IB Maths AA HL
Topic 2 — Functions
Paper 1 & 2
~8 min read
Composite Transformations of Graphs
When two or more transformations are applied at once, the order matters — apply them in the wrong sequence and the answer will be off. The good news: vertical and horizontal transformations don’t interfere with each other, so the only sequencing you need to think about is within each direction. Vertical follows order of operations; horizontal works backwards.
📘 What you need to know
- Vertical composite: y = af(x) + b — apply vertical stretch SF a first, then translate by (0, b). Order = order of operations on y.
- Horizontal composite: y = f(ax + b) — apply translate by (−b, 0) first, then horizontal stretch SF 1/a. Order = reverse of order of operations on x.
- Negative a in either form means there’s also a reflection — in the x-axis for −f(x), in the y-axis for f(−x).
- Vertical and horizontal are independent: you can do all the horizontal work first, then all the vertical work — or vice versa. As long as the within-direction order is right.
- Point mapping: for y = af(bx + c) + d, a point (p, q) on f maps to ((p − c)/b, aq + d).
- State the transformations explicitly: examiners want each transformation named (translation by what vector, stretch by what SF, in which direction).
Vertical composite — af(x) + b
Order: stretch first, translate second
y = f(x) → y = af(x) → y = af(x) + b
Step 1: vertical stretch with scale factor a. Step 2: translate by a‘s output by b. Doing the translation first would give the wrong answer (af(x) + ab instead of af(x) + b).
🤔 Why stretch before translate?
Read the equation as if you’re computing y: take f(x), multiply by a, then add b. The transformations follow that same order — stretch first (multiplication), then translate (addition).
Effect on a point: (p, q) on y = f(x) maps to (p, aq + b) on y = af(x) + b. The x-coordinate stays put — it’s a vertical-only transformation.
Horizontal composite — f(ax + b)
Order: translate first, stretch second
y = f(x) → y = f(x + b) → y = f(ax + b)
Step 1: translate by (−b, 0). Step 2: horizontal stretch with scale factor 1/a. The order is the reverse of the order of operations on x — that’s the part that trips students up.
🤔 Why is horizontal order reversed?
Horizontal transformations work on the input to f. To “undo” the operations you read left-to-right, you have to apply the transformations right-to-left. So in f(ax + b), the “+ b” gets handled first (giving the translation), then the “a×” handles afterwards (giving the stretch).
Effect on a point: (p, q) on y = f(x) maps to ((p − b)/a, q) on y = f(ax + b). The y-coordinate is untouched.
Factoring shortcut: rewrite f(ax + b) as f(a(x + b/a)). Now you can do the stretch first (SF 1/a) and then the translation by (−b/a, 0) — both orders give the same final graph. Pick whichever feels easier.
Mixing horizontal and vertical
Full composite
y = af(bx + c) + d — (p, q) → (p − cb, aq + d)
Horizontal work
new x = (p − c) ÷ b
handles b and c only — needs reverse order
Vertical work
new y = aq + d
handles a and d only — natural order
Treat the horizontal and vertical work as two separate problems. Solve the x-side using “(p − c) ÷ b” and the y-side using “aq + d“. Then combine. No need to worry about which goes first — they’re independent.
🧭 Recipe — applying composite transformations
- Identify the form: af(x) + b (vertical only), f(ax + b) (horizontal only), or af(bx + c) + d (both).
- Vertical part: stretch SF a first, then translate by b (or d). Apply (p, q) → (p, aq + b).
- Horizontal part: translate by −b (or −c) first, then stretch SF 1/a. Apply (p, q) → ((p − b)/a, q).
- Combine the two — horizontal and vertical never interfere.
- Negative coefficients: include the matching reflection (in x-axis if vertical SF is negative; in y-axis if horizontal SF is negative).
Worked examples
WE 1Find the equation after a vertical composite
The graph of f(x) = x2 − 4 is transformed by a vertical stretch with scale factor 2 followed by a translation by the vector 0
5. Find the equation of the new graph.
Apply y → 2y + 5 to f(x)
y = 2f(x) + 5
y = 2(x² − 4) + 5
y = 2x² − 8 + 5
y = 2x² − 3
y = 2x² − 3
stretch first (everything ×2), then translate (+5). Translating first would have shifted the −4 along with everything else and given the wrong constant
WE 2Apply a vertical composite to key points
The graph of y = f(x) has a maximum at A(3, 7) and a minimum at B(−1, −2). Find the new coordinates of A and B on the graph of y = 4f(x) − 6.
Apply y → 4y − 6 to each y-coordinate
A(3, 7) → (3, 4(7) − 6) = (3, 22)
B(−1, −2) → (−1, 4(−2) − 6) = (−1, −14)
max (3, 22); min (−1, −14)
x-coordinates are untouched because it’s a vertical-only transformation
WE 3Apply a horizontal composite — the tricky one
The graph of y = f(x) passes through the point C(8, 5) and has a vertical asymptote at x = −3. Find the corresponding feature on the graph of y = f(2x + 4).
Apply x → (x − b)/a with a = 2, b = 4
new x = (old x − 4)/2
Image of C(8, 5)
new x = (8 − 4)/2 = 2
y unchanged: 5
point: (2, 5)
Image of VA x = −3
new x = (−3 − 4)/2 = −7/2
VA: x = −7/2
horizontal-only — y-coordinate of the point doesn’t move; verify by step-by-step: translate by (−4, 0) gives (4, 5) and x = −7, then horizontal stretch SF 1/2 halves the x-coords
WE 4Describe the sequence of transformations
The graph of y = f(x) is transformed to give the graph of y = 3f(x) − 4. Describe the sequence of transformations.
Identify the form: af(x) + b with a = 3, b = −4
vertical-only composite
Order: stretch first, then translate
step 1: vertical stretch parallel to the y-axis, scale factor 3
step 2: translation by the vector (0, −4)
vertical stretch SF 3, then translation by (0, −4)
naming both the stretch direction (“parallel to y-axis”) and the translation vector picks up full marks
WE 5Describe a horizontal composite
The graph of y = f(x) is transformed to give the graph of y = f(3x − 9). Describe the sequence of transformations.
Identify the form: f(ax + b) with a = 3, b = −9
horizontal-only composite
Order: translate first, then stretch
step 1: translation by (−b, 0) = (9, 0) — right by 9
step 2: horizontal stretch parallel to the x-axis, scale factor 1/3
translation by (9, 0), then horizontal stretch SF 1/3
factoring as f(3(x − 3)) gives an equivalent description: stretch SF 1/3 first, then translation by (3, 0). Both are correct — the final graph is the same.
WE 6Full composite — horizontal and vertical together
The graph of y = f(x) passes through the point (10, 4). Find the image of this point on the graph of y = 12f(x + 6) − 3.
Identify: af(bx + c) + d with a = 1/2, b = 1, c = 6, d = −3
Horizontal: new x = (p − c)/b = (10 − 6)/1 = 4
10 → 4
Vertical: new y = aq + d = (1/2)(4) − 3 = 2 − 3 = −1
4 → −1
image: (4, −1)
treat horizontal and vertical separately — solve each independently, then combine. b = 1 means no horizontal stretch, just the translation by (−6, 0)
💡 Top tips
- Vertical follows order of operations on y: af(x) + b means multiply (stretch), then add (translate).
- Horizontal reverses the order of operations on x: f(ax + b) means deal with the +b first (translate), then the a× (stretch).
- Use the point map: (p, q) → ((p − c)/b, aq + d) is the fastest route when you need the image of a point.
- Factor out the coefficient inside the bracket when in doubt: f(2x + 6) = f(2(x + 3)) makes the order obvious.
- Horizontal and vertical are independent — treat them as two separate problems and combine at the end.
- For a sketch question: pick 2 or 3 labelled points, transform each one, plot, then connect with the same shape as the original.
- State each transformation with full IB language: “translation by the vector (…)” and “stretch parallel to the x/y-axis with scale factor (…)”.
⚠ Common mistakes
- Doing horizontal in the natural order: translating before factoring out the coefficient gives the wrong answer.
- Translating before stretching for vertical composites: af(x) + b ≠ a(f(x) + b). Distributing the a changes the constant.
- Forgetting the reciprocal in the horizontal stretch: f(2x + 4) involves SF 1/2, not 2.
- Writing “translation of (5, 0)” instead of “translation by the vector (5, 0)” — examiners want the right wording.
- Mixing horizontal and vertical shifts in a single calculation. Handle them separately to avoid sign errors.
- Skipping a transformation in the description: every coefficient or constant in the equation corresponds to one transformation. Account for each.
- Incorrect order in the description: even if the final graph is right, examiners deduct marks for stating the wrong sequence.
And that closes Section 2.5 — Transformations of Graphs. The four types — translations, reflections, stretches, and composites — work for every function you’ve met so far and every one you’ll meet later. The next section, Polynomial Functions, dives back into algebra: factor and remainder theorems, repeated roots, and the tools for handling cubics, quartics, and beyond.
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