IB Maths AI HL Transformations of Graphs Paper 1 & 2 Scale factors, asymptotes ~8 min read

Stretches of Graphs

A stretch scales the graph along one axis. y = af(x) stretches vertically by scale factor a; y = f(x/a) stretches horizontally by scale factor a. Watch the trap: y = f(ax) is a horizontal stretch by scale factor 1/a, NOT a.

📘 What you need to know

Vertical stretch: y = af(x) stretches by scale factor a parallel to the y-axis. y-coordinates multiply by a; x-coordinates stay.
Horizontal stretch: y = f(x/a) stretches by scale factor a parallel to the x-axis. x-coordinates multiply by a; y-coordinates stay.
The trap: y = f(ax) is a horizontal stretch by scale factor 1a — the bracket inverts the factor.
Scale factor > 1: graph grows away from the axis. 0 < SF < 1: graph compresses toward the axis (don’t call it a “squash” in the exam).
Points on the axis of stretch are unchanged (on x-axis for vertical stretch; on y-axis for horizontal stretch).
Asymptotes: horizontal asymptote y = k → y = ak under vertical stretch; vertical asymptote x = k → x = ak under horizontal stretch.

Vertical vs horizontal stretch

Multiplying the function output by a constant — y = af(x) — scales every y-coordinate by a. The curve stretches vertically: points above the x-axis move further up, points below move further down. Multiplying the input by 1/a — y = f(x/a) — scales every x-coordinate by a. The curve stretches horizontally. The two transformations are independent: y-stretch leaves x-values alone and vice versa. Crucially, when you see f(ax) with a > 1, the input is being multiplied by a, which actually shrinks the curve horizontally (scale factor 1/a).

Stretching points and asymptotes

For vertical stretch y = af(x): a point (p, q) moves to (p, aq). The vertex’s y-value scales by a; x-intercepts (where f = 0) stay put because a × 0 = 0. Horizontal asymptotes follow: y = k becomes y = ak; vertical asymptotes don’t move. For horizontal stretch y = f(x/a): a point (p, q) moves to (ap, q). The vertex’s x-value scales by a; y-intercept (where x = 0) stays put. Vertical asymptotes follow: x = k becomes x = ak; horizontal asymptotes don’t move.

Vertical stretch SF 2 doubles every y-coordinate (vertex drops to (0, −8); roots stay at ±2). Horizontal stretch SF 2 doubles every x-coordinate (vertex stays at (0, −4); roots move to ±4).

Stretches at a glance y = af(x) ↔ vertical SF a · y = f(x/a) ↔ horizontal SF a trap: y = f(ax) ↔ horizontal SF 1a (input multiplied means curve compressed)

Spotting the f(ax) trap

The most common exam slip is reading y = f(2x) as “horizontal stretch by 2.” It’s actually a horizontal stretch by 1/2 — the curve gets narrower, not wider. Think of it this way: a point (4, 7) on f lands on f(2x) at the x-value where 2x = 4, i.e. x = 2. So (4, 7) moves to (2, 7) — halved, not doubled. The reciprocal rule is mandatory examiner knowledge.

One-axis rule (again): a stretch only changes the coordinate matching its name. Vertical stretch → y-coords scale. Horizontal stretch → x-coords scale. Asymptotes obey the same rule.

🧭 Recipe — stretching a graph

Identify direction and SF: outside the function → vertical; inside the bracket → horizontal. Read the SF carefully.
Apply the trap rule: if you see f(ax), the horizontal SF is 1a. If you see f(x/a), the SF is a.
Stretch every key point: vertical → multiply y by SF; horizontal → multiply x by SF. Points on the axis of stretch don’t move.
Stretch the asymptotes: horizontal asymptote scales under vertical stretch; vertical asymptote scales under horizontal stretch.
Sketch the new curve using the stretched points, with the same general shape.

Worked examples

WE 1

Describe each stretch

For the graph of y = f(x), state the stretch type and scale factor:
(a) y = 3f(x)
(b) y = f(x/4)
(c) y = 12f(x)
(d) y = f(5x)

read the structure of each (a) coefficient outside → vertical SF 3 (b) divide inside by 4 → horizontal SF 4 (c) ½ outside → vertical SF ½ (d) multiply inside by 5 → horizontal SF 1/5 (the trap!) 3 vertical, 4 horizontal, ½ vertical, 1/5 horizontal remember: f(ax) has SF 1/a, not a.

WE 2

Image of a point

The point P(6, −2) lies on the graph of y = f(x). Find the corresponding point on:
(a) y = 4f(x)
(b) y = f(x/3)
(c) y = f(2x)
(d) y = 15f(x)

multiply matching coord by SF (a) vertical SF 4: (6, 4·−2) = (6, −8) (b) horizontal SF 3: (3·6, −2) = (18, −2) (c) horizontal SF ½: (½·6, −2) = (3, −2) (d) vertical SF 1/5: (6, −2/5) = (6, −0.4) (6, −8) · (18, −2) · (3, −2) · (6, −0.4)

WE 3

Stretch a quadratic

Given f(x) = x² − 6x + 8. (a) Find y = 3f(x) expanded. (b) Find y = f(2x) expanded. (c) State the vertex of each new graph.

vertex of f: complete square x² − 6x + 8 = (x − 3)² − 1, vertex (3, −1) (a) 3f(x) = 3(x² − 6x + 8) 3x² − 18x + 24, vertex (3, −3) (y-coord × 3: −1 × 3 = −3 ✓) (b) f(2x) = (2x)² − 6(2x) + 8 4x² − 12x + 8, vertex (3/2, −1) (x-coord × ½: 3 × ½ = 3/2 ✓) vertical stretch scales y; horizontal stretch SF ½ halves x.

WE 4

Stretches of asymptotes

The graph of y = f(x) has vertical asymptote x = 6 and horizontal asymptote y = −4. State the asymptotes of:
(a) y = 5f(x)
(b) y = f(x/2)
(c) y = f(3x)
(d) y = 14f(x)

apply matching axis rule (a) vertical SF 5: VA x = 6 (unchanged); HA y = 5·−4 = −20 (b) horizontal SF 2: VA x = 2·6 = 12; HA y = −4 (unchanged) (c) horizontal SF 1/3: VA x = (1/3)·6 = 2; HA y = −4 (unchanged) (d) vertical SF ¼: VA x = 6 (unchanged); HA y = ¼·−4 = −1 (a) x=6, y=−20 · (b) x=12, y=−4 · (c) x=2, y=−4 · (d) x=6, y=−1

WE 5

Reverse: write the equation

The graph of y = f(x) is transformed. Write the equation of the new graph in each case:
(a) Vertical stretch, scale factor 4.
(b) Horizontal stretch, scale factor 2.
(c) Vertical stretch, scale factor 1/3.
(d) Horizontal stretch, scale factor 1/5.

apply the formula for each (a) y = 4f(x) (b) y = f(x/2) (c) y = (1/3)f(x) (d) horizontal SF 1/5 ↔ f(ax) with a = 5 (d) y = f(5x) part (d) is the classic trap: SF 1/5 ↔ f(5x).

WE 6

Applied: spring with new amplitude and period

A spring’s displacement (cm) at time t (s) is modelled by f(t) = sin t. A stiffer, shorter spring has displacement d(t) = 3f(2t). (a) Describe the transformations. (b) State the new amplitude and period. (c) Find d(π4).

(a) read 3 outside & 2 inside 3f(t): vertical stretch SF 3 (amplitude × 3) f(2t): horizontal stretch SF ½ (period × ½) vertical SF 3, horizontal SF ½ (b) apply to original original sin t: amplitude 1, period 2π new amplitude 3 cm, new period π s (c) substitute t = π/4 d(π/4) = 3 sin(2·π/4) = 3 sin(π/2) = 3·1 d(π/4) = 3 cm (max)

💡 Top tips

Outside → vertical, inside → horizontal. Treat coefficient placement like a switch.
Avoid the words “squash” or “compress” in formal answers — say “stretch by scale factor 12” instead.
Reciprocal rule: any time you see f(ax), flip a to get the SF.
Points on the axis of stretch don’t move — use that as a quick check.
Sketch the original first, mark a few key points, then scale to produce the new graph.

⚠ Common mistakes

Treating f(2x) as a stretch by 2 — it’s actually scale factor 1/2 (curve gets narrower).
Scaling the wrong coordinate — vertical stretch doesn’t move x-values; horizontal stretch doesn’t move y-values.
Forgetting that x-intercepts stay put under vertical stretch (since a × 0 = 0).
Forgetting that y-intercepts stay put under horizontal stretch (since the value at x = 0 doesn’t change).
Moving the wrong asymptote — vertical stretch only affects horizontal asymptotes, and vice versa.

Next up: Composite Transformations — combining stretches, reflections, and translations, plus the order rules af(x) + b and f(ax + b).

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