IB Maths AI SL Topic 3 — Modelling with Functions Paper 1 & 2 Optimisation contexts ~7 min read

Quadratic Models

Quadratic models — f(x) = ax² + bx + c — describe anything that rises and then falls (or vice versa): a thrown ball reaching maximum height before coming down, a company’s profit peaking at the right price, a bridge arch curving from one footing to the other. The vertex IS the maximum or minimum, found at x = −b/(2a) (in the formula booklet). That’s why quadratics are the standard tool for AI SL optimisation problems — no calculus required.

📘 What you need to know

Standard form: f(x) = ax² + bx + c. a > 0: ∪ shape with a MINIMUM. a < 0: ∩ shape with a MAXIMUM.
Initial value: f(0) = c. This is the output before anything happens (initial height, initial cost, etc.). May be 0, positive, or negative depending on context.
Vertex (max or min): x = −b/(2a) (from the formula booklet). Substitute back for the y-coordinate — that’s the maximum or minimum value of the model.
When to use a quadratic: a quantity that rises then falls (or vice versa) with a SINGLE turning point. Projectile motion, profit/price, area for fixed perimeter, dose-response.
The a-value sets the curvature: large |a| means a steep, narrow curve; small |a| means a wide, flat curve. The sign decides max vs min.
Limitations: a quadratic is SYMMETRIC about its vertex — real data may not be. It also has only one extremum, so one “tail” runs off to ±∞. Restrict the domain to keep predictions sensible.

Interpreting the parameters

Every quadratic model has three numbers a, b, c, and each has a real-world meaning that exam questions love to ask about.

c = initial value. This is the function evaluated at x = 0 — e.g. the launch height of a stone, the fixed costs before selling anything, the height of a bridge at one footing.

a controls the shape. Negative a ⇒ downward-opening parabola with a MAX (most “maximise the X” problems). Positive a ⇒ upward-opening with a MIN (cost minimisation, etc.). Larger absolute value ⇒ sharper curve.

The vertex IS the answer to the optimisation question. Whatever the context, “find the maximum” or “find the minimum” is the same calculation: x = −b/(2a) for the optimal input, then substitute back for the optimal output.

Quadratic model — formula booklet f(x) = ax² + bx + c ⇒ max/min at x = − b2a

Each labelled point answers a different exam question. The y-intercept gives the initial launch height (c = 25). The vertex gives the maximum height (peak). The positive root gives when the ball lands.

Setting up a quadratic model

If you’re given the equation, just identify a, b, c and apply the vertex formula. If you have to build the model from features, choose a form that matches what you know:

Factored form f(x) = a(x − p)(x − q) — use when you know both roots (the zeros). Sub in one other point to find a.

Vertex form f(x) = a(x − h)² + k — use when you know the maximum/minimum point (h, k). Sub in one other point to find a.

Why c can be negative: in a profit model, fixed costs are often paid before any sales, so P(0) is a loss (negative). That’s the initial-value interpretation — not a mistake.

Domain in a real-world model

The mathematical quadratic continues to ±∞, but real quantities don’t. A ball doesn’t pass through the ground; a price can’t be negative; a fence has fixed total length. Restrict the domain to the meaningful range — usually starting at 0 and ending at where the model crosses a natural boundary (lands, becomes zero, hits a constraint).

🧭 Recipe — using a quadratic model

Read off a, b, c from the equation. Check sign of a to decide if you’re looking for a max (a < 0) or a min (a > 0).
Initial value: f(0) = c. State its meaning in context (initial height, fixed cost, etc.).
Find the vertex: x_opt = −b/(2a) gives the input. Substitute back to find the maximum/minimum output.
For “when does it hit zero” (e.g. ball lands, break-even): solve f(x) = 0 using the GDC or factorising. Reject any negative or non-physical root.
If building the model: pick factored form (you know roots) or vertex form (you know max/min point); substitute an extra point to find a; expand to standard form if asked.

Worked examples

WE 1

Profit model — interpret and maximise

A bookshop’s daily profit (in $) is modelled by P(x) = −2x² + 80x − 150, where x is the number of books sold. (a) Find P(0) and explain its meaning. (b) Find the number of books that gives the maximum profit. (c) State the maximum profit.

(a) initial value P(0) = −150 means: a $150 LOSS before any books sold (fixed daily costs) (b) axis of symmetry x = −b/(2a) a = −2, b = 80 x = −80/(2×−2) = −80/−4 = 20 (c) substitute x = 20 P(20) = −2(400) + 80(20) − 150 = −800 + 1600 − 150 = 650 (a) $150 loss at zero sales · (b) 20 books · (c) max profit $650 a < 0 means the parabola opens DOWN, so the vertex is a MAXIMUM — matching the question of “maximum profit”. Selling more than 20 books would presumably mean discounting or staffing costs eat into profit.

WE 2

Projectile from a cliff — full feature set

A stone is thrown upward from the top of a 25 m cliff. Its height above the ground (in m) at time t (s) is modelled by h(t) = −5t² + 20t + 25. Find: (a) the initial height, (b) the maximum height and when it occurs, (c) the time the stone hits the ground.

(a) initial height = h(0) h(0) = 25 m (matches the cliff height) (b) max at t = −b/(2a) t = −20/(2×−5) = 2 s h(2) = −5(4) + 20(2) + 25 = −20 + 40 + 25 = 45 max height 45 m at t = 2 s (c) lands when h = 0 −5t² + 20t + 25 = 0 ⇒ t² − 4t − 5 = 0 (t − 5)(t + 1) = 0 t = 5 or t = −1 (reject −1: time can’t be negative) (a) 25 m · (b) 45 m at t = 2 s · (c) lands at t = 5 s domain in context: 0 ≤ t ≤ 5. Outside that range the model has no physical meaning — the stone has either not been thrown yet or has already landed.

WE 3

Revenue model — optimum price & break-even

A garden centre’s daily revenue (in $) from selling tomato plants at price p dollars each is R(p) = 200p − 8p². (a) Find the price that maximises revenue. (b) State the maximum daily revenue. (c) Find the two break-even prices (where R = 0) and interpret each.

Rewrite in standard form: R(p) = −8p² + 200p a = −8, b = 200, c = 0 (a) optimum price at p = −b/(2a) p = −200/(2×−8) = −200/−16 = 12.5 (b) max revenue R(12.5) = 200(12.5) − 8(156.25) = 2500 − 1250 = 1250 (c) break-even: R(p) = 0 200p − 8p² = 0 ⇒ 8p(25 − p) = 0 p = 0 or p = 25 (a) $12.50 · (b) max $1250 · (c) p = 0 (free, no money in) and p = 25 (too expensive, no sales) a classic pricing problem: too cheap and you get nothing per unit; too expensive and no one buys. The sweet spot is exactly halfway between the two break-even prices: (0 + 25)/2 = 12.5 ✓.

WE 4

Build a model — bridge arch

A bridge has a parabolic arch. The arch is 40 m wide at ground level (where the height is 0) and reaches a maximum height of 20 m exactly at the midpoint. (a) Find a quadratic model h(x) for the height, where x is the horizontal distance from one footing. (b) Find the height of the arch 15 m from one footing.

(a) Step 1 — use factored form (we know the roots) roots at x = 0 and x = 40 h(x) = a·x(x − 40) Step 2 — use the vertex (20, 20) to find a at midpoint x = 20, h = 20: 20 = a(20)(20 − 40) = a(20)(−20) = −400a a = 20/(−400) = −0.05 Step 3 — equation h(x) = −0.05x(x − 40) = −0.05x² + 2x (b) h(15) h(15) = −0.05(225) + 2(15) = −11.25 + 30 = 18.75 (a) h(x) = −0.05x² + 2x · (b) 18.75 m factored form is the fastest way in when you know the roots: y = a(x − p)(x − q). Then ONE more point (any point) is enough to pin down a.

WE 5

Minimisation — truck costs

A delivery firm models its weekly cost (in $) of operating a truck as C(s) = 0.05s² − 8s + 500, where s is the average speed in km/h. (a) Find C(60) and interpret. (b) Find the speed that minimises the weekly cost. (c) State the minimum cost.

(a) substitute s = 60 C(60) = 0.05(3600) − 8(60) + 500 = 180 − 480 + 500 = 200 at 60 km/h, the truck costs $200/week to run (b) a = 0.05 > 0 ⇒ this is a MIN problem s = −b/(2a) = −(−8)/(2×0.05) = 8/0.1 = 80 (c) min cost = C(80) C(80) = 0.05(6400) − 8(80) + 500 = 320 − 640 + 500 = 180 (a) C(60) = $200/week · (b) speed s = 80 km/h · (c) min cost $180/week interpretation: very slow speeds are inefficient (long hours, low throughput); very fast speeds increase fuel costs disproportionately. The “sweet spot” balances the two — a classic real-world minimum.

WE 6

Build a model — football kicked upward

A football is kicked vertically upward from the ground. The ball reaches a maximum height of 10 m after 2 s. The height (m) at time t (s) is modelled by h(t) = at² + bt. (a) Find a and b. (b) State the time the ball lands back on the ground.

(a) Step 1 — max at t = 2 means −b/(2a) = 2 b = −4a Step 2 — max value h(2) = 10 a(4) + b(2) = 10 4a + 2(−4a) = 10 4a − 8a = 10 −4a = 10 ⇒ a = −2.5 b = −4(−2.5) = 10 h(t) = −2.5t² + 10t (b) lands when h = 0 −2.5t² + 10t = 0 −2.5t(t − 4) = 0 t = 0 or t = 4 (a) a = −2.5, b = 10 · (b) lands at t = 4 s SYMMETRY shortcut: a quadratic is symmetric about its vertex. Starts at the ground (t = 0), peaks at t = 2, so by symmetry lands at t = 2 + 2 = 4. Saves the factorisation.

💡 Top tips

Always state the meaning of c in context: not just “c = −150″ but “the initial loss is $150 before any books are sold”. Interpretation marks add up.
Decide max vs min from the sign of a FIRST: a < 0 = max problem; a > 0 = min problem. Catches a lot of “wait, which one?” confusion.
Use the right form for the data you have: standard for y-intercept; factored for known roots; vertex form for known max/min point.
The vertex is automatically symmetric between the roots: x_vertex = (p + q)/2 for roots p, q. Faster than the formula when both roots are clean.
Restrict the domain to physical sense: a thrown ball model only applies from launch to landing; a price model is undefined for negative prices. Always state the meaningful range.

⚠ Common mistakes

Stopping at the vertex x-coordinate: “what’s the max profit?” needs the y-value — substitute back. The x-value is just where it occurs.
Forgetting to reject one root: “when does the ball land?” gives two roots; only the positive one is physical. State which you reject and why.
Sign error in −b/(2a): for b negative the answer becomes positive. E.g. for C(s) = 0.05s² − 8s + 500, b = −8, so s = −(−8)/(2×0.05) = 80, NOT −80.
Calling the vertex the “global max” of a real system: quadratics ARE globally maximised at the vertex (within their domain), but only because the model is restricted. Outside the domain (negative time, etc.) the model gives meaningless values.
Confusing “a negative” with “function negative”: a < 0 means the parabola opens downward (shape), NOT that the outputs are negative everywhere. A profit model with a < 0 still gives positive profits between its two break-even prices.

Up next: Cubic Models. Cubics describe quantities with TWO turning points (a local max and a local min) within their domain — useful for things like a bungee jumper’s depth that rises, falls, rises again, or a volume that peaks then drops. The vertex formula doesn’t apply, but the GDC’s max/min tools handle them directly.

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