IB Math AI HL · Topic 2
Functions
At SL this topic is about choosing a model. At HL it is also about building on one: composite and inverse functions, transformations, richer models — logistic, sinusoidal, piecewise — and the logarithmic scaling that turns a curved data set into a straight line you can read.
What the syllabus covers
Ten assessed sections: SL 2.1 to 2.6, shared with AI SL, and AHL 2.7 to 2.10, which are HL only.
How it is examined
Modelling questions are about choosing, fitting and then answering in the context's own terms.
The GDC fits the parameters and finds intersections and extremes. The marks are for the choice of model, the values you took from the screen, and the sentence that returns the answer to the context.
Logistic and other bounded models have an asymptote that means something: the carrying capacity. Questions ask for it, for when growth is fastest, and for what the model says in the long run.
Sinusoidal models use radians at HL. Amplitude, period and the principal axis come from f(x) = a sin(b(x − c)) + d, with the period 2π/b.
Scaling data (AHL 2.10) is a reading skill. If ln y against ln x is a straight line the relationship is y = a xⁿ; if ln y against x is straight it is y = a bˣ. The marks go to turning gradient and intercept back into the parameters.
Composite and inverse functions are examined with exact values and with domains: the order of composition, the inverse found by swapping x and y, and the domain a composite is defined on.
Transformations are described one at a time and completely: the type, the direction and the size.
A marked question, mark by mark
A linearisation question where the gradient is read correctly and the intercept is not turned back into a parameter.
Practise modelling end to end, on fresh exercises
From SL 2.1 to AHL 2.10, every generated exercise comes back with each mark awarded or withheld and a reason — including the interpretation sentence that closes a modelling question.
Start your trialNew to the codes? M1, A1, R1 explained →
Linearising data · AHL 2.10
Seed 4M18
Two variables are modelled by y = a xⁿ. A graph of ln y against ln x is a straight line with gradient 3 and vertical intercept 1.5.
- (a)Find the value of n.
- (b)Find the value of a, correct to three significant figures.
Your work
ln y = n ln x + ln a
n = 3
a = 1.5
Marks awarded
- M1Mark awarded. Took logarithms to reach ln y = n ln x + ln a, a straight line in ln x.
- A1Mark awarded. n = 3, the gradient.
- A1Mark not awarded. The intercept is ln a, not a. Here ln a = 1.5, so a = e^1.5 = 4.48.
Where marks are actually lost
The graphs are rarely the difficulty. Reading them back into parameters and context is.
Taking the intercept of a log–log or semi-log line as the parameter itself.
CostThe intercept is the logarithm of the parameter. Without undoing the logarithm the accuracy mark is lost.
Leaving the GDC in degree mode for a sinusoidal model.
CostThe period and every value from the model come out wrong together, and the accuracy marks go.
Composing functions in the wrong order — (f ∘ g)(x) written as g(f(x)).
CostEverything downstream is built on the wrong function. Follow-through can rescue later marks, but not the composition itself.
Stating a model's prediction as a bare number, without units or a sentence in context.
CostIn a modelling question the interpretation is marked. A number alone does not answer what was asked.
Other Math AI HL topics
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