Graph Transformations

This lesson covers graph transformations as required by the AQA A-Level Mathematics specification (7357). You must be able to apply translations, stretches, and reflections to the graphs of functions, and understand the effect of combining transformations. This topic is fundamental and connects to almost every other area of the specification.

The Standard Transformations

Given a function y = f(x), the following transformations can be applied:

Translations

Transformation	Effect	Description
y = f(x) + a	Vertical translation by a	Shift up by a (or down if a < 0)
y = f(x + a)	Horizontal translation by −a	Shift left by a (or right if a < 0)
y = f(x − a)	Horizontal translation by +a	Shift right by a

Key point: y = f(x + a) shifts left (the opposite direction to what you might expect). Think of it as: the curve reaches each y-value a units sooner.

Example: The curve y = x² has vertex at (0, 0).

y = x² + 3 shifts up 3: vertex at (0, 3).
y = (x − 2)² shifts right 2: vertex at (2, 0).
y = (x + 1)² − 4 shifts left 1 and down 4: vertex at (−1, −4).

Translation as a vector: y = f(x − a) + b represents a translation by the vector (a, b).

Stretches

Transformation	Effect	Description
y = af(x)	Vertical stretch, scale factor a	Multiply all y-coordinates by a
y = f(ax)	Horizontal stretch, scale factor 1/a	Multiply all x-coordinates by 1/a

Example: Given y = sin x:

y = 3 sin x stretches vertically by factor 3 (amplitude becomes 3).
y = sin 2x stretches horizontally by factor 1/2 (period becomes π instead of 2π).

Exam Tip: For y = f(ax), the scale factor is 1/a, not a. This is a very common source of errors. If a = 2, the graph is compressed horizontally by factor 1/2 (it becomes "narrower", not wider).

Reflections

Transformation	Effect	Description
y = −f(x)	Reflection in the x-axis	All y-coordinates are negated
y = f(−x)	Reflection in the y-axis	All x-coordinates are negated

Example: Given y = x³:

y = −x³ reflects in the x-axis.
y = (−x)³ = −x³ reflects in the y-axis. (For this particular function, both reflections give the same result because x³ is an odd function.)

Example: Given y = eˣ:

y = −eˣ reflects in the x-axis (graph goes below the x-axis).
y = e⁻ˣ reflects in the y-axis (graph now decays instead of growing).

Combining Transformations

When multiple transformations are applied, the order matters. The general approach is:

Operations Inside the Function Argument (Affecting x)

For y = f(ax + b), complete the square on the argument:

y = f(a(x + b/a))

This represents:

Horizontal translation by −b/a.
Horizontal stretch by scale factor 1/a.

The order in which you apply these depends on whether you work from the "outside in" or "inside out".

Standard Order of Combined Transformations

For y = p·f(a(x − h)) + k, applied to y = f(x):

Starting from y = f(x):

Horizontal translation by h (replace x with x − h).
Horizontal stretch by factor 1/a (replace x with ax).
Vertical stretch by factor p (multiply f by p).
Vertical translation by k (add k).

However, many textbooks and mark schemes accept different valid orders provided the final equation is correct.

Example: Describe the transformations that map y = sin x onto y = 2 sin(3x − π) + 1.

Rewrite: y = 2 sin(3(x − π/3)) + 1.

Starting from y = sin x:

Translate right by π/3: y = sin(x − π/3).
Horizontal stretch, scale factor 1/3: y = sin(3(x − π/3)) = sin(3x − π).
Vertical stretch, scale factor 2: y = 2 sin(3x − π).
Translate up by 1: y = 2 sin(3x − π) + 1.

Transformations of Specific Curves

Quadratics

The general form y = a(x − h)² + k tells us:

Vertex is at (h, k).
Vertical stretch by factor a (reflection in x-axis if a < 0).

Example: y = −2(x + 3)² + 5 has vertex at (−3, 5), opens downward (a = −2), and is stretched vertically by factor 2.

Reciprocal Function

Starting from y = 1/x:

y = 1/(x − 2) + 3 has asymptotes at x = 2 and y = 3 (translated right 2 and up 3).

Exponential and Logarithmic Functions

Starting from y = eˣ:

y = eˣ⁻¹ + 2 translates right 1 and up 2. Asymptote shifts from y = 0 to y = 2.

Starting from y = ln x:

y = ln(x + 3) − 1 translates left 3 and down 1. Asymptote shifts from x = 0 to x = −3.

Effect on Key Features

When applying transformations, track how key features change:

Feature	Translation (a, b)	Vertical stretch (factor k)	Reflection in x-axis
Coordinates (x, y)	(x + a, y + b)	(x, ky)	(x, −y)
x-intercepts	Shift by a	Unchanged (if y = 0, ky = 0)	Unchanged
y-intercept	Changes	Multiplied by k	Negated
Turning point	Shifts	y-coordinate multiplied by k	y-coordinate negated
Asymptotes	Shift accordingly	Horizontal: multiplied by k; Vertical: unchanged	Horizontal: negated

Worked Exam-Style Example

The curve y = f(x) passes through (0, 3), (2, 0), and (5, −1). Sketch the curve y = 2f(x − 1) + 3, stating the coordinates of the transformed points.

Starting from y = f(x):

Step 1: Replace x with x − 1 (translate right 1):

(0, 3) → (1, 3), (2, 0) → (3, 0), (5, −1) → (6, −1).

Step 2: Multiply f by 2 (vertical stretch by factor 2):

(1, 3) → (1, 6), (3, 0) → (3, 0), (6, −1) → (6, −2).

Step 3: Add 3 (translate up 3):

(1, 6) → (1, 9), (3, 0) → (3, 3), (6, −2) → (6, 1).

Transformed points: (1, 9), (3, 3), (6, 1).

Summary

y = f(x) + a: vertical translation (up if a > 0).
y = f(x + a): horizontal translation (left if a > 0).
y = af(x): vertical stretch by factor a.
y = f(ax): horizontal stretch by factor 1/a.
y = −f(x): reflection in the x-axis.
y = f(−x): reflection in the y-axis.
The order of transformations matters when combining.
Track key points (intercepts, turning points, asymptotes) through each transformation.

Exam Tip: When describing transformations, use precise mathematical language. Say "translation by vector (3, 0)" or "horizontal stretch with scale factor 1/2 about the y-axis", not "move right" or "squash". When applying combined transformations, process them step by step and clearly show how each key point transforms. Always state the new coordinates of key features.

A-Level Deep Dive: Graph Transformations

Spec mapping

AQA 7357 specification, Paper 1 — Pure Mathematics, Section B (Algebra and Functions): the specification requires students to "understand the effect of simple transformations on the graph of $y = f(x)$ , including sketching associated graphs" for the four atomic transformations $y = f(x) + a$ , $y = f(x - a)$ , $y = kf(x)$ and $y = f(kx)$ , plus reflections $y = -f(x)$ and $y = f(-x)$ . The same content reappears synoptically across Section D (Coordinate geometry, transforming circles and conics), Section F (Trigonometry, sketching $y = a\sin(bx + c) + d$ ), Section G (Exponentials and logarithms, transforming $y = e^x$ to $y = e^{x - 1} + 2$ ) and Section L (Differentiation, where transformed graphs preserve stationary-point structure under affine maps). The AQA formula booklet does not list transformation rules — they must be memorised, and the direction of $f(x - a)$ in particular trips up large numbers of candidates every series.

Worked example with full mark scheme

Question (8 marks):

The function $f$ is defined by $f(x) = x^2 - 4x + 1$ for all real $x$ .

(a) Express $f(x)$ in completed-square form. (2)

(b) The curve $y = f(x)$ is transformed in two stages: first by the translation $y = f(x - 3)$ , and then by the stretch $y = 2f(x - 3)$ . Sketch each stage, marking the coordinates of the vertex and the $y$ -intercept on every diagram. (6)

Solution with mark scheme:

(a) Complete the square: $x^2 - 4x + 1 = (x - 2)^2 - 4 + 1 = (x - 2)^2 - 3$ .

M1 — correct halving of the linear coefficient and subtraction of the squared term. A1 — fully simplified completed-square form $(x - 2)^2 - 3$ .

(b) Stage 1 — translation $y = f(x - 3)$ .

Replacing $x$ by $(x - 3)$ inside the function shifts the graph 3 units to the right (not left — see misconception 1 below). The new function is $g(x) = ((x - 3) - 2)^2 - 3 = (x - 5)^2 - 3$ .

The vertex of $y = f(x)$ at $(2, -3)$ moves to $(5, -3)$ . The $y$ -intercept changes: substituting $x = 0$ into $g$ gives $g(0) = (0 - 5)^2 - 3 = 25 - 3 = 22$ .

M1 — correct shift direction (right, not left). A1 — vertex $(5, -3)$ and $y$ -intercept $(0, 22)$ both correctly identified on the sketch.

Stage 2 — stretch $y = 2g(x) = 2f(x - 3)$ .

Multiplying the output by $2$ produces a vertical stretch of scale factor $2$ from the $x$ -axis. Every $y$ -coordinate doubles; $x$ -coordinates are unchanged. So the vertex $(5, -3)$ moves to $(5, -6)$ , and the $y$ -intercept $(0, 22)$ moves to $(0, 44)$ . The new function is $h(x) = 2(x - 5)^2 - 6$ .

M1 — correct identification of vertical stretch from the $x$ -axis (not horizontal). A1 — vertex $(5, -6)$ correctly placed. A1 — $y$ -intercept $(0, 44)$ correctly placed; both diagrams show the parabola opening upward with the new stationary point clearly marked. B1 — sketches drawn in correct sequence with each stage labelled $y = f(x - 3)$ and $y = 2f(x - 3)$ respectively.

Total: 8 marks (M3 A4 B1).

Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): The graph of $y = f(x)$ has a single stationary point at $(4, 9)$ and crosses the $y$ -axis at $(0, 1)$ .

(a) State the coordinates of the stationary point and the $y$ -intercept on $y = f(2x) - 5$ . (3)

(b) State the coordinates of the stationary point and the $y$ -intercept on $y = -f(x + 1)$ . (3)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — recognising $f(2x)$ as a horizontal stretch of scale factor $\tfrac{1}{2}$ , halving $x$ -coordinates while leaving $y$ -coordinates unchanged: $(4, 9) \to (2, 9)$ .
M1 (AO1.1b) — applying the vertical translation $-5$ after the stretch: $(2, 9) \to (2, 4)$ .
A1 (AO1.1b) — $y$ -intercept: at $x = 0$ , $f(2 \cdot 0) - 5 = f(0) - 5 = 1 - 5 = -4$ , giving $(0, -4)$ .

(b)

M1 (AO1.1a) — $f(x + 1)$ shifts the graph 1 unit left, so the stationary point moves from $(4, 9)$ to $(3, 9)$ .
M1 (AO1.1b) — applying the negation $y \mapsto -y$ reflects in the $x$ -axis: $(3, 9) \to (3, -9)$ .
A1 (AO1.1b) — $y$ -intercept: at $x = 0$ , $-f(0 + 1) = -f(1)$ . Without more information $f(1)$ is unknown, so the candidate must state "insufficient information" or note that the $y$ -intercept depends on $f(1)$ . (Examiners accept either provided the reasoning is shown.)

Total: 6 marks split AO1 = 6. AQA transformations questions are almost entirely AO1 — they test fluent recall of the four rules and the ability to compose them in sequence.

Synoptic links

Connects to:

Section B — Functions (composite and inverse functions): $f(x - a)$ is the composite $f \circ g$ where $g(x) = x - a$ . Recognising transformations as function composition unifies the topic with the wider functions strand and prepares the ground for inverse-function reflection in $y = x$ .
Section F — Trigonometry: sketching $y = 3\sin(2x + \pi/3) - 1$ requires composing a horizontal stretch (factor $\tfrac{1}{2}$ ), a horizontal translation (the bracket factorises as $2(x + \pi/6)$ , so shift left by $\pi/6$ ), a vertical stretch (factor $3$ ) and a vertical translation (down $1$ ). Every standard trig sketch is a composite transformation problem in disguise.
Section J — Modulus function: the graph $y = |f(x)|$ reflects the part of $y = f(x)$ below the $x$ -axis upward, and $y = f(|x|)$ replaces the left half with a mirror of the right half. Both depend on the same axis-based reasoning as the negation transformations $y = -f(x)$ and $y = f(-x)$ .
Section L — Differentiation: stationary points are preserved under translations and stretches in a predictable way. If $f'(c) = 0$ then $g(x) = f(x - a)$ has $g'(c + a) = 0$ , and $h(x) = kf(x)$ has $h'(c) = 0$ at the same $x$ . This means transformation problems and stationary-point problems share a common scaffold.
Section M — Modelling: real-world functions almost always require translation and scaling to fit data — population growth modelled as $P(t) = P_0 e^{kt}$ is a vertical stretch of $y = e^{kt}$ by factor $P_0$ . Recognising that a fitted model is a transformed standard curve is half the modelling job.

Mark-scheme literacy

Graph-transformation questions on 7357 split AO marks heavily toward AO1:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	75–85%	Stating direction of shift, identifying stretch factor, applying transformations in correct order, computing image coordinates of named points
AO2 (reasoning / interpretation)	10–20%	Justifying which transformation acts first when composing two operations; recognising that $f(2(x - a))$ is not the same as $f(2x - a)$
AO3 (problem-solving)	0–10%	Open-ended sketching with multiple acceptable orderings; rare at AS level, more common in Year 2 within trigonometric or exponential contexts

Graph Transformations

Graph Transformations

The Standard Transformations

Translations

Stretches

Reflections

Combining Transformations

Operations Inside the Function Argument (Affecting x)

Standard Order of Combined Transformations

Transformations of Specific Curves

Quadratics

Reciprocal Function

Exponential and Logarithmic Functions

Effect on Key Features

Worked Exam-Style Example

Summary

A-Level Deep Dive: Graph Transformations

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 1 format

Synoptic links

Mark-scheme literacy

More in Mathematics