Optimisation with Quadratics (AI SL)
Many real quadratic models exist to answer one question: what's the best (or worst) it gets? A ball thrown in the air has a greatest height, a fenced enclosure has a largest possible area, a cost function has a lowest point. This page covers how to pull that maximum or minimum value out of a quadratic model, with worked examples and the mistakes that cost marks. It's part of the broader Quadratic Models topic.
32 questions on this sub-topic.
Finding the turning point
Covered under IB syllabus reference SL2.5: modelling with quadratic functions \(f(x)=ax^2+bx+c\ (a\neq0)\), including the axis of symmetry, vertex, zeros and intercepts. Technology can be used to find these features directly.
Vertex (turning point)
\(x = -\dfrac{b}{2a}\)
Not in the formula booklet, but easy to derive: it's the axis of symmetry of the parabola. Substitute this \(x\) back into \(f(x)\) to get the maximum or minimum value.
Completed square form
\(f(x) = a(x-p)^2+q\)
Once in this form, the vertex is simply \((p,q)\) - useful when a question explicitly asks you to complete the square rather than jump straight to the GDC.
Need the full syllabus table and GDC walkthroughs? See Quadratic Models.
Worked examples
A ball's height is \(h(t) = -5t^2 + 20t\) metres.
Find the maximum height.
Worked solution
In \(h(t)=-5t^2+20t\) the coefficient of \(t^2\) is \(a=-5<0\), so the parabola opens downward and its vertex is the maximum point. M1
Use \(t=-\dfrac{b}{2a}\) with \(a=-5,\ b=20\): \(t=-\frac{20}{2(-5)}=\frac{20}{10}=2\text{ s}.\) A1
\(h(2)=-5(2)^2+20(2)=-20+40=20\text{ m}.\) A1
The height of a flare is \(h(t)=-5t^2+30t.\)
(a)(i) Write \(h\) in the form \(-5(t-p)^2+q.\) State \(p.\)
(a)(ii) State \(q.\)
(b) State the maximum height and when it occurs.
Worked solution
(a)(i) Complete the square. Factor \(-5\) from the \(t\)-terms:
\(h(t)=-5\big(t^2-6t\big)=-5\big((t-3)^2-9\big)=-5(t-3)^2+45.\) A1
(a)(ii) So \(p=3,\ q=45\). A1
(b) Interpret. Since the squared term is multiplied by \(-5<0\), the maximum of \(h\) occurs when \((t-3)^2=0\), i.e. \(t=3\) s, giving \(h=45\) m. A1 The maximum height is \(45\) m at \(t=3\) s.
Common mistakes
- Reading off the wrong coordinate. \(t=-\frac{b}{2a}\) only gives the input at the turning point - it is not the maximum or minimum value itself. You still need to substitute it back into the model to get the actual height, area, or cost.
- Sign errors with a negative leading coefficient. When \(a<0\), students sometimes treat the vertex as a minimum out of habit. Check the sign of \(a\) first: negative means the parabola opens downward and the vertex is a maximum.
- Skipping a sensible domain. A model like \(h(t)=-5t^2+20t\) is only physically meaningful while \(h(t)\ge0\) - don't quote an optimum time that falls outside the range the context allows.
Ready to practise properly?
32 optimisation-with-quadratics questions, marked instantly like the real exam.
Quick answers
How do you find the maximum or minimum value of a quadratic model?
Find the vertex. Either use \(t=-\tfrac{b}{2a}\) to get the input value and substitute back in, or graph the model on your GDC and use the maximum/minimum tool to read the turning point directly.
Do I need to complete the square to optimise a quadratic model?
Not always - the vertex formula and the GDC's graphing tools usually get there faster. Completing the square is mainly used when a question explicitly asks you to write the model in vertex form first.