Functions as Models is 10% of the exam by category, but modeling stems also carry much of the Representations weight (30%). The longest word problems on the exam live here — the official samples include a ~75-word continuous-interest problem answered by typing a number rounded to the nearest tenth. Modeling computations (logs, e^x, intersections) are Section 1 territory, where the online graphing calculator is available.
With equally spaced inputs, run the diagnostics:
| Pattern in outputs | Model family |
|---|---|
| constant first differences | linear |
| constant second differences | quadratic |
| approximately constant ratios | exponential |
| repeating cycle around a middle value | sinusoidal |
| approaches a fixed value | needs a horizontal asymptote (exponential approach) |
From context, the structural clues: constant amount per time step → linear; one peak with symmetric rise and fall → quadratic; constant percent per time step → exponential; "repeats every T hours" → sinusoidal; "approaches but never reaches" → horizontal asymptote. "It grows fast" is not evidence — quadratics grow fast too. Differences vs. ratios decide.
One comparison fact the exam tests directly: an increasing exponential eventually exceeds any linear function, regardless of starting values or coefficients. Constant-percent growth applies to an ever-larger base, so its per-step gain grows without bound while a linear function's stays fixed. An early lead by the linear function is always temporary.
All exponential models share one skeleton — initial value × (factor)^(elapsed time / time per factor):
| Phrase | Model |
|---|---|
| "grows p% per unit time" | y = a(1 + p/100)^t |
| "loses p% per unit time" | y = a(1 − p/100)^t |
| "doubles every T units" | y = a·2^(t/T) |
| "half-life T" | y = a·(1/2)^(t/T) |
Two habits keep these honest. First, percent decrease keeps the complement: losing 3.5% per hour means keeping 96.5%, so the base is 0.965 — never 0.035. Second, test t = T: at one doubling time the exponent should equal exactly 1. Writing 2^(12t) for "doubles every 12 years" fails that test instantly (it doubles twelve times a year).
From two points: for y = a·b^t through (0, 12) and (3, 96): a = 12 (the t = 0 value), and b³ = 96/12 = 8, so b = 2, giving y = 12·2^t. The ratio over k steps is b^k — take the k-th root, don't average. Careful: a line through the same two points also exists (here y = 12 + 28t); two points alone never prove the family. The problem's context or additional data does.
Compounding percents multiply, never add. Growth of 6% per year for 10 years multiplies by (1.06)^10 ≈ 1.79 — about 79% total, not 60%. "p% per year for n years = np%" is the exam's favorite wrong answer.
Compound annually: A = P(1 + r)^t
Continuous: A = Pe^(rt)
Solving for time takes a logarithm. The official CLEP sample: money in an account earning 5% compounded continuously — how many years to increase by 50%? Set Pe^(0.05t) = 1.5P:
e^(0.05t) = 1.5 → 0.05t = ln 1.5 → t = ln 1.5 / 0.05 ≈ 8.1 years
This is a numeric-entry item on the real exam: no choices, type 8.1 into the box, rounded to the nearest tenth as instructed. Notes on the routine: divide out P first (the answer never depends on the starting amount); "increase by 50%" means a factor of 1.5, not 0.5 and not 2; and the natural log matches the base e — using log₁₀ without adjusting gives a wrong answer that looks plausible.
Doubling at 8% compounded annually: (1.08)^t = 2 → t = ln 2/ln 1.08 ≈ 9.0 years. (The rule-of-72 estimate 72/8 = 9 confirms it.)
Every periodic context — Ferris wheel, tide, daylight hours — asks for the same four numbers:
Worked model: a Ferris wheel with center 25 m above ground and diameter 40 m turns once every 60 s; riders board at the bottom at t = 0. Midline = center height 25; amplitude = radius 20 (half the 40 m diameter); b = 2π/60 = π/30; bottom start → −cos:
h(t) = 25 − 20 cos(πt/30)
Certify it in ten seconds: h(0) = 25 − 20 = 5 m (boarding platform ✓); h(30) = 25 + 20 = 45 m (top, half a revolution later ✓). The two standard wrong turns: b = period (writing cos(60t)) instead of 2π/period, and midline = ground level instead of hub height.
[GRAPH: One cycle of h(t) = 25 − 20cos(πt/30): t-axis 0 to 60 s, h-axis 0 to 50 m. Curve starts at (0, 5), rises through (15, 25), peaks at (30, 45), falls through (45, 25), returns to (60, 5). Midline y = 25 dashed; amplitude arrows ±20; period bracket 60 s.]
Every parameter answers to its units. In a tide model D(t) = 6 + 3cos(π(t − 2)/6) with D in meters and t in hours: 6 is the average water depth in meters (the midline — not the depth at t = 0, which is 7.5 m); 3 is the amplitude (high tide 9 m, low tide 3 m); the period is 2π/(π/6) = 12 hours; the peak sits at t = 2. In a linear model B(t) = 12t + 340, the 12 is a rate — "12 units per month" — and 340 is the starting value.
Models also have expiration dates. A quadratic fit to sales data from weeks 1–10 that predicts −1,200 units in week 20 hasn't malfunctioned — it has been extrapolated past its supported region. The right conclusion is "not appropriate for week-20 predictions," not "sales will be negative" (trusting the model over reality) and not "the regression was wrong" (the fit to weeks 1–10 may be excellent). Exponential models expire too: unconstrained percent growth eventually exceeds any physical capacity, and cooling objects approach room temperature — a horizontal asymptote at the ambient value, which is what "approaches but never reaches 21°C" demands of a model.
Q1. B. A constant ratio between consecutive outputs over equal input steps is the defining property of exponential growth — each step multiplies by the same factor (here ≈ 1.15, i.e., 15% growth per step). A) would show constant differences, not constant ratios. C) would show constant second differences. D) would show outputs cycling around a middle value, not steadily multiplying. E) rational models are signaled by asymptote behavior or "ratio of quantities" contexts, not constant output ratios. Fix: differences constant → linear; second differences constant → quadratic; ratios constant → exponential.
Q2. B. 24 hours is 24/8 = 3 half-lives: 400 → 200 → 100 → 50 mg; equivalently 400·(1/2)^(24/8) = 400/8 = 50. A) stops after two halvings, miscounting 24/8. C) runs four halvings. D) divides 400 by 3, treating half-life decay as division by the number of half-lives. E) halves only once. Fix: count half-lives as elapsed time ÷ half-life, and halve exactly that many times — the exponent t/T does the counting for you.
Q3. D. Doubling every 12 years needs the factor 2 with exponent t/12: at t = 12 the exponent is exactly 1, one doubling ✓. A) has exponent 12t — the population would double twelve times in the first year. B) is a half-life model; the population would shrink. C) adds the growth instead of multiplying, and at t = 0 gives 5,001, not 5,000. E) uses 12 as the base and 2 as the time scale, swapping the roles of the two numbers. Fix: "doubles every T" → base 2, exponent t/T; test t = T and demand exponent 1.
Q4. B. Pe^(0.05t) = 1.5P → e^(0.05t) = 1.5 → t = ln(1.5)/0.05 ≈ 0.4055/0.05 ≈ 8.1 years. This mirrors the official numeric-entry sample: the P cancels, and "increase by 50%" means factor 1.5. A) uses simple-interest reasoning (50%/5% per year = 10 years), ignoring compounding. C) computes log₁₀(1.5)/0.05 ≈ 3.5 — wrong logarithm base for base e. D) is ln(2)/0.05 ≈ 13.9, solving for doubling instead of a 50% increase. E) is ln(1.5)/0.01 ≈ 40.5, a decimal slip converting 5% to 0.01. Fix: "increase by p%" → target factor 1 + p/100; with e^rt, always invert with the natural log, and write r as a decimal before dividing.
Q5. A. a = f(0) = 12; then b³ = 96/12 = 8 gives b = 2, so f(t) = 12·2^t. B) uses the 3-step ratio 8 as the 1-step base, skipping the cube root. C) takes a square root of 8's neighbor 16, or halves the exponent — either way the wrong root for a 3-step gap. D) anchors the model at the wrong point, starting from 96 and decaying. E) is the line through the same two points — it matches both data values but is not of the required exponential form a·b^t. Fix: the ratio across k steps is b^k — take the k-th root; and remember two points fit a line just as well, so honor the family the problem specifies.
Q6. E. Losing 3.5% per hour means keeping 96.5%: b = 1 − 0.035 = 0.965. A) uses the loss rate itself as the base — the model would collapse by 96.5% every hour. B) misplaces the decimal on that same error. C) is 3.5% growth, the sign reversed. D) uses the raw percent number as a base, growing 250% per hour. Fix: b = 1 + (signed rate as a decimal); decay bases sit slightly below 1, not near 0.
Q7. C. Midline = center height 25; amplitude = radius 20 (half the 40 m diameter); b = 2π/60 = π/30; boarding at the bottom → −cos. Check: h(0) = 5 (platform), h(30) = 45 (top) ✓. A) boards at the top (+cos starts at the max: h(0) = 45). B) swaps midline and amplitude, giving impossible negative heights (minimum −5 m). D) boards at the midline (−sin starts at 25 m heading down), not the bottom. E) uses the full diameter as the amplitude, sending the rider 15 m underground. Fix: midline = hub height, amplitude = radius (half the diameter), b = 2π/period, and −cos for a bottom start — then verify t = 0 and the half-period.
Q8. D. In a·cos-form models, the constant added outside is the midline — the average value the depth oscillates around: (max + min)/2 = (9 + 3)/2 = 6 m. A) the maximum is midline + amplitude = 9 m. B) the time between high tides is the period, 2π/(π/6) = 12 hours. C) the amplitude is the 3, not the 6. E) the depth at midnight is D(0) = 6 + 3cos(−π/3) = 7.5 m — the midline is not the initial value unless the model starts at a midline crossing. Fix: outside-added constant = midline = average value; check any "value at t = 0" claim by actually evaluating the model at 0.
Q9. E. "Approaches 21 but never reaches it" is precisely a horizontal asymptote at y = 21 — the long-run value of an exponential-decay-toward-ambient model. A) a zero at 21 would mean the temperature output hits 0 when something equals 21 — the wrong feature entirely. B) a vertical asymptote would make the temperature blow up near some time, not level off. C) a vertex describes a quadratic's turning point; the coffee never turns around and reheats. D) the y-intercept is the starting temperature — hot coffee starts well above 21. Fix: "approaches / levels off / never reaches" → horizontal asymptote at the approached value; "starts at" → y-intercept.
Q10. A. g grows by 10% of an ever-larger amount, so its per-step gain increases without bound, while f gains a fixed 5,000 per step; once g's gain passes 5,000, g overtakes f permanently. This holds for any increasing exponential vs. any linear function. B) confuses "ahead now" with "ahead always" — the large coefficient only delays the crossover. C) their growth rates never equalize; the exponential's rate keeps increasing. D) reverses the true long-run comparison. E) starting values shift when the crossover happens, never whether it happens. Fix: family beats coefficients in the long run — exponential (base > 1) eventually exceeds any linear function, regardless of starting values.
Q11. C. Solve 300(1.08)^t = 600 → (1.08)^t = 2 → t = ln 2/ln 1.08 ≈ 9.0 years (and 1.08⁹ ≈ 1.999 confirms it; the rule-of-72 estimate 72/8 = 9 agrees). A) computes 50/8, dividing the percent gain by the rate without logarithms. B) misapplies a rule-of-83 style estimate or slips on the log ratio. D) is the simple-interest answer: 300 + 300(0.08)t = 600 gives t = 12.5, ignoring compounding. E) doubles the doubling-time estimate. Fix: doubling under annual compounding → t = ln 2/ln(1 + r); simple-interest arithmetic always overestimates the time because it ignores growth on growth.
Q12. C. Week 20 is double the fitted range; an impossible prediction (negative sales) is the signature of extrapolation beyond the data's support — the model may fit weeks 1–10 perfectly and still fail out there. A) trusts the model over reality; negative unit sales cannot occur. B) an absurd extrapolated value says nothing about the regression arithmetic, which is checked against the fitted weeks. D) a linear fit could extrapolate just as badly (or worse) — family substitution doesn't cure extrapolation. E) impossible predictions don't impugn the observed data. Fix: judge a model only inside (or near) its fitted range; impossible outputs beyond that range mean "stop extrapolating," not "start over."
Q1. B. A constant ratio between consecutive outputs over equal input steps is the defining property of exponential growth — each step multiplies by the same factor (here ≈ 1.15, i.e., 15% growth per step). A) would show constant differences, not constant ratios. C) would show constant second differences. D) would show outputs cycling around a middle value, not steadily multiplying. E) rational models are signaled by asymptote behavior or "ratio of quantities" contexts, not constant output ratios. Fix: differences constant → linear; second differences constant → quadratic; ratios constant → exponential.
Q2. B. 24 hours is 24/8 = 3 half-lives: 400 → 200 → 100 → 50 mg; equivalently 400·(1/2)^(24/8) = 400/8 = 50. A) stops after two halvings, miscounting 24/8. C) runs four halvings. D) divides 400 by 3, treating half-life decay as division by the number of half-lives. E) halves only once. Fix: count half-lives as elapsed time ÷ half-life, and halve exactly that many times — the exponent t/T does the counting for you.
Q3. D. Doubling every 12 years needs the factor 2 with exponent t/12: at t = 12 the exponent is exactly 1, one doubling ✓. A) has exponent 12t — the population would double twelve times in the first year. B) is a half-life model; the population would shrink. C) adds the growth instead of multiplying, and at t = 0 gives 5,001, not 5,000. E) uses 12 as the base and 2 as the time scale, swapping the roles of the two numbers. Fix: "doubles every T" → base 2, exponent t/T; test t = T and demand exponent 1.
Q4. B. Pe^(0.05t) = 1.5P → e^(0.05t) = 1.5 → t = ln(1.5)/0.05 ≈ 0.4055/0.05 ≈ 8.1 years. This mirrors the official numeric-entry sample: the P cancels, and "increase by 50%" means factor 1.5. A) uses simple-interest reasoning (50%/5% per year = 10 years), ignoring compounding. C) computes log₁₀(1.5)/0.05 ≈ 3.5 — wrong logarithm base for base e. D) is ln(2)/0.05 ≈ 13.9, solving for doubling instead of a 50% increase. E) is ln(1.5)/0.01 ≈ 40.5, a decimal slip converting 5% to 0.01. Fix: "increase by p%" → target factor 1 + p/100; with e^rt, always invert with the natural log, and write r as a decimal before dividing.
Q5. A. a = f(0) = 12; then b³ = 96/12 = 8 gives b = 2, so f(t) = 12·2^t. B) uses the 3-step ratio 8 as the 1-step base, skipping the cube root. C) takes a square root of 8's neighbor 16, or halves the exponent — either way the wrong root for a 3-step gap. D) anchors the model at the wrong point, starting from 96 and decaying. E) is the line through the same two points — it matches both data values but is not of the required exponential form a·b^t. Fix: the ratio across k steps is b^k — take the k-th root; and remember two points fit a line just as well, so honor the family the problem specifies.
Q6. E. Losing 3.5% per hour means keeping 96.5%: b = 1 − 0.035 = 0.965. A) uses the loss rate itself as the base — the model would collapse by 96.5% every hour. B) misplaces the decimal on that same error. C) is 3.5% growth, the sign reversed. D) uses the raw percent number as a base, growing 250% per hour. Fix: b = 1 + (signed rate as a decimal); decay bases sit slightly below 1, not near 0.
Q7. C. Midline = center height 25; amplitude = radius 20 (half the 40 m diameter); b = 2π/60 = π/30; boarding at the bottom → −cos. Check: h(0) = 5 (platform), h(30) = 45 (top) ✓. A) boards at the top (+cos starts at the max: h(0) = 45). B) swaps midline and amplitude, giving impossible negative heights (minimum −5 m). D) boards at the midline (−sin starts at 25 m heading down), not the bottom. E) uses the full diameter as the amplitude, sending the rider 15 m underground. Fix: midline = hub height, amplitude = radius (half the diameter), b = 2π/period, and −cos for a bottom start — then verify t = 0 and the half-period.
Q8. D. In a·cos-form models, the constant added outside is the midline — the average value the depth oscillates around: (max + min)/2 = (9 + 3)/2 = 6 m. A) the maximum is midline + amplitude = 9 m. B) the time between high tides is the period, 2π/(π/6) = 12 hours. C) the amplitude is the 3, not the 6. E) the depth at midnight is D(0) = 6 + 3cos(−π/3) = 7.5 m — the midline is not the initial value unless the model starts at a midline crossing. Fix: outside-added constant = midline = average value; check any "value at t = 0" claim by actually evaluating the model at 0.
Q9. E. "Approaches 21 but never reaches it" is precisely a horizontal asymptote at y = 21 — the long-run value of an exponential-decay-toward-ambient model. A) a zero at 21 would mean the temperature output hits 0 when something equals 21 — the wrong feature entirely. B) a vertical asymptote would make the temperature blow up near some time, not level off. C) a vertex describes a quadratic's turning point; the coffee never turns around and reheats. D) the y-intercept is the starting temperature — hot coffee starts well above 21. Fix: "approaches / levels off / never reaches" → horizontal asymptote at the approached value; "starts at" → y-intercept.
Q10. A. g grows by 10% of an ever-larger amount, so its per-step gain increases without bound, while f gains a fixed 5,000 per step; once g's gain passes 5,000, g overtakes f permanently. This holds for any increasing exponential vs. any linear function. B) confuses "ahead now" with "ahead always" — the large coefficient only delays the crossover. C) their growth rates never equalize; the exponential's rate keeps increasing. D) reverses the true long-run comparison. E) starting values shift when the crossover happens, never whether it happens. Fix: family beats coefficients in the long run — exponential (base > 1) eventually exceeds any linear function, regardless of starting values.
Q11. C. Solve 300(1.08)^t = 600 → (1.08)^t = 2 → t = ln 2/ln 1.08 ≈ 9.0 years (and 1.08⁹ ≈ 1.999 confirms it; the rule-of-72 estimate 72/8 = 9 agrees). A) computes 50/8, dividing the percent gain by the rate without logarithms. B) misapplies a rule-of-83 style estimate or slips on the log ratio. D) is the simple-interest answer: 300 + 300(0.08)t = 600 gives t = 12.5, ignoring compounding. E) doubles the doubling-time estimate. Fix: doubling under annual compounding → t = ln 2/ln(1 + r); simple-interest arithmetic always overestimates the time because it ignores growth on growth.
Q12. C. Week 20 is double the fitted range; an impossible prediction (negative sales) is the signature of extrapolation beyond the data's support — the model may fit weeks 1–10 perfectly and still fail out there. A) trusts the model over reality; negative unit sales cannot occur. B) an absurd extrapolated value says nothing about the regression arithmetic, which is checked against the fitted weeks. D) a linear fit could extrapolate just as badly (or worse) — family substitution doesn't cure extrapolation. E) impossible predictions don't impugn the observed data. Fix: judge a model only inside (or near) its fitted range; impossible outputs beyond that range mean "stop extrapolating," not "start over."