Example 1 — identify a linear model
Values 12, 17, 22, 27 have constant difference +5. A linear rule is appropriate for this pattern.
Example 2 — identify a quadratic model
Values 2, 7, 16, 29 have first differences 5, 9, 13 and constant second difference 4. This is evidence of a quadratic relation.
Example 3 — identify an exponential model
Values 80, 96, 115.2, 138.24 multiply by 1.2. This represents 20% growth per interval, so an exponential model is appropriate.
Example 4 — compound interest
$5000 earns 4% p.a. compounded annually for 3 years. A=5000(1.04)³=$5624.32.
Interpretation: under the stated fixed-rate/no-transaction assumptions, the balance after 3 annual compounding periods is $5624.32.
Example 5 — depreciation
A $24,000 asset loses 15% of its current value each year. It retains 85%, so V=24000(0.85)ᵗ. After 2 years: $17,340.
Example 6 — compare linear and exponential change
Start at 100. Model L adds 20 per year: L=100+20t. Model E grows 20% per year: E=100(1.2)ᵗ. At t=5, L=200 while E≈248.83. Equal-looking early change does not imply equal long-term behaviour.
Example 7 — threshold and whole periods
P=200(1.15)ᵗ. At t=4, P≈349.8; at t=5, P≈402.3. If the question asks when P first exceeds 400 at a whole period, the answer is 5 periods.
Example 8 — half-life model
If a quantity halves every 10 years, Q=Q₀(1/2)^(t/10). After 30 years, three half-lives have passed, so Q=Q₀/8.
Example 9 — model selection from data
A table has outputs 4, 9, 16, 25 for consecutive inputs 1,2,3,4. First differences are 5,7,9 and second differences are 2, so a quadratic model is indicated. Calling it exponential merely because growth accelerates would be incorrect.
Example 10 — critique extrapolation
A wildlife model P=600(1.08)ᵗ assumes 8% growth indefinitely. Habitat capacity, food, disease and management can change the rate. The model may be useful over an evidence-supported interval but unreliable far beyond it.