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

Straight-Line Depreciation and Appreciation: Linear Graphs

F1.2 A1About 70 minutes

Learning goals

I can recognize straight-line (linear) depreciation and appreciation from a description, a table, an equation or a graph.
I can write an equation V = initial value + (rate of change) × t and explain what each part means.
I can read and use a linear graph to find a value at a given time, or the time for a given value.
I can compare two linear depreciation situations and find when their values are equal.

Words to know

straight-line (linear) depreciation
Losing the same number of dollars of value every year. Its graph is a straight line going down.

rate of change
How much the value changes per year. For a straight line it is the slope: rise ÷ run. It is negative for depreciation and positive for appreciation.

initial value
The value at time 0 (the purchase price). On the graph it is the vertical intercept.

horizontal intercept
The point where the graph meets the time axis: the time when the value reaches $0.

salvage value
The value expected at the end of an item's useful life (what it can still be sold for, even as scrap).

useful life
How many years an item is expected to be used.

Hook: the school laptop

A $1 500 laptop is expected to be worth $0 after 5 years of use. A simple way to model this is to say it loses the same amount every year: 1500 ÷ 5 = $300 per year. This is straight-line depreciation, and its graph is a straight line.

You are studying linear relations in the Algebra strand of this course. Today you will see that the same ideas (initial value, rate of change, intercepts) describe how some things gain or lose value.

The equation of straight-line depreciation

V = initial value − (amount lost per year) × t

For the laptop: V = 1500 − 300t, where t is the number of years since purchase and V is the value in dollars.

1500 is the initial value (the vertical intercept).
−300 is the rate of change: the value drops $300 each year (the slope).
The horizontal intercept is where V = 0: 1500 − 300t = 0, so t = 5 years.

t (years)	0	1	2	3	4	5
V ($)	1 500	1 200	900	600	300	0

A table of values for V = 1500 − 300t. The value goes down by $300 every year: the first differences are all −300, a sign of a linear relation.

A straight-line depreciation graph. For every 1 year to the right, the line falls $300: slope = −300 ÷ 1 = −300 dollars per year.

Worked example

Example 1: Use the equation both ways

The laptop's value is V = 1500 − 300t.

(a) Find its value after 3 years. (b) When is it worth $450?

(a) Substitute t = 3: V = 1500 − 300(3) = 1500 − 900 = $600.
(b) Substitute V = 450 and solve: 450 = 1500 − 300t.
Subtract 1500 from both sides: −1050 = −300t.
Divide by −300: t = 3.5. Check: 1500 − 300(3.5) = 1500 − 1050 = 450. ✓

(a) $600; (b) after 3.5 years (3 years and 6 months).

Worked example

Example 2: Find the equation from two values

A bike was worth $900 new and $540 after 3 years. Assume straight-line depreciation. Write an equation and find its value after 5 years.

Rate of change = (540 − 900) ÷ (3 − 0) = −360 ÷ 3 = −120 dollars per year.
Initial value = 900.
Equation: V = 900 − 120t.
After 5 years: V = 900 − 120(5) = 900 − 600 = $300.

V = 900 − 120t; after 5 years the bike is worth about $300.

Straight-line depreciation with a salvage value

Businesses often assume equipment keeps a salvage value at the end of its useful life:

amount lost per year = (cost − salvage value) ÷ useful life

A food-truck oven costs $8 000, is used for 5 years, and can then be sold for $500. It loses (8000 − 500) ÷ 5 = $1 500 per year, so V = 8000 − 1500t for 0 ≤ t ≤ 5.

Linear appreciation

Some things gain the same amount every year. A good example is money earning simple interest: $2 000 at 4% simple interest gains 2000 × 0.04 = $80 a year, so A = 2000 + 80t. Its graph is a straight line going up, with a positive rate of change.

Simple interest is linear appreciation: the same $80 is added every year, so the points lie on a straight line rising from $2 000.

Worked example

Example 3: Two items, one graph

Laptop A: V = 1500 − 300t. Laptop B (a cheaper model): V = 900 − 100t. When are they worth the same amount? Which is worth more after that?

Set the values equal: 1500 − 300t = 900 − 100t.
Add 300t to both sides: 1500 = 900 + 200t.
Subtract 900: 600 = 200t, so t = 3.
Value at t = 3: 1500 − 300(3) = 600, and 900 − 100(3) = 600. ✓
After t = 3, A keeps dropping faster (−$300 a year compared with −$100), so B is worth more.

They are both worth $600 after 3 years. After that, laptop B is worth more.

Laptop A starts higher but has a steeper (more negative) slope, so the lines cross at t = 3. The steeper line loses value faster.

Is straight-line realistic?

Straight-line depreciation is a model. It is simple and is often used in business planning. Many real items, like cars, actually lose more in the first years and less later, which gives a curved graph. You will study that model next lesson.

Hands-on activity

Graph a depreciation model at home

You need: graph paper (or a spreadsheet), a ruler, a pencil, and an item with a known price (for example a bike, a hockey stick, a tablet, a lawn mower or a snowblower).

Choose an item. Estimate how many years it will be useful and what it might sell for at the end (salvage value). Ask a grown-up for their best guess.
Find the amount lost per year: (price − salvage) ÷ years.
Write the equation V = price − (amount per year) × t.
Make a table and a graph. Label the intercept and the slope.
Use your graph to answer: when is it worth half of its price?

Model answer: snowblower $1 200, 8 years, salvage $200. Amount per year = (1200 − 200) ÷ 8 = $125. V = 1200 − 125t. Half the price is $600: 1200 − 125t = 600 gives t = 4.8 years.

Just for fun

Why did the straight-line graph refuse to go to the party? It heard everyone there was going off on a tangent.

Your turn

For V = 1200 − 150t, what is the initial value and what is the rate of change? What do they mean?

Find the value after 4 years for V = 1200 − 150t.

When is the item in the previous question worth $0?

Use the laptop graph (V = 1500 − 300t) to find the value after 2 years.

A bike is worth $900 new and $540 after 3 years (straight-line). Write its equation.

An $8 000 oven has a salvage value of $500 after 5 years. What is it worth after 2 years (straight-line)?

A collectible's value is V = 150 + 25t. Is this appreciation or depreciation? Find its value after 6 years.

More practice: warm-up

V = 2000 − 250t. Find V when t = 2.

Is the slope of a depreciation line positive or negative?

What is the initial value of V = 800 − 100t?

Find V when t = 4 for V = 800 − 100t.

Does a linear appreciation graph go up or down from left to right?

More practice: on your own

A phone costs $1 100 and is expected to be worth $200 after 3 years (straight-line). Write the equation and find the value after 2 years.

Item A: V = 2400 − 400t. Item B: V = 1600 − 200t. When are they worth the same, and what is that value?

$2 000 earns 4% simple interest: A = 2000 + 80t. When will it be worth $2 400?

A straight-line graph passes through (0, 24 000) and (6, 6 000). Find the rate of change and the value at t = 4.

The laptop loses $300 a year. What percent of its original value is that? What percent of its value at the start of year 4 ($600) is it?

Challenge

When is the $1 500 laptop (V = 1500 − 300t) worth 25% of its original price?

A $1 200 snowblower has a salvage value of $200 after 8 years (straight-line). During which year does it first become worth less than half its price?

A $30 000 car is modelled two ways: straight-line, losing $3 000 a year, or losing 15% of its value each year. Which model gives the lower value after 2 years? What does this suggest?

Check your understanding

Quiz

In V = 1500 − 300t, what does 300 represent?

What is the value after 3 years if V = 1500 − 300t?

A straight-line depreciation graph is…

A bike is worth $900 new and $540 after 3 years (straight-line). What is the rate of change?

When does V = 800 − 100t reach $0?

Which situation is an example of linear appreciation?

For parents and teachers

Big idea: F1.2 asks students to use graphs of appreciation and depreciation to answer questions. This lesson uses the straight-line model, which connects directly to linear relations in the Algebra strand of this course (initial value, rate of change, intercepts, solving linear equations). It also builds A1 (representing, connecting and reasoning).

Common difficulties:

Sign errors when finding the rate of change. A loss is negative.
Solving for t. Students may subtract in the wrong order. Encourage substituting the answer back to check.
Confusing the total change with the rate per year (−360 over 3 years is −120 per year).

How to help: Pick something your family owns and estimate its useful life and resale value together. Have your teen write the straight-line equation and explain what the numbers mean. Remind them that straight-line is a model; real values can differ.