Four one-line stories. Name the function family:
Model selection is translation: behavior described in words → feature owned by a family. The exam gives the story; you supply the family and defend the choice with rates of change.
With equally spaced inputs, run the diagnostics from Lesson 2:
| Pattern in outputs | Model |
|---|---|
| constant first differences | linear |
| constant second differences | quadratic |
| constant third differences | cubic |
| roughly constant ratios | exponential (Unit 2) |
| approaches a fixed value | rational or exponential (asymptote) |
Real data are noisy — the honest claim is "approximately constant second differences, so a quadratic model is appropriate." (Hedged language earns points; certainty about noisy data loses them.)
Structural clues in the story:
Three construction routes, by what you're given:
a(x − h)² + k; zeros → a(x − r₁)(x − r₂)…; then one more point pins a.a; systems of equations for the rest).[GRAPH: Scatterplot of rise-then-fall data points (days 1–9 vs. attendance) with a concave-down parabola fit through them; the parabola extended as a dashed curve past day 13 dipping below the x-axis, region labeled "model breakdown — negative attendance".]
Problem: The average cost per phone of producing x phones is A(x) = (50,000 + 30x)/x. Why is a rational model structurally right here, and what does its horizontal asymptote mean?
Solution: Average cost is total cost ÷ quantity — a ratio of two functions of x, which is what a rational function is. Degrees are equal (1 = 1), so lim x→∞ A(x) = 30/1 = 30: as production grows, average cost per phone approaches $30 (the per-unit cost, once the $50,000 fixed cost is spread thin). The asymptote is the floor: no production level makes the average dip below $30.
Interpretation: "Per," "average," and "concentration" are rational-function trigger words.
Problem: A fountain's water arc peaks 8 feet above the ground at a horizontal distance of 3 feet from the nozzle, and the water lands 7 feet from the nozzle. Model the arc's height h as a function of horizontal distance x.
Strategy: Vertex given → vertex form; landing point pins a.
Solution: Vertex (3, 8): h(x) = a(x − 3)² + 8. Landing at (7, 0):
0 = a(16) + 8 → a = −1/2
h(x) = −(1/2)(x − 3)² + 8
Sanity: a < 0 ✓ (arc opens downward); nozzle height h(0) = −4.5 + 8 = 3.5 ft ✓ plausible.
Interpretation: Choose the form the givens hand you — a vertex means vertex form; don't fight your way there from standard form.
Problem: Values: f(1) = 5, f(2) = 12, f(3) = 23, f(4) = 38. Find a model and predict f(6).
Solution: Δ = 7, 11, 15; Δ² = 4, 4 → quadratic with a = Δ²/(2h²) = 4/2 = 2. Then f(x) = 2x² + bx + c: f(1) = 2 + b + c = 5 and f(2) = 8 + 2b + c = 12 → subtract: 6 + b = 7 → b = 1, c = 2.
f(x) = 2x² + x + 2 f(6) = 72 + 6 + 2 = 80
Check f(4): 32 + 4 + 2 = 38 ✓
Interpretation: Exact data → algebra recovers the model precisely; regression is for noise.
Problem: A biologist tracks an invasive plant's coverage (m²): (week, area) = (0, 2.1), (2, 4.8), (4, 12.2), (6, 23.9), (8, 40.1). Fit a quadratic model and estimate coverage in week 9.
Solution (workflow):
1. Enter weeks in L1, areas in L2.
2. QuadReg → A(t) ≈ 0.559t² + 0.284t + 2.071 (coefficients to three decimals; store the regression equation to Y1 unrounded).
3. Evaluate Y1(9) — using full precision — ≈ 49.9 m².
Write-up: "Using quadratic regression, A(t) ≈ 0.559t² + 0.284t + 2.071. A(9) ≈ 49.9, so the model predicts about 49.9 m² of coverage in week 9."
Interpretation: The graded moves: three-decimal coefficients, full-precision evaluation, contextual sentence with units. (Week 9 is just past the data's edge — near-extrapolation is acceptable with a caveat; week 30 would not be.)
1. (C). Single peak with symmetric rise and fall → one extremum → quadratic. (D) would need more turning points; (B) needs asymptotic behavior.
2. (A). Constant third differences ↔ degree 3. (B) is second differences.
3. (D). Average cost = total ÷ quantity: a ratio → rational (Example 1). (A) linear; (B) quadratic; (C) exponential.
4. (A). From Example 3: f(x) = 2x² + x + 2, f(6) = 80. (B)/(D) arise from wrong b or c; (C) from a = 2.5.
B(t) = 12t + 340 (books, t in months). The 12 means5. (B). Slope of a linear model = constant rate: 12 books per month. (A) confuses slope with intercept (340).
6. (C). n extrema needs degree ≥ n + 1 → at least 4. (A) forgets the +1; (B)/(D) overshoot the minimum.
7. (D). Week 20 is double the data window; the impossible prediction signals extrapolation failure, not a computing error. (A) trusts the model over reality; (B)/(C) misdiagnose — the quadratic may fit weeks 1–10 perfectly well.
8. (B). "Approaches but never reaches 21" is a horizontal asymptote at y = 21. (A) would mean the cocoa hits 21°C when the variable equals... a zero is an output of 0; (C) confuses axes.
P(x) = a(x − 200)(x − 1400) requires9. (A). Rise-peak-fall = concave down → a < 0. The zeros alone don't fix the sign (that's the trap in (D)) — but the described shape does.
h(t) = −4.9t² + 24t + 1.2 meters. The FIRST time the ball reaches a height of 20 meters is approximately10. (B). Solve −4.9t² + 24t + 1.2 = 20 → 4.9t² − 24t + 18.8 = 0 → t = (24 ± √(576 − 368.48))/9.8 = (24 ± √207.52)/9.8 → t ≈ 0.979 or 3.919. First time: 0.979 s. (A) is the second crossing (on the way down); (C) is the vertex time (24/9.8); (D) halves the correct answer.
11. (D). Δoutput/Δinput = liters per minute. (B) inverts; (C) multiplies.
12. (FRQ-style, 6 points) (i) [2 pts] First differences: 4.5, 6.7, 8.9, 11.4 — not constant, so linear is inappropriate; second differences: 2.2, 2.2, 2.5 — approximately constant, consistent with a quadratic. (Hedged language for noisy data.) (ii) [2 pts] QuadReg: A(t) ≈ 1.143t² + 3.289t + 10.026 (three decimals; store full precision). (Any correctly executed regression on these data earning coefficients within rounding of these values receives credit.) (iii) [2 pts] A(6) ≈ 70.9 m² (using stored precision). Limitation: week 6 lies outside the observed weeks 0–4, so this extrapolation assumes the growth pattern continues unchanged — algae growth could saturate (the lake is finite), making the quadratic unreliable beyond the data.
12. (FRQ-style) 📱 The table shows a lake's algae coverage:
| week t | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| area (m²) | 10 | 14.5 | 21.2 | 30.1 | 41.5 |
(i) Explain, using ratios or differences, why a quadratic model is more appropriate than a linear one. (ii) Use quadratic regression to find a model A(t); write coefficients to three decimal places. (iii) Use the model to predict coverage in week 6, and state one limitation of that prediction.
Daily attendance at a county fair, in hundreds of visitors, was recorded on five days:
| day d | 1 | 3 | 5 | 7 | 9 |
|---|---|---|---|---|---|
| attendance N (hundreds) | 36.5 | 60.5 | 72.5 | 72.5 | 60.5 |
(a) (i) Find the average rate of change of N over the interval [3, 7]. (ii) Interpret its meaning in context, with units.
(b) (i) Explain why a quadratic model is appropriate for these data, using rates of change over equal-length intervals. (ii) Find a quadratic regression model for N(d).
(c) (i) Use the model to predict attendance on day 11. (ii) The fair's organizers want to use the model to plan for day 14. Explain one reason the model should not be used for that prediction.
(a) [2 pts] (i) [1 pt] (N(7) − N(3))/(7 − 3) = (72.5 − 60.5)/4 = 3 hundred visitors per day. (ii) [1 pt] From day 3 to day 7, daily attendance increased by an average of about 300 visitors per day. (Value + units + context.)
(b) [2 pts] (i) [1 pt] Over equal 2-day intervals, the rates of change are (60.5−36.5)/2 = 12, (72.5−60.5)/2 = 6, (72.5−72.5)/2 = 0, (60.5−72.5)/2 = −6 hundred visitors/day: the rate of change decreases by a constant 6 each interval. Rates changing at a constant rate is the signature of a quadratic. (ii) [1 pt] Quadratic regression: N(d) = −1.5d² + 18d + 20 (here the data fit exactly; on the real exam report a ≈ −1.500, b ≈ 18.000, c ≈ 20.000 and store the full-precision model).
(c) [2 pts] (i) [1 pt] N(11) = −1.5(121) + 18(11) + 20 = −181.5 + 198 + 20 = 36.5 → about 3,650 visitors on day 11. (ii) [1 pt] Any one: the model predicts N(14) = −294 + 252 + 20 = −22 (negative attendance — impossible, the model breaks down); day 14 is far outside the observed days 1–9, so extrapolation is unreliable; the fair may not even run 14 days. (One clearly stated limitation earns the point.)
1. (C). Single peak with symmetric rise and fall → one extremum → quadratic. (D) would need more turning points; (B) needs asymptotic behavior.
2. (A). Constant third differences ↔ degree 3. (B) is second differences.
3. (D). Average cost = total ÷ quantity: a ratio → rational (Example 1). (A) linear; (B) quadratic; (C) exponential.
4. (A). From Example 3: f(x) = 2x² + x + 2, f(6) = 80. (B)/(D) arise from wrong b or c; (C) from a = 2.5.
5. (B). Slope of a linear model = constant rate: 12 books per month. (A) confuses slope with intercept (340).
6. (C). n extrema needs degree ≥ n + 1 → at least 4. (A) forgets the +1; (B)/(D) overshoot the minimum.
7. (D). Week 20 is double the data window; the impossible prediction signals extrapolation failure, not a computing error. (A) trusts the model over reality; (B)/(C) misdiagnose — the quadratic may fit weeks 1–10 perfectly well.
8. (B). "Approaches but never reaches 21" is a horizontal asymptote at y = 21. (A) would mean the cocoa hits 21°C when the variable equals... a zero is an output of 0; (C) confuses axes.
9. (A). Rise-peak-fall = concave down → a < 0. The zeros alone don't fix the sign (that's the trap in (D)) — but the described shape does.
10. (B). Solve −4.9t² + 24t + 1.2 = 20 → 4.9t² − 24t + 18.8 = 0 → t = (24 ± √(576 − 368.48))/9.8 = (24 ± √207.52)/9.8 → t ≈ 0.979 or 3.919. First time: 0.979 s. (A) is the second crossing (on the way down); (C) is the vertex time (24/9.8); (D) halves the correct answer.
11. (D). Δoutput/Δinput = liters per minute. (B) inverts; (C) multiplies.
12. (FRQ-style, 6 points) (i) [2 pts] First differences: 4.5, 6.7, 8.9, 11.4 — not constant, so linear is inappropriate; second differences: 2.2, 2.2, 2.5 — approximately constant, consistent with a quadratic. (Hedged language for noisy data.) (ii) [2 pts] QuadReg: A(t) ≈ 1.143t² + 3.289t + 10.026 (three decimals; store full precision). (Any correctly executed regression on these data earning coefficients within rounding of these values receives credit.) (iii) [2 pts] A(6) ≈ 70.9 m² (using stored precision). Limitation: week 6 lies outside the observed weeks 0–4, so this extrapolation assumes the growth pattern continues unchanged — algae growth could saturate (the lake is finite), making the quadratic unreliable beyond the data.
🎯 Exam tip: FRQ 1 has a rhythm: compute → interpret → model → predict → doubt. The first four are mechanical if you've practiced the calculator workflow. The fifth — articulating a limitation — is free points for anyone who memorizes the two magic sentences: "outside the observed data" and "the model predicts impossible values beyond x = …".