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2.6PureStretch

Modelling with Quadratics

Quadratics model anything with a single turning point — the path of a thrown ball, the area of a field, profit against price. The maths is the same as before; the new skill is translating a real situation into a quadratic and interpreting the answer.

30 min Quadratics
Edexcel AS level Maths: 2.6 Modelling with QuadraticsCurated by Kepler, not produced by us. Watch it before the notes below.
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What you'll be able to do

  • Set up a quadratic model from a worded problem
  • Use the turning point to find a maximum or minimum
  • Interpret roots and intercepts in context
  • Decide which solutions are physically sensible
1

From words to a quadratic

Define your variables clearly, then write the relationship as a quadratic. Projectile-motion problems are the classic example, where height depends on time.

h=−5t2+ut+h0h = -5t^{2} + ut + h_{0}
A typical height model: uu is the launch speed and h0h_0 the start height.
2

Maximum and minimum values

Because a quadratic has exactly one turning point, completing the square (or the line of symmetry) finds the greatest or least value of the model — the maximum height, minimum cost, and so on.

Example 1A ball’s height is h=−5t2+20th = -5t^2 + 20t. Find its maximum height.
1Complete the square: −5(t2−4t)=−5[(t−2)2−4]=−5(t−2)2+20-5(t^2 - 4t) = -5\left[(t - 2)^2 - 4\right] = -5(t - 2)^2 + 20.
2The maximum of hh is the constant 2020, reached when t=2t = 2.
Answer
maximum height 2020 m at t=2t = 2 s

Tip — A maximum/minimum question is really a “find the vertex” question in disguise.

3

Interpreting the roots

In context, the roots and intercepts have real meanings. For a height-time model, a root is when the object is at height zero — usually when it lands.

Example 1For h=−5t2+20th = -5t^2 + 20t, when does the ball hit the ground?
1Set h=0h = 0: −5t(t−4)=0-5t(t - 4) = 0.
2So t=0t = 0 (launch) or t=4t = 4.
Answer
it lands at t=4t = 4 s
4

Choosing sensible answers

A quadratic gives two solutions, but the context often rules one out. Negative times, negative lengths or values beyond the model’s range should be discarded with a brief reason.

Tip — Always sanity-check: a time of −2 s or a width of −3 m is not physically possible, so reject it.

Formula recap

h=−5t2+ut+h0h = -5t^{2} + ut + h_{0}
Standard projectile-height model.
max/min at the vertex (−p,q)\text{max/min at the vertex } (-p, q)
From completed-square form.
h=0⇒object at ground levelh = 0 \Rightarrow \text{object at ground level}
Roots give contextual events.

Common mistakes to avoid

Keeping both algebraic solutions even when one is impossible.
Reject solutions that make no physical sense (e.g. negative time) and say why.
Confusing the time of maximum height with the maximum height itself.
The vertex gives both: the t-coordinate is when, the y-coordinate is how high.

Key takeaways

  • Define variables, then translate the situation into a quadratic.
  • The vertex gives the maximum or minimum value of the model.
  • Roots and intercepts carry real-world meaning (e.g. landing time).
  • Discard solutions that are not physically sensible.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Check your understanding

Work through questions on Modelling with Quadratics. Use the feedback to decide what needs another look.

5 past-paper questions · 11 marks

Practise this topic

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