A single maximum or minimum → think quadratic: A quadratic model fits when the quantity rises to a peak then falls (or falls to a trough then rises).
The graph is a parabola.
The key feature is one turning point — either a maximum or a minimum.
Signal words in IB questions that suggest a quadratic model:
- "Maximum height", "highest point", "peak" — parabola opens downward (a < 0)
- "Minimum cost", "lowest point", "bottom of a dip" — parabola opens upward (a > 0)
- Object in projectile motion: always quadratic in time
- Area optimisation problems: usually quadratic
- Revenue = price × quantity where price depends on quantity: usually quadratic
| Scenario | Quadratic? | Clue |
|---|---|---|
| Dolphin jump: h = −0.5d² + 3d | Yes (max) | Negative leading coefficient → has a maximum height |
| Carpet area: A = x(20 − πx/2) | Yes (max) | Area optimisation → quadratic in x |
| Population growing 15% per year | No — exponential | Percentage growth → exponential |
| Ball rolling at constant speed | No — linear | Constant speed → linear distance-time |
- Parabola opens upward — vertex is a MINIMUM
- Parabola opens downward — vertex is a MAXIMUM
- x = −b/(2a)
- Substitute x = −b/(2a) back into f(x)
A dolphin jumps and its height is h(d) = −0.5d² + 3d, where d is horizontal distance in metres.
Find the maximum height and the value of d at which it occurs.
Step by step
- Identify a = −0.5, b = 3. Since a < 0, the vertex is a maximum.
- Find the d-coordinate of the vertex.
- Find the maximum height.
Final answer
Maximum height = 4.5 m, occurring at a horizontal distance of 3 m from take-off.
[Diagram: math-graph-intersection] - Available in full study mode
GDC method is equally valid: Enter f(x) in Y1, set a suitable window (e.g. Xmin=0, Xmax=7, Ymin=−1, Ymax=6), then use Calc → Maximum.
The GDC confirms (3, 4.5).
Both methods earn full marks.
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Trap 1 — confusing x at vertex with the vertex y-value: If asked "find the maximum height", give the y-coordinate of the vertex, not the x-coordinate.
These are different numbers.
State both clearly: "Maximum height = 4.5 m occurs at d = 3 m."
Trap 2 — forgetting the domain: The dolphin re-enters the water where h(d) = 0.
Solve −0.5d² + 3d = 0 → d(−0.5d + 3) = 0 → d = 0 or d = 6.
So the model is only valid for 0 ≤ d ≤ 6.
Answers outside this range are meaningless.
For h(d) = −0.5d² + 3d, find where the dolphin re-enters the water.
Step by step
- Set h(d) = 0.
- Factorise.
- Solve.
Final answer
The dolphin re-enters the water at d = 6 m. (d = 0 is the starting point.)
Trap 3 — sign of "a": Always check: if the real-world situation has a maximum (peak), then a must be negative.
If you get a positive a when modelling a maximum situation, recheck your equation.
The model is a tool — use it to answer the actual question: Once you have the quadratic model, typical IB follow-up questions ask: find the maximum/minimum, find zeros (where object lands, when profit = 0), or find f(x) for a given x.
A company's profit P (thousands of $) when selling x hundred units is P(x) = −2x² + 12x − 10.
(a) Find the maximum profit.
(b) Find the number of units for which there is no profit.
Step by step
- Part (a): find the vertex. a = −2, b = 12.
- Part (b): set P = 0 and solve.
- Divide by −2.
- Factorise.
Final answer
(a) Maximum profit = $8 000 when selling 300 units. (b) Zero profit when selling 100 units or 500 units.