A quadratic that has to be read twice
Reading one quadratic three ways: for its roots, for its maximum, and for what those mean in the situation.
Scenario
A community pool models its weekly membership revenue with R = -2m^2 + 120m, where m is the monthly membership price in dollars and R is the weekly revenue in dollars.
The model was fitted from two years of pricing experiments and holds for prices between $0 and $60.
Answers in order, explanations withheld, and a report on where the chain broke.
- 3.1Factored equationsStandard
At which two prices does the model predict zero revenue?
- A$0 and $60
- B$0 and $120
- C$30 and $60
- D$0 and $30
Show the answer and reasoningHide the reasoning
$0 and $60
Factor: R = -2m(m - 60), which is zero when m = 0 or m = 60. At $0 nobody is charged, and at $60 the model predicts nobody joins.
- 3.2Quadratic functionsChallenge
Builds on: The two roots found in question 1 — the maximum is halfway between them, so no calculus or formula is needed.
Which price does the model predict will produce the greatest revenue?
- A$30
- B$60
- C$45
- D$15
Show the answer and reasoningHide the reasoning
$30
A parabola is symmetric about the midpoint of its roots, so the maximum sits halfway between $0 and $60, at $30. Using x = -b / (2a) gives the same answer: -120 / (2 x -2) = 30.
- 3.3Quadratic functionsChallenge
Builds on: The revenue-maximising price of $30 from question 2.
What weekly revenue does the model predict at that price?
- A$1,800
- B$3,600
- C$900
- D$1,200
Show the answer and reasoningHide the reasoning
$1,800
Substitute m = 30: R = -2(900) + 120(30) = -1800 + 3600 = $1,800.
- 3.4Interpreting modelsChallenge
Builds on: The shape established in questions 1 to 3 — which is exactly what makes the tempting wrong answer, extrapolating past the roots, look reasonable.
The pool is considering a $70 membership. What does the model say about that price?
- ANothing reliable, because $70 is outside the range the model was fitted over.
- BRevenue would be negative, so the pool would lose money.
- CRevenue would be higher than at $60.
- DRevenue would be the same as at $50.
Show the answer and reasoningHide the reasoning
Nothing reliable, because $70 is outside the range the model was fitted over.
The stimulus states the model holds between $0 and $60. Substituting 70 does produce a negative number, but reading that as a prediction of losses treats an extrapolation as a result. Recognising where a model stops applying is the point of the stated range.