year before, because 4% is taken of a larger value. The dashed line adds the first year's $20 000 every year; the curve pulls further and further above it.

Which model should I use?

Cars, electronics and many machines: a percent model usually fits better, because the biggest drops come first.
Business planning or simple estimates: straight-line is often used because it is easy.
Homes and investments: a percent model is common, but no model can promise future prices. Always say "if the trend continues".

Hands-on activity

Ratio or difference? Test real data

You need: a grown-up, a used-car website or a phone resale website, a calculator or spreadsheet.

Choose one car or phone model. Find typical selling prices for the same model at 1, 2, 3 and 4 years old (one ad per age is fine).
Make a table of age and price.
Calculate the differences and the ratios. Which are closer to being constant?
Estimate the yearly percent of depreciation from the average ratio. (A ratio of 0.82 means about 18% per year.)
Use your model to predict the price at 6 years old. Do you trust the prediction? Why?

Model answer: prices $24 000, $20 500, $17 200, $14 600. Differences: −3 500, −3 300, −2 600 (not constant). Ratios: 0.854, 0.839, 0.849 (fairly constant, about 0.85), so the percent model fits better: about 15% a year.

Just for fun

Percent depreciation is like a cookie that loses a bite equal to 15% of what is left every time you look away. You keep losing cookie, but you never quite run out.

Your turn

Write an equation for a $1 200 phone that depreciates 25% per year. Find its value after 3 years.

Using the car table, how much value is lost in year 1 and in year 4? Why are they different?

A $500 000 home appreciates 4% a year. What is it worth after 5 years?

Is each table linear or percent? Table A: 1000, 900, 800, 700. Table B: 1000, 900, 810, 729.

Use the car graph or table: during which year is the car first worth less than $20 000?

A hockey card worth $40 appreciates 8% per year. Estimate its value after 5 years.

More practice: warm-up

What do you multiply by each year for 15% depreciation?

What do you multiply by each year for 4% appreciation?

$100 depreciates 10% for one year. What is it worth?

$100 appreciates 10% for two years. What is it worth?

Describe the shape of a percent-depreciation graph.

More practice: on your own

A $900 TV depreciates 20% a year. Find its value after 4 years.

A $450 000 home appreciates 3.5% a year. Find its value after 5 years.

A $20 000 car depreciates 18% a year. Find its value after 3 years and the total dollars lost.

Use the Rule of 72 to estimate how long the $500 000 home (4% a year) takes to double. Check with the formula.

Compare the two car models after 2 years and after 8 years: straight-line (32 000 − 3 200t) and percent (32 000 × 0.85ᵗ).

Challenge

A car worth $32 000 is worth $23 120 two years later. Assuming a constant percent, what is the yearly rate of depreciation?

Will V = 32 000(0.85)ᵗ ever reach $0? Explain.

Home A: $500 000, appreciating 4% a year. Home B: $560 000, appreciating 2% a year. After how many whole years is A first worth more than B?

Check your understanding

Quiz

Which equation models a $20 000 car that loses 15% of its value each year?

What does a percent-depreciation graph look like?

A table shows 800, 720, 648, 583.20. What kind of model is it?

A $500 000 home appreciates 4% a year. About what is it worth after 10 years?

With percent depreciation, the dollars lost each year…

Which item is usually modelled best by percent depreciation?

For parents and teachers

Big idea: F1.2 asks students to identify situations involving appreciation and depreciation and use the graphs to answer questions. Here students meet the percent (exponential) model, which gives a curved graph, and compare it with the straight-line model from the last lesson. This connects to non-linear relations in the Algebra strand of the course and to exponential functions in later high-school math.

Common difficulties:

Using 0.15 instead of 0.85 for 15% depreciation. Ask, "What fraction of the value is kept?"
Expecting the dollar loss to stay the same each year.
Treating predictions as facts, especially for home prices.

How to help: Look up used prices for the same car or phone model at different ages and test whether the ratios are roughly constant (the activity in this lesson). Talk about why no one can promise what a house or a collectible will be worth in the future.