Week 8, Session 3: Function Modeling — Choosing the Right Model

The SAT® asks you to select the right mathematical model and interpret its parameters in real-world contexts. This is equally a reading skill and a math skill — and it tests both simultaneously.

Week 8, Session 3 Learning Goals
By the end of this session you will:

1. Classify a data pattern as linear, exponential, or quadratic by examining differences or ratios.
2. Build the model equation from a context description (initial value, rate, doubling/half-life).
3. Interpret every parameter (slope, intercept, base, exponent) in context with correct units.
4. Solve a break-even problem by setting two model expressions equal.
5. Distinguish between a percentage rate and a multiplicative factor.

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Part 1 — The Classification Decision Tree

Look at equally spaced x-values (or time steps) and examine the y-values:

Step 1: Compute first differences (y₂−y₁, y₃−y₂, ...):

• Constant differences → LINEAR model: y = mx+b.

Step 2: If differences aren't constant, compute ratios (y₂/y₁, y₃/y₂, ...):

• Constant ratios → EXPONENTIAL model: y = a·bˣ.

Step 3: If differences increase by the same amount each time (second differences constant):

• QUADRATIC model: y = ax²+bx+c.

Example 1 — Classify from a table:

First differences: 5, 10, 20 (not constant).
Ratios: 10/5=2, 20/10=2, 40/20=2 (constant ratio = 2).
→ Exponential: y = 5·2ˣ.

Example 2 — Second differences for quadratic:

First differences: 2, 4, 6.
Second differences: 2, 2 (constant!).
→ Quadratic model.

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Part 2 — Building the Model Equation

Linear: y = mx+b.

• Slope m = rate of change (units: [y units per x unit]).
• Intercept b = initial value (value when x=0).

Exponential: y = a·bᵗ.

• a = initial amount (when t=0).
• b = growth factor per period. Growth: b>1. Decay: 0<b<1.
• Rate r relates to factor: b = 1+r (growth) or b = 1−r (decay).
• Doubling time T: b^T = 2, i.e., T = log(2)/log(b).
• Half-life T: b^T = 1/2.

Example 3 — Interpreting parameters:
P(t) = 2000·(1.06)^t (annual, t in years).

• 2000 = initial population.
• 1.06 = growth factor → 6% annual growth rate.
• After 10 years: P(10) = 2000·(1.06)¹⁰ ≈ 3582.

Example 4 — Half-life model:
Substance halves every 4 hours. Start = 200g.
M(t) = 200·(1/2)^(t/4).
After 12 hours (3 half-lives): M(12) = 200·(1/8) = 25g.

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Part 3 — Reading a Compound Exponential

B(t) = 500·(2)^(t/3):

• 500 = initial bacteria count.
• 2 = doubles (factor).
• t/3 = counts the number of 3-hour periods → doubling period = 3 hours.

Rewriting for a different base:
B(t) = 500·(2^(1/3))^t = 500·(1.26)^t (approx. 26% hourly growth).

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Part 4 — Break-Even Analysis

Set two cost/revenue/population models equal and solve.

Example 5:
Company A: Revenue R_A = 5000+200t.
Company B: Revenue R_B = 1000·(1.05)^t.
Break-even: set R_A = R_B and solve (often requires Desmos for non-linear).

Example 6 — Linear break-even:
Plan A costs: C_A = 50+10x.
Plan B costs: C_B = 20x.
Break-even: 50+10x = 20x → 50 = 10x → x = 5.
For x > 5: Plan A is cheaper (lower rate of increase).

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Part 5 — Percentage Rate vs. Multiplicative Factor

"Population grows by 8% per year" → factor = 1.08 (NOT 0.08).
"Population decays by 15% per year" → factor = 0.85 (NOT 0.15).

After n years at rate r: A = A₀(1+r)ⁿ (growth) or A = A₀(1−r)ⁿ (decay).

Compound interest: A = P(1 + r/n)^(nt) where n = compounding periods per year.

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Common Errors Table

Before Week 9, Session 1, answer at least 10 of the 12 lesson-practice questions correctly without notes. For each miss, record one precise cause: linear vs. exponential vs. quadratic classification, initial value identification, growth factor vs. rate confusion, half-life exponent structure, parameter interpretation in context, break-even setup, rewriting the base, or percentage/factor conversion.