Introduction to Linear Polynomials · Lesson 4 of 7
Linear Growth and Linear Decay
“Some quantities rise, some fall, and linear graphs refuse to overreact.”
• Distinguish linear growth from linear decay. • Identify the initial value and constant rate of change in a linear model. • Build and interpret value tables. • Write equations for real growth and decay situations. • Explain why constant change over equal intervals produces linear behaviour.
A linear pattern is a pattern in which the quantity changes by the same amount each time the stage number increases by one. This means that as we move from one stage to the next, the difference between consecutive values remains constant. The pattern may be shown using shapes, numbers, objects, or a table, but the important idea is the way the quantity changes.
So, when we study a linear pattern, we should not focus only on how the picture looks. We should look carefully at the values and ask: how much is being added or subtracted at each step? If this change stays the same throughout the pattern, then the relationship is linear. This constant change is what allows us to describe the pattern using a linear expression.
Linear growth occurs when a quantity increases by a fixed amount over equal intervals.
Linear decay occurs when a quantity decreases by a fixed amount over equal intervals.
Linear Growth: Journey Cost
Suppose the cost of a journey is C(d) = 100 + 60d, where d is distance in kilometres. The starting cost is ₹100. Every extra kilometre adds ₹60.
| Distance d (km) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Cost C(d) ₹ | 100 | 160 | 220 | 280 | 340 | 400 |
Problem
What is the cost for 15 km? For what distance is the cost ₹700?
- 1.For 15 km: C(15) = 100 + 60(15) = 1000.
- 2.For a cost of ₹700: 100 + 60d = 700.
- 3.Then 60d = 600, so d = 10.
- 4.Thus the cost is ₹1000 for 15 km, and ₹700 for 10 km.
Linear Decay: Water Level
Suppose a tank starts with water height 3 m and loses 0.5 m of height each month. The height after t months is h(t) = 3 − 0.5t.
| Month t | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Height h (m) | 3 | 2.5 | 2 | 1.5 | 1 |
Problem
Find the height after 5 months.
- 1.Substitute t = 5.
- 2.h(5) = 3 − 0.5(5).
- 3.h(5) = 3 − 2.5 = 0.5 m.
Initial Value and Constant Change
Every model of the form y = ax + b has two key ingredients. The value b is the value when x = 0, so it is the starting or initial value. The coefficient a is the amount by which y changes when x increases by 1.
| Model | Initial value | Change per unit | Type |
|---|---|---|---|
| 100 + 60d | 100 | +60 | Growth |
| 3 − 0.5t | 3 | −0.5 | Decay |
| 1.75 + 0.5t | 1.75 | +0.5 | Growth |
| 10000 − 800t | 10000 | −800 | Decay |
Plant Growth
Problem
A plant is 1.75 ft tall and grows 0.5 ft each month. Write a model and find its height after 7 months.
- 1.Initial height = 1.75 ft.
- 2.Monthly growth = 0.5 ft.
- 3.So h(t) = 1.75 + 0.5t.
- 4.h(7) = 1.75 + 3.5 = 5.25 ft.
Depreciation as Linear Decay
Problem
A phone costs ₹10,000 and loses ₹800 in value every year. Write the model and find its value after 3 years.
- 1.Initial value = ₹10,000.
- 2.Annual change = −₹800.
- 3.So v(t) = 10000 − 800t.
- 4.v(3) = 10000 − 2400 = ₹7600.
Population Growth
Problem
A village begins with population 750 and gains 50 people each year. Find the population after 6 years.
- 1.Model: P(t) = 750 + 50t.
- 2.P(6) = 750 + 300 = 1050.
- 3.So the population after 6 years is 1050.
When Does a Linear Quantity Reach Zero?
Problem
A prepaid balance of ₹600 decreases by ₹15 each day. After how many days does it become zero?
- 1.Model: b(x) = 600 − 15x.
- 2.Set b(x) = 0.
- 3.600 − 15x = 0.
- 4.15x = 600.
- 5.x = 40.
- 6.The balance runs out after 40 days.
Practice Problems
- A plant begins at 1.75 ft and grows 0.5 ft per month. Make a table from month 0 to month 10 and write the equation.
- A phone worth ₹10,000 loses ₹800 per year. Make a table for years 0 to 8 and write the equation.
- A village starts with population 750 and gains 50 people each year. Make a table for years 0 to 10 and write the equation.
- A prepaid balance starts at ₹600 and decreases by ₹15 each day. Write the equation, determine when it becomes zero, and list the first 10 daily values.
Key Takeaways
• Linear growth adds the same amount over equal intervals. • Linear decay subtracts the same amount over equal intervals. • In y = ax + b, b is the initial value and a is the constant rate of change. • Positive a gives growth and negative a gives decay. • Tables, equations and verbal descriptions are different views of the same linear model.
Next, we use two observed data points to find the linear relationship y = ax + b between two variables.