Cubic Models (AI HL)
A cubic model \(f(x) = ax^3 + bx^2 + cx + d\) can bend twice, which makes it the right tool for data that rises, dips, then rises again (or the reverse) - something a straight line or parabola can't capture. This page covers reading the model's shape, finding its turning points, and the mistakes that most often cost marks. It's part of the broader Linear, Quadratic & Cubic Models topic.
11 questions on this sub-topic.
The model and its turning points
Covered under IB syllabus reference SL2.5, which sets out linear, quadratic and cubic modelling as one family of function types. The cubic form itself is in the formula booklet; finding turning points is a calculus skill you apply to it.
Cubic model
\(f(x) = ax^3 + bx^2 + cx + d\)
In the formula booklet. Recognise it from context clues like "an S-shaped trend" or data with two changes of direction.
Turning points of a cubic
Differentiate, set \(f'(x)=0\), and solve. Each solution is a candidate maximum or minimum - check which by the sign of \(f'\) either side, or the second derivative.
Need the full syllabus wording, the formula reference table, and GDC regression steps? See Linear, Quadratic & Cubic Models.
Worked examples
A cubic model \(f(x) = a(x+1)(x-2)(x-4)\) passes through \((0, 16)\).
(a) Find \(a\).
(b)(i) State the \(x\)-intercept with \(x<0\).
(b)(ii) State the \(x\)-intercept with \(0
Worked solution
(a) \(f(0) = a(1)(-2)(-4) = 8a = 16.\) M1
\(a = 2.\) A1
(b)(i) \(x\)-intercepts where each factor is zero: M1
\(x = -1.\) A1
(b)(ii) \(x = 2.\) A1
(b)(iii) \(x = 4.\) A1
A cubic model is \(f(x) = x^3 - 6x^2 + 9x + 1.\)
(a) Find \(f'(x).\)
(b)(i) Find the \(x\)-coordinate of the turning point with \(x<2\).
(b)(ii) Find the \(x\)-coordinate of the turning point with \(x>2\).
Worked solution
(a) \(f'(x) = 3x^2 - 12x + 9.\) M1
\(f'(x) = 3x^2-12x+9.\) A1
(b)(i) \(3(x^2 - 4x + 3) = 3(x-1)(x-3) = 0.\) M1
\(3(x-1)(x-3) = 0.\) A1
\(x = 1\) (local max). A1
(b)(ii) \(x = 3\) (local min). A1
Common mistakes
- Looking for a single vertex, like a quadratic. A cubic can have zero or two turning points, never one fixed "vertex" - and no axis of symmetry to exploit, so each turning point has to be found separately from \(f'(x)=0\).
- Stopping once \(f'(x)=0\) is solved. Solving for \(x\) only locates the candidate turning points - you still need to state whether each is a maximum or minimum, usually from the sign of \(f'\) either side.
- Misreading end behaviour from the leading coefficient. If \(a<0\), the graph falls to \(-\infty\) as \(x\to+\infty\) and rises to \(+\infty\) as \(x\to-\infty\) - the opposite of what a positive \(a\) gives, so check the sign of \(a\) before describing either end.
Ready to practise properly?
11 cubic-model questions, marked instantly like the real exam.
Quick answers
What is the general form of a cubic model?
\(f(x) = ax^3 + bx^2 + cx + d\), with \(a \ne 0\). A cubic can have up to two turning points and, unlike a quadratic, has no fixed axis of symmetry.
How do I find the turning points of a cubic?
Differentiate to get \(f'(x)\), set \(f'(x) = 0\), and solve. Each solution is a candidate maximum or minimum - check which by testing the sign of \(f'(x)\) either side of it, or with the second derivative. For calculator work, see Using your GDC on the full topic page.