The shape of an arch
One quadratic, read four ways: as zeros, as a vertex, as a constraint, and as an equivalent expression.
Scenario
A footbridge arch is modelled by h(x) = -0.08x^2 + 1.6x, where x is the horizontal distance in metres from the left base and h(x) is the height of the arch in metres above the deck.
The model holds only between the two points where the arch meets the deck.
Answers in order, explanations withheld, and a report on where the chain broke.
- 12.1Quadratic functionsFoundation
Where does the arch meet the deck?
- AAt x = 0 and x = 20
- BAt x = 0 and x = 16
- CAt x = 0 only
- DAt x = 1.6 and x = 20
Show the answer and reasoningHide the reasoning
At x = 0 and x = 20
Set h(x) = 0 and factor: x(-0.08x + 1.6) = 0, so x = 0 or -0.08x = -1.6, giving x = 20. The arch spans 20 metres.
- 12.2Quadratic functionsStandard
Builds on: The zeros from question 1. Without them you would reach for the formula x = -b / 2a; with them the midpoint is quicker and harder to get wrong.
How high is the arch at its highest point, and where is that point?
- A8 metres, 10 metres from the left base
- B16 metres, 10 metres from the left base
- C8 metres, 8 metres from the left base
- D1.6 metres, 20 metres from the left base
Show the answer and reasoningHide the reasoning
8 metres, 10 metres from the left base
A parabola is symmetric about its vertex, so the maximum sits midway between the zeros, at x = 10. Then h(10) = -0.08(100) + 16 = -8 + 16 = 8 metres.
- 12.3Interpreting modelsChallenge
Builds on: Question 1 fixed the span, which is what "centred" means here, and question 2 supplied the symmetry. The third choice is the trap: clearance is decided at the tightest point, not the peak.
A lorry 4 metres tall and 12 metres wide is driven under the arch, centred on the span. Does it pass?
- AYes — the tightest point is 5.12 metres, leaving 1.12 metres of clearance.
- BNo — the arch is lower than 4 metres at the edges of the lorry.
- CYes, because the arch reaches 8 metres at its peak.
- DIt cannot be decided without knowing how long the lorry is.
Show the answer and reasoningHide the reasoning
Yes — the tightest point is 5.12 metres, leaving 1.12 metres of clearance.
Centred on a 20-metre span, a 12-metre lorry runs from x = 4 to x = 16. The arch is lowest over that stretch at its two edges: h(4) = -1.28 + 6.4 = 5.12 metres, and h(16) is the same by symmetry. Since 5.12 > 4, it passes.
- 12.4Equivalent expressionsChallenge
Builds on: The vertex (10, 8) from question 2 — vertex form is that pair of numbers written into the expression, which is why reading it off is faster than expanding all four choices.
Which expression is equivalent to h(x) and shows the maximum height as a constant?
- Ah(x) = -0.08(x - 10)^2 + 8
- Bh(x) = -0.08(x - 8)^2 + 10
- Ch(x) = -0.08x(x - 20)
- Dh(x) = -0.08(x + 10)^2 + 8
Show the answer and reasoningHide the reasoning
h(x) = -0.08(x - 10)^2 + 8
Expanding gives -0.08(x^2 - 20x + 100) + 8 = -0.08x^2 + 1.6x - 8 + 8, which is h(x). The second choice swaps the two vertex coordinates and gives 9.68 rather than 8 at x = 10. The third is equivalent but factored, so it shows the zeros instead. The fourth puts the vertex at x = -10.