IB Maths AI HLTransformations of GraphsPaper 1 & 2af(x)+b, f(ax+b), order rules~10 min read
Composite Transformations of Graphs
Combining stretches, reflections and translations is mostly mechanical, except for the order. af(x) + b: stretch first, then translate (natural order). f(ax + b): translate first, then stretch (reverse order). Horizontals and verticals are independent — their relative order doesn’t matter.
๐ What you need to know
Vertical composite af(x) + b: stretch by SF a FIRST, then translate by (0; b).
Horizontal composite f(ax + b): translate by (−b; 0) FIRST, then stretch by SF 1/a. (Reverse of order of operations!)
Why reversed for horizontals? Because the operations act on the input, so the last operation written on x is the first one done to the graph.
Independence: horizontal and vertical transformations can swap orders freely. Only the relative order within H or within V matters.
Negative coefficient → include a reflection: −f(x) reflects in x-axis; f(−x) reflects in y-axis.
Point image: under af(x) + b, point (p, q) → (p, aq + b). Under f(ax + b), (p, q) → ((p − b)/a, q).
The two order rules
For vertical composites af(x) + b, the order matches the order of operations on the output: first multiply by a (stretch), then add b (translate). For horizontal composites f(ax + b), the order reverses: even though the bracket reads “multiply by a then add b” on the input x, the graph is built by translating first, then stretching. The reason: horizontal transformations act on the input variable, and the input goes through the operations from inside out — the last operation written on x is the first one applied to the graph.
Both composites end with the desired equation, but only in the orders shown. Swapping the order changes the constants — see WE 3 and WE 4.
Order rules at a glanceaf(x) + b: stretch SF a → translate (0; b)
f(ax + b): translate (−b; 0) → stretch SF 1/aH and V are independent — relative order doesn’t matter
Why the order matters
Try doing the operations in the wrong order on y = 3f(x) + 2 starting from f(x) = x2 at x = 2 (so f(2) = 4). Doing stretch then translate gives 3(4) + 2 = 14; doing translate then stretch gives 3(4 + 2) = 18. Different answers! The first matches 3f(x) + 2; the second would give 3f(x) + 6 — the constant got multiplied by the stretch. Order matters whenever two transformations are in the same family (both vertical or both horizontal), because each one’s effect compounds with the other’s.
The output rule: for vertical composites, do operations in the natural order they appear in af(x) + b — multiply, then add. The input rule: for horizontal composites, do them in reverse — even though ax + b “looks like” multiply-then-add, the graph must be translated first (because the +b on the input is the last thing the input sees, which means it’s the first thing applied to the graph).
Combining horizontal and vertical
When a problem mixes horizontal and vertical transformations, the relative order between an H and a V doesn’t matter — you can interleave them however you like. What does matter is the order within each family. So for y = 2f(3x + 6) − 1, do verticals in order (stretch SF 2, then translate (0; −1)) and horizontals in order (translate (−6; 0), then stretch SF 1/3) — but you can do all four in any interleaving. A point (p, q) on f ends up at ((p − 6)/3, 2q − 1) on the new graph.
๐งญ Recipe — composite transformations
Identify horizontals and verticals: anything inside the bracket of f is horizontal; anything outside is vertical.
Verticals first rule: for af(x) + b, stretch by SF a first, then translate by (0; b).
Horizontals reversed rule: for f(ax + b), translate by (−b; 0) first, then stretch by SF 1/a.
Apply to key points: (p, q) → ((p − b)/a, a‘q + b‘) under a‘f(ax + b) + b‘.
Sketch step-by-step: do one transformation at a time, redraw between each — especially helpful for complex composites.
Worked examples
WE 1
Vertical composite af(x) + b
For y = 4f(x) − 3: (a) Describe the transformations in order. (b) Find the image of the point P(2, 5) under this transformation.
(a) read coefficient and constanta = 4, b = โ3verticals: stretch FIRST, then translatevertical stretch SF 4, then translate (0; โ3)(b) apply to P(2, 5)stretch SF 4: (2, 5) โ (2, 20)translate (0; โ3): (2, 20) โ (2, 17)image: (2, 17)check: new y = 4ยท5 โ 3 = 17 โ
WE 2
Horizontal composite f(ax + b)
For y = f(2x + 6): (a) Describe the transformations in order. (b) Find the image of the point Q(8, −1) under this transformation.
(a) read coefficientsa = 2, b = 6horizontals: translate FIRST (by โb)translate (โ6; 0), then horizontal stretch SF 1/2(b) apply to Q(8, โ1)translate (โ6; 0): (8, โ1) โ (2, โ1)stretch SF 1/2: (2, โ1) โ (1, โ1)image: (1, โ1)check: at new x = 1, new equation gives f(2ยท1 + 6) = f(8) = original y = โ1 โ
WE 3
Why order matters — vertical
Consider the desired equation y = 3f(x) + 2. (a) Apply “stretch SF 3, then translate (0; 2)” to f. (b) Apply “translate (0; 2), then stretch SF 3” to f. (c) Which gives 3f(x) + 2?
(a) stretch then translatef(x) โ 3f(x) โ 3f(x) + 2y = 3f(x) + 2 โ(b) translate then stretchf(x) โ f(x) + 2 โ 3ยท[f(x) + 2]y = 3f(x) + 6 โ(c) compareonly (a) gives 3f(x) + 2 — the stretch acts on the constant if done second, doubling it from 2 to 6.
WE 4
Why order matters — horizontal
Consider the desired equation y = f(2x − 4). (a) Apply “translate (4; 0), then stretch SF 1/2” to f. (b) Apply “stretch SF 1/2, then translate (4; 0)” to f. (c) Which gives f(2x − 4)?
(a) translate then stretchf(x) โ f(x โ 4) โ f(2x โ 4)y = f(2x โ 4) โ(b) stretch then translatef(x) โ f(2x) โ f(2(x โ 4)) = f(2x โ 8)y = f(2x โ 8) โ(c) compareonly (a) gives f(2x โ 4) — the stretch doubles the โ4 if applied to the translated form, hence โ8.
WE 5
Mixing H and V — full composite
The graph of y = f(x) undergoes: vertical stretch SF 3, vertical translation (0; −1), horizontal translation (4; 0), horizontal stretch SF 1/2. (a) Write the equation of the new graph. (b) Find the image of A(2, 1) on the new graph.
An idealised spring’s displacement (cm) at time t (s) is f(t) = sin t. A real spring has triple the amplitude, half the period, and oscillates about an equilibrium 0.5 cm above the original. (a) Write the new displacement function d(t). (b) State the new amplitude, period, max and min. (c) Find d(π4).
(a) build the compositetriple amplitude: vertical SF 3 โ 3 sin thalve period: horizontal SF 1/2 โ 3 sin(2t)shift equilibrium up 0.5: โ 3 sin(2t) + 0.5d(t) = 3 sin(2t) + 0.5(b) propertiesamplitude: 3 cm; period: 2ฯ/2 = ฯ smax: 3 + 0.5 = 3.5; min: โ3 + 0.5 = โ2.5amp 3 cm ยท period ฯ s ยท max 3.5 cm ยท min โ2.5 cm(c) substitute t = ฯ/4d(ฯ/4) = 3 sin(ฯ/2) + 0.5 = 3ยท1 + 0.5d(ฯ/4) = 3.5 cm (max)
๐ก Top tips
Verticals follow the formula reading order: af(x) + b = stretch then translate.
Horizontals are reversed: f(ax + b) = translate then stretch — remember this is reverse!
H and V are independent: interleave them however you like, as long as the H sequence and V sequence stay correct internally.
Apply to key points step by step rather than trying to compute the final image directly — less error-prone.
Negative coefficients add a reflection: −2f(x) = vertical stretch SF 2 + reflection in x-axis.
โ Common mistakes
Doing horizontal stretch before translate on f(ax + b) — gives the wrong constant (off by a factor of a).
Doing vertical translate before stretch on af(x) + b — gives a translation of ab instead of b.
Forgetting that f(ax) has SF 1/a, not a — carried over from single stretches.
Mixing up the order WITHIN the H sequence when there are multiple horizontal operations — treat them as a chain.
Not factorising ax + b when computing a point image — the formula is ((p − b)/a, q), not ((p/a) − b, q).
Chapter complete — you now have all four Transformations of Graphs sub-topics: Translations, Reflections, Stretches, and Composite Transformations. Ready for any AI HL Paper 1 or 2 transformation question.
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