Quadratic Models (AI HL)
A quadratic model \(f(x) = ax^2 + bx + c\) is the natural choice whenever a real-world quantity rises to a single peak (or falls to a single trough) and then reverses - projectile height, revenue against price, area against a fence length. This page covers the vertex, the axis of symmetry, and how to set up optimisation problems that lead to a quadratic. It's part of the broader Linear, Quadratic & Cubic Models topic.
14 questions on this sub-topic.
The model and its vertex
Covered under IB syllabus reference SL2.5, which sets out linear, quadratic and cubic modelling as one family of function types. Both the model and its axis of symmetry are in the formula booklet.
Quadratic model and axis of symmetry
\(f(x)=ax^2+bx+c,\ x=-\dfrac{b}{2a}\)
In the formula booklet. The axis of symmetry gives the vertex's \(x\)-coordinate directly - substitute it back into \(f(x)\) for the \(y\)-coordinate.
Setting up an optimisation model
Write the quantity to maximise or minimise as a function of one variable, using any given constraint to eliminate a second one. Once it's a quadratic in a single variable, the vertex formula gives the optimum straight away.
Need the full syllabus wording, the formula reference table, and GDC regression steps? See Linear, Quadratic & Cubic Models.
Worked examples
A company's revenue is modelled by \(R(x) = -2x^2 + 40x\) (in \$1000s), where \(x\) is the price.
(a) Find the price that maximises revenue.
(b) Find the maximum revenue.
Worked solution
(a) Vertex at \(x = -\dfrac{b}{2a} = -\dfrac{40}{2(-2)}\) M1
\(= 10.\) A1
(b) \(R(10) = -2(100) + 400\) M1
\(= 200\) ($200 000). A1
A rectangular pen against a wall uses 60 m of fence on three sides.
Find the dimensions giving maximum area.
Worked solution
Width \(x\), length \(\ell = 60 - 2x.\) M1
Area \(A = x(60 - 2x) = 60x - 2x^2.\) A1
\(A' = 60 - 4x = 0 \Rightarrow x\) M1
\(= 15.\) A1
\(\ell = 30.\) A1
Max area \(= 450\) m². A1
Common mistakes
- Giving the vertex's \(x\)-coordinate as the final answer. \(x=-b/2a\) only locates where the maximum or minimum happens - if the question asks for the maximum value itself, you still have to substitute back into \(f(x)\) to get it.
- Dropping the minus sign in \(-b/2a\). It's easy to compute \(b/2a\) on autopilot under time pressure and end up with the vertex reflected to the wrong side of the \(y\)-axis - always write the formula out with its sign before substituting.
- Optimising with two variables still in play. Before differentiating or using the vertex formula, the quantity being optimised has to be written as a function of a single variable - use the given constraint to eliminate the second one first.
Ready to practise properly?
14 quadratic-model questions, marked instantly like the real exam.
Quick answers
How do I find the vertex of a quadratic model?
For \(f(x) = ax^2 + bx + c\), the axis of symmetry (and the vertex's \(x\)-coordinate) is \(x = -\dfrac{b}{2a}\). Substitute that value back into \(f(x)\) to get the vertex's \(y\)-coordinate - the maximum or minimum value of the model.
How do I set up a quadratic optimisation problem?
Write the quantity you're maximising or minimising (area, revenue, profit) as a function of one variable, using any given constraint to eliminate a second variable. Once it's a single quadratic in one variable, the vertex formula gives the optimum directly. For calculator steps, see Using your GDC on the full topic page.