Lesson 20: Algebra II Capstone: Modeling and Review

This course has built a toolkit family by family: complex numbers, composition and inverses, polynomials, rational functions, exponentials and logarithms, sequences and series, matrices, radicals, conics, trigonometry, probability, inequalities, regression and limits. Real problems do not announce which family they belong to. This capstone runs one business through the whole toolkit — modelling, solving, checking, interpreting — and closes with the review map you should use to revise: every earlier lesson, named by the job it does.

The capstone: StreamWave, a subscription startup

Subscribers (exponential). StreamWave launches with 400 subscribers and grows 40% per month: S(t) = 400(1.4)^t, t in months. Revenue (linear in subscribers). At $12 per subscriber per month, R(t) = 12·S(t) = 4,800(1.4)^t. Costs (linear). C(t) = 9,000 + 1,500t (platform fees plus growing support costs). The story so far:

Break-even (solving a mixed equation). Profit changes sign between months 3 and 4. Solving 4,800(1.4)^t = 9,000 + 1,500t exactly is impossible with this course's algebra (the unknown is in both an exponent and a linear term — even calculus complains), so solve numerically: at t = 3.1, revenue ≈ $13,622 vs costs $13,650 (still short); at t = 3.2, revenue ≈ $14,088 vs costs $13,800 (ahead). Break-even at t ≈ 3.1 months. Knowing when to switch from exact methods to numerical ones is a capstone skill in itself.

Average cost (rational). The cost per subscriber, C(t)/S(t), is a rational-style model: at t = 0 it is 9,000/400 = $22.50; at t = 4 it is 15,000/1,537 ≈ $9.76. It falls toward the long-run behaviour set by the race between exponential growth and linear costs — the same "asymptote thinking" as the manufacturer's average cost in Lesson 3, now with an exponential in the denominator.

Pricing (quadratic). A market test says monthly demand at price p is D(p) = 900 − 30p subscribers. Revenue R(p) = p(900 − 30p) = −30p² + 900p is a downward parabola with zeros at p = 0 and p = 30 and vertex at p = 900/(2·30) = $15, where R(15) = 15 · 450 = $6,750. Above $30 the price kills demand entirely — the model itself marks its own domain of sense, 0 ≤ p ≤ 30.

Seasonality (sinusoid). Viewing demand swells in winter: a multiplier 1 + 0.25 sin(2π(t − 2)/12) with period 12 months, peaking (multiplier 1.25) when the sine equals 1 — at t = 5 — and dipping to 0.75 at t = 11. Multiplying model components like this (growth × season) is how professional forecasts are actually assembled.

Reach (logarithm). Cumulative social-media reach grows like L(t) = 5,000 ln(t + 1): at t = 9 months, L = 5,000 ln 10 ≈ 11,513 people — fast early growth flattening relentlessly, the mirror image of the exponential. When a quantity grows by *ratios*, reach for exponentials; when it grows by *diminishing increments*, think logarithm.

The review map: what each tool is for

Study method: how to revise this course

Worked examples teach more than summaries: redo the worked examples in this lesson set with the solutions covered, then the Try it boxes, then mixed sets where you choose the tool — tool selection under uncertainty is the actual exam skill, and it is why this capstone mixes families without labels. For each topic keep one error from the Common Mistakes list that you personally have made, and drill its correction. Finally, verify like the builders of these lessons did: substitute back, test a number, check the units and the story. An answer that fits the equation but not the situation is wrong.

Exam Experience

1. If f(x) = 2x + 1 and g(x) = x², then (g ∘ f)(3) equals

• (A) 19
• (B) 49
• (C) 37
• (D) 9

Answer: (B). f(3) = 7, g(7) = 49. (A) is (f ∘ g)(3) = f(9) = 19 — order, again.

2. The horizontal asymptote of f(x) = (3x − 1)/(x + 2) is

• (A) y = 3
• (B) y = 1/3
• (C) y = −2
• (D) y = 0

Answer: (A). Equal degrees: ratio of leading coefficients 3/1 = 3. (C) is the vertical asymptote — the standard swap.

3. The solution of log_2(x − 1) = 4 is

• (A) x = 9
• (B) x = 17
• (C) x = 15
• (D) x = 8

Answer: (B). x − 1 = 2^4 = 16, so x = 17. Check the domain: x − 1 = 16 > 0 ✓. (C) forgets to add the 1 back.

Constructed response. A promoter sells concert tickets at price p dollars. Attendance is A(p) = 1,200 − 20p, and the event's fixed costs are $8,000, so profit is P(p) = p(1,200 − 20p) − 8,000. (a) Find the ticket price that maximises revenue, and the maximum revenue. (b) Find all prices for which the promoter at least breaks even, to the nearest cent. (c) At the price from (a), how many tickets are sold, and what is the profit? (d) Explain, using the structure of P, why there are two break-even prices rather than one.

Scoring guide. (a) 2 points: 1 for R(p) = −20p² + 1,200p with vertex at p = 1,200/40 = $30; 1 for R(30) = 30 · (1,200 − 600) = $18,000. (b) 3 points: 1 for −20p² + 1,200p − 8,000 ≥ 0 → p² − 60p + 400 ≤ 0 (dividing by −20 and flipping); 1 for the roots (60 ± √2,000)/2 = (60 ± 44.72)/2 ≈ 7.64 and 52.36; 1 for the interval statement: break-even or better for prices between about $7.64 and $52.36. (c) 1 point: A(30) = 600 tickets; profit = 18,000 − 8,000 = $10,000. (d) 1 point: P is a downward-opening quadratic in p — profit is negative for very low prices (too little revenue per ticket to cover fixed costs) and for very high prices (attendance collapses), and positive between its two zeros; two break-even prices are the two crossings of the parabola with the price axis.

Going deeper: a second scenario — the food truck's loan and larder

StreamWave showed growth; a food truck shows borrowing. The truck borrows $10,000 at 1% monthly interest and repays $250 per month. The balance follows the recurrence B_{n+1} = 1.01·B_n − 250: after month 1, $9,850; month 2, $9,698.50; month 3, $9,545.49 (compute: 9,698.50 · 1.01 = 9,795.49, minus 250). The interest portion shrinks as the balance falls. The geometric-series machinery of Lesson 7 sums the whole schedule into a closed form, and setting the balance to zero gives the payoff time n = ln(P/(P − rB_0))/ln(1 + r), where P is the payment and rB_0 = 0.01 · 10,000 = $100 is the first month's interest: n = ln(250/150)/ln(1.01) ≈ 0.5108/0.00995 ≈ 51.3 months — so the loan clears with a final smaller payment in month 52. The recurrence is the honest model; the closed form is Lessons 5 and 7 fused.

Its ingredient costs are a matrix product (Lesson 8): daily usage of buns, patties and cheese across three menu items, times supplier prices — the same S·P shape as the sports shop. And its weekend demand follows a sinusoid over the festival season (Lesson 13) while its review ratings climb logarithmically (Lesson 5). The truck and the startup share no story details and almost every mathematical organ. That is the capstone's real content: the toolkit is small, the costumes are endless, and the professional skill is the recognition, not the recall.

Choosing the tool: three rapid problems

(1) A culture starts with 500 bacteria and doubles every 20 minutes. How many after 2 hours? *Family: exponential (doubling).* Six doublings: 500 · 2^6 = 32,000. (2) Find the sum of the first 50 even numbers. *Family: arithmetic series.* S = (50/2)(2 + 100) = 25 · 102 = 2,550. (3) Solve √(x + 9) = x − 3. *Family: radical equation — isolate, square, check.* Squaring: x + 9 = x^2 − 6x + 9 → x^2 − 7x = 0 → x = 0 or x = 7. Check x = 0: √9 = 3 ≠ −3, extraneous. Check x = 7: √16 = 4 = 7 − 3 ✓. Answer x = 7. Three problems, three families, no labels given — the recognition step in each (doubling → powers of 2; equal steps → series; variable under a root → square and check) is the whole game, and it is trainable exactly this way: mixed sets, family named before computing.