Understand · explore · practise
Build quadratic models from a context or data, find sensible roots and maxima, and explain the assumptions and limits of the answer.
Before you startQuadratic equations, turning points and units
01 / Build a model
A model translates selected features of a situation into mathematics. State what the input measures, its units and its allowed values. An exact algebraic answer can still be only an estimate of reality.
A community growing bed uses an existing wall for one side and 24 m of edging for the other three. Let w metres be its width away from the wall and l metres its length along the wall.
2w + l = 24 ⇒ l = 24 − 2w
A = wl = w(24 − 2w)
A(w) = 24w − 2w²
A is in square metres. Both lengths must be positive, so 0 < w < 12. We assume a rectangular bed, a straight wall and negligible edging thickness.
The largest area is not a square here: the wall supplies one long side without using any edging.
A rectangular layout has height 1 unit and width r units, with r > 1. Remove a 1-by-1 square from one end. Suppose the remaining rectangle has the same proportions as the original after a quarter-turn.
The original long-to-short ratio is r/1. The remaining rectangle has long side 1 and short side r − 1, so similarity requires:
r = 1/(r − 1)
r² − r − 1 = 0
r = (1 ± √5)/2
Only r = (1 + √5)/2 is greater than 1. This exact proportion is called the golden ratio. Its remaining width is less than 1, consistent with the sides used in the ratio.
Width 6 m, length 12 m. Area 72 m², the largest possible area with 24 m of edging.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Find a maximum
Complete the square, then check that the turning point lies inside the physical domain.
A(w) = 24w − 2w²
= 72 − 2(w − 6)²
The subtracted square is smallest at w = 6, giving maximum area 72 m². The length is 24 − 2(6) = 12 m. Width 6 m is allowed because 0 < w < 12.
If the width is also restricted to 0 < w ≤ 4, the vertex is unavailable. The area increases over that interval, so the maximum is at w = 4: A = 64 m².
For a target area, solve the quadratic and interpret every root. Two different shapes can have the same area.
24w − 2w² = 54
Rearrange and divide by −2.
w² − 12w + 27 = 0
(w − 3)(w − 9) = 0
Both algebraic roots lie in 0 < w < 12.
w = 3 ⇒ l = 18
w = 9 ⇒ l = 6
Both use 24 m of edging.
3 × 18 = 9 × 6 = 54
A second constraint, such as available space, could choose between them.
03 / A model in time
A simple model for a foam ball launched from a platform is h(t) = 1 + 8t − 5t². Here t is seconds after launch and h is height in metres above the ground. This is an illustrative model, assuming constant downward acceleration and ignoring air resistance.
h(0) = 1 tells us the launch height. The constant is not the maximum: that occurs later.
Ground contact means h = 0. The two algebraic roots are (4 ± √21)/5, but the negative one is before launch. The model describes this flight only from t = 0 until the positive root, approximately 1.72 s.
After impact the same quadratic would predict negative height. That is a reason to stop using the flight model, not a prediction that the ball travels through the ground.
h(t) = 1 + 8t − 5t²
Factor out −5 from the terms in t.
h(t) = 21/5 − 5(t − 4/5)²
Expand to check the constants.
Maximum: 4.2 m at t = 0.8 s
The vertex occurs during the modelled flight.
h = 3 ⇒ t = (4 ± √6)/5
The ball passes 3 m on the way up and again on the way down, at about 0.310 s and 1.29 s.
04 / Fit to data
A quadratic model y = ax² + bx + c has three coefficients. Three data points with different x values give three simultaneous equations.
In a fictional workshop trial, a device records outputs 7, 11 and 19 at settings x = 0, 1 and 2. Treating these readings as exact determines the model in the worked example.
The fitted model predicts y = 31 at x = 3. That setting lies outside the measured interval: it is an extrapolation. A perfect fit to three readings does not establish that the device follows a quadratic elsewhere.
If the predicted output is 15, solve 2x² + 2x + 7 = 15. This gives x = (−1 ± √17)/2. Only the positive root, approximately 1.56, lies in the trial interval 0 ≤ x ≤ 2.
At x = 0: c = 7
Use the point that removes both variable terms first.
At x = 1: a + b + 7 = 11
At x = 2: 4a + 2b + 7 = 19
Subtract 7 from each equation.
a + b = 4
4a + 2b = 12
Subtract twice the first equation from the second: 2a = 4.
a = 2, b = 2, c = 7
y = 2x² + 2x + 7
Substitute all three input values to check the fit.
05 / Limits & decisions
A print studio estimates that a monthly subscription priced at £p will attract n = 90 − 3p customers, over a trial range £10 ≤ p ≤ £25. Revenue is price multiplied by number of customers.
R(p) = p(90 − 3p)
= 675 − 3(p − 15)²
Within that estimate, the highest monthly revenue is £675 at p = 15, with 45 customers. This is revenue, not profit: costs have not been subtracted.
Suppose the studio can serve at most 36 customers and wants a price at which estimated demand does not exceed capacity. Then 90 − 3p ≤ 36 requires p ≥ 18. Within 18 ≤ p ≤ 25, the maximum predicted revenue is £648 at p = 18.
This assumes the linear demand estimate remains valid, all predicted customers buy, and capacity is handled by selecting a suitable price. Discrete customers and uncertain demand limit the precision of the prediction.
u₀ = 0, uₙ₊₁ = √(2 + uₙ)
This produces a growing nest of square roots.
0 ≤ uₙ ≤ 2 for every n
If 0 ≤ uₙ ≤ 2, then 0 ≤ √(2 + uₙ) ≤ 2. This proves the bound by induction.
uₙ₊₁ ≥ uₙ
For 0 ≤ uₙ ≤ 2, 2 + uₙ − uₙ² = (2 − uₙ)(uₙ + 1) ≥ 0.
L = √(2 + L)
(L − 2)(L + 1) = 0 ⇒ L = 2
An increasing bounded sequence converges. Continuity gives the limit equation; non-negativity excludes −1. Solving the quadratic alone would not prove convergence.
06 / Your turn
Define quantities clearly, keep units and restrictions, and distinguish what follows from the model from what has been observed.
A rectangular display has height x cm and width (x + 5) cm. Its area is 84 cm². Find its dimensions.
Area = height × width. A length must be positive.
x(x + 5) = 84
x² + 5x − 84 = 0
(x + 12)(x − 7) = 0
Reject x = −12. The display is 7 cm by 12 cm.
An illustrative height model is h(t) = 2 + 6t − 4t² metres, from launch until first ground contact. Find the launch height, maximum height and flight time.
Use h(0), complete the square, then solve h = 0 and reject a negative time.
h(0) = 2
h(t) = 17/4 − 4(t − 3/4)²
Maximum height 4.25 m at 0.75 s.
4t² − 6t − 2 = 0
t = (3 ± √17)/4
Flight time is (3 + √17)/4 ≈ 1.78 s. The negative root is outside the modelled time interval.
A fictional output q has readings q(0) = 4, q(1) = 9 and q(2) = 18. Fit q(x) = ax² + bx + c and predict q(3). Explain one limitation.
Use x = 0 to find c, then eliminate b from the remaining two equations.
c = 4
a + b = 5; 4a + 2b = 14
2a = 4 ⇒ a = 2, b = 3
q(x) = 2x² + 3x + 4
q(3) = 31
The prediction at 3 extrapolates beyond the measured settings. Three points fitting a quadratic do not establish a physical law.
A model predicts Q(s) = −2s² + 20s + 7 for 0 ≤ s ≤ 4. Find the maximum allowed output.
Check whether the vertex input is allowed.
Q(s) = 57 − 2(s − 5)²
The vertex s = 5 is excluded. The curve increases throughout 0 ≤ s ≤ 4, so the maximum allowed output is Q(4) = 55 at s = 4.
A decorative arch is modelled by h(x) = a − bx² metres for −3 ≤ x ≤ 3, where x is horizontal distance from its centre. The centre is 5 m high and each end is 2 m high. Find a, b and the width over which h ≥ 4 m.
Use h(0) and h(3), then solve the height inequality. The width extends on both sides of the centre.
a = 5; 2 = 5 − 9b ⇒ b = 1/3
5 − x²/3 ≥ 4 ⇒ x² ≤ 3
−√3 ≤ x ≤ √3
The width is 2√3 m, approximately 3.46 m. This interval is within the modelled arch.
A quadratic model predicts a maximum revenue of £900. A student calls this “a guaranteed profit of £900”. Explain two errors.
Think about costs and about whether model predictions are certainties.
Revenue is money received before costs, so it is not profit. A model prediction depends on its assumptions and estimates, so the amount is not guaranteed. Check that the input giving the maximum is also feasible.
07 / Recap
Section 1 of 7 · Build a model