Transformation of Graphs · Study Notes · Edexcel IGCSE 4MA1 Higher

Translate, stretch and reflect graphs using function notation. A* topic — counter-intuitive but rule-based.

At a glance

Vertical translation: $f(x) + a$ moves UP by $a$ .
Horizontal translation: $f(x + a)$ moves LEFT by $a$ (sign flips!).
Reflection in $x$ -axis: $-f(x)$ .
Reflection in $y$ -axis: $f(-x)$ .
Vertical stretch: $kf(x)$ — by factor $k$ .
Horizontal stretch: $f(kx)$ — by factor $1/k$ (inverse).

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

3.6 — Translate graphs vertically and horizontally.
3.6 — Reflect graphs in the $x$ -axis and $y$ -axis.
3.6 — Stretch graphs in horizontal and vertical directions.
3.6 — Combine multiple transformations.

Translations

▼

$+a$ shifts UP. $f(x + a)$ shifts LEFT (sign flips for horizontal).

Vertical translation: $f(x) + a$ .

Adds $a$ to every $y$ -value.
Graph moves UP by $a$ (or DOWN if $a$ negative).

Example. $f(x) = x^{2}$ .

$f(x) + 3 = x^{2} + 3$ . Vertex moves from $(0,0)$ to $(0, 3)$ . Up by $3$ .
$f(x) - 2 = x^{2} - 2$ . Vertex at $(0, -2)$ . Down by $2$ .

Horizontal translation: $f(x + a)$ .

Replaces $x$ with $x + a$ INSIDE the function.
Graph moves LEFT by $a$ (counter-intuitive!).
$f(x - a)$ moves RIGHT by $a$ .

Example. $f(x) = x^{2}$ .

$f(x + 2) = (x + 2)^{2}$ . Vertex at $(-2, 0)$ . LEFT by $2$ .
$f(x - 3) = (x - 3)^{2}$ . Vertex at $(3, 0)$ . RIGHT by $3$ .

Why does horizontal translation work backwards?

Think: 'where is $f(0)$ in the new graph?' For $f(x + 2)$ : at $x + 2 = 0$ , i.e. $x = -2$ . So the value originally at $x = 0$ is now at $x = -2$ . Graph shifted LEFT.

Outside the bracket moves the graph up; inside the bracket moves it the opposite way — left.

Combined translation.

$f(x + a) + b$ : shift LEFT by $a$ AND UP by $b$ .

Vector notation: translation by $\binom{-a}{b}$ .

Worked qualitative. Sketch $y = (x - 2)^{2} + 3$ .

This is $f(x) = x^{2}$ translated by $\binom{2}{3}$ .
Vertex at $(2, 3)$ .
Up-opening parabola.

Edexcel tip. Always state translation as a VECTOR. Mark schemes credit the explicit vector.

$f(x) + a$ : UP by $a$ .
$f(x + a)$ : LEFT by $a$ (counter-intuitive).
$f(x - a)$ : RIGHT by $a$ .
Vector: $\binom{-a}{b}$ for $f(x + a) + b$ .

Reflections

▼

$-f(x)$ : flip in $x$ -axis. $f(-x)$ : flip in $y$ -axis.

Reflection in $x$ -axis: $y = -f(x)$ .

Negate every $y$ -value.
Graph flips upside down.

Example. $f(x) = x^{2} + 1$ .

$-f(x) = -(x^{2} + 1) = -x^{2} - 1$ .
Down-opening parabola, vertex at $(0, -1)$ .

Reflection in $y$ -axis: $y = f(-x)$ .

Negate every $x$ -value.
Graph flips left-right.

−f(x) flips the curve in the x-axis (output negated); f(−x) flips it in the y-axis (input negated).

Example. $f(x) = (x - 3)^{2}$ .

$f(-x) = (-x - 3)^{2} = (x + 3)^{2}$ .
Vertex moves from $(3, 0)$ to $(-3, 0)$ .

Symmetry test.

If $f(-x) = f(x)$ , the function is EVEN — symmetric about $y$ -axis. (E.g. $x^{2}, \cos x$ .)

If $f(-x) = -f(x)$ , the function is ODD — symmetric about origin. (E.g. $x, x^{3}, \sin x$ .)

Worked qualitative. Is $f(x) = x^{2} + 1$ symmetric about the $y$ -axis?

$f(-x) = (-x)^{2} + 1 = x^{2} + 1 = f(x)$ . Yes — even function.

Edexcel tip. Memorise: ' $-f$ ' negates OUTPUT (X-axis flip). ' $f(-x)$ ' negates INPUT (Y-axis flip).

$-f(x)$ : flip in $x$ -axis.
$f(-x)$ : flip in $y$ -axis.
Even functions: symmetric about $y$ -axis.
Odd functions: symmetric about origin.

Stretches

▼

$kf(x)$ : vertical stretch by $k$ . $f(kx)$ : horizontal stretch by $1/k$ .

Vertical stretch: $y = kf(x)$ .

Multiplies every $y$ -value by $k$ .
Stretches graph VERTICALLY by factor $k$ .

Example. $f(x) = \sin x$ .

$2f(x) = 2\sin x$ . Amplitude doubles.

Horizontal stretch: $y = f(kx)$ .

Replaces $x$ with $kx$ INSIDE the function.
Stretches HORIZONTALLY by factor $1/k$ (inverse!).
$f(2x)$ : COMPRESSES by factor $2$ (stretches by $1/2$ ).

Example. $f(x) = \sin x$ has period $360°$ .

$f(2x) = \sin(2x)$ has period $180°$ . (Compressed.)
$f(x/2) = \sin(x/2)$ has period $720°$ . (Stretched.)

Why is horizontal stretch by $1/k$ ?

Think: 'where does the graph hit $y =$ value?' For $f(2x) = \sin(2x)$ : hit at $2x = 90°$ , i.e. $x = 45°$ . Originally hit at $x = 90°$ . So compressed (less $x$ to reach the same $y$ ).

Common stretches:

Function	Effect
$2f(x)$	Vertical stretch ×2
$\dfrac{1}{2}f(x)$	Vertical compression by 2 (or stretch ×1/2)
$f(2x)$	Horizontal compression by 2 (period halves)
$f(x/2)$	Horizontal stretch ×2 (period doubles)
$-f(x)$	Reflection (no stretch)
$f(-x)$	Reflection

Edexcel tip. When a question shows graph $y = f(x)$ and asks to draw $y = 2f(x)$ , multiply each $y$ -value by $2$ . Mark schemes credit accuracy at key points.

$kf(x)$ : vertical stretch by $k$ .
$f(kx)$ : horizontal stretch by $1/k$ .
Horizontal is the INVERSE.
Stretches change shape; reflections + translations don't.

Quick recap

$f(x) + a$ : UP by $a$ .
$f(x + a)$ : LEFT by $a$ .
$-f(x)$ : $x$ -axis reflection.
$f(-x)$ : $y$ -axis reflection.
$kf(x)$ : vertical stretch by $k$ .
$f(kx)$ : horizontal stretch by $1/k$ .

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Vertical translation: $f(x) + a$ .
Horizontal translation: $f(x + a)$ — LEFT by $a$ .
$x$ -axis reflection: $-f(x)$ .
$y$ -axis reflection: $f(-x)$ .
Vertical stretch: $kf(x)$ .
Horizontal stretch: $f(kx)$ — by $1/k$ .

How it’s examined

Graph transformations on Higher Tier (4-6 marks). Often given $y = f(x)$ as a graph; asked to draw various transformations. Examiner reports flag (1) wrong direction for horizontal translation, (2) confusing reflection axes, (3) wrong factor for horizontal stretch.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H May/Jun 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on transformation of graphs, written exactly the way a tutor would explain them at the board.

1Vertical translation $y = f(x) + a$
Higher• translation
▼
Question
If $f(x) = x^{2}$ , sketch $y = f(x) + 3$ and describe the transformation.
Step-by-step solution
1. Step 1
  $y = f(x) + 3 = x^{2} + 3$ .
2. Step 2
  This is the graph of $y = x^{2}$ shifted UP by $3$ .
3. Step 3
  Vertex moves from $(0, 0)$ to $(0, 3)$ .
Answer
Translation by $\binom{0}{3}$ (up by $3$ ).
2Horizontal translation $y = f(x + a)$
Higher• Adapted from 4MA1/1H May/Jun 2024 Q20• translation
▼
Question
If $f(x) = x^{2}$ , sketch $y = f(x + 2)$ .
Step-by-step solution
1. Step 1
  $y = (x + 2)^{2}$ .
2. Step 2
  Translation by $\binom{-2}{0}$ (LEFT by $2$ , NOT right).
3. Step 3
  Vertex moves from $(0, 0)$ to $(-2, 0)$ .
Answer
Translation by $\binom{-2}{0}$ (left by $2$ ).
Examiner tip
$f(x + a)$ shifts LEFT by $a$ (counter-intuitive!). $f(x - a)$ shifts RIGHT.
3Reflections
Higher• reflection
▼
Question
$f(x) = x^{2} + 1$ . Describe (a) $y = -f(x)$ , (b) $y = f(-x)$ .
Step-by-step solution
1. Step 1
  (a) $-f(x) = -(x^{2} + 1) = -x^{2} - 1$ . Reflection in the $x$ -axis (flip upside down).
2. Step 2
  (b) $f(-x) = (-x)^{2} + 1 = x^{2} + 1 = f(x)$ . (Same — symmetric about $y$ -axis.)
Answer
(a) Reflection in $x$ -axis. (b) Same graph (symmetric).
Examiner tip
$-f(x)$ : reflect in X-AXIS. $f(-x)$ : reflect in Y-AXIS.
4Stretches
Higher• stretch
▼
Question
$f(x) = \sin x$ . Describe (a) $y = 2f(x)$ , (b) $y = f(2x)$ .
Step-by-step solution
1. Step 1
  (a) $y = 2\sin x$ . Vertical stretch by factor $2$ . Amplitude doubles.
2. Step 2
  (b) $y = \sin(2x)$ . Horizontal stretch by factor $\dfrac{1}{2}$ (or compression). Period halves.
Answer
(a) Vertical stretch ×2. (b) Horizontal stretch by × $\dfrac{1}{2}$ .
Examiner tip
$kf(x)$ : VERTICAL stretch by $k$ . $f(kx)$ : HORIZONTAL stretch by $1/k$ (counter-intuitive — INVERSE for horizontal).
5Image of a point under $y = f(-x)$ [1 mark]
Getting started• Paper 1H/2H style — 1 mark• transformations, reflection
▼
Question
The point $(5, -4)$ lies on the graph of $y = f(x)$ .

Write down the coordinates of the image of this point on the graph of $y = f(-x)$ . [1]
Step-by-step solution
1. Step 1
  $y = f(-x)$ is $y = f(ax)$ with $a = -1$ : a reflection in the $y$ -axis. The $x$ -coordinate changes sign and the $y$ -coordinate stays the same.
2. Step 2
  B1 Change the sign of $x$ . Check: at $x = -5$ , $f(-(-5)) = f(5) = -4$ .
  $(5,\,-4) \rightarrow (-5,\,-4)$
Answer
$(-5, -4)$
Examiner tip
Examiner tip: The minus sign is inside the bracket, so it acts on $x$ . A minus sign outside, $y = -f(x)$ , would act on $y$ .

Common mistake: Reflecting in the $x$ -axis instead, giving $(5, 4)$ . That is the image on $y = -f(x)$ .
6Equation and turning point after a stretch parallel to the y-axis [2 marks]
Getting started• Paper 1H/2H style — 2 marks• transformations, stretch, quadratic graphs
▼
Question
The graph of $y = x^{2} + 1$ is stretched parallel to the $y$ -axis with scale factor $3$ .

(a) Write down an equation of the new graph. [1]

(b) Write down the coordinates of the turning point of the new graph. [1]
Step-by-step solution
1. Step 1
  B1 (a) A stretch parallel to the $y$ -axis with scale factor $3$ is $y = 3f(x)$ . Multiply the whole expression by $3$ .
  $y = 3(x^{2} + 1) = 3x^{2} + 3$
2. Step 2
  B1 (b) The turning point of $y = x^{2} + 1$ is $(0, 1)$ . Multiply its $y$ -coordinate by $3$ .
  $(0,\,1) \rightarrow (0,\,3)$
Answer
(a) $y = 3x^{2} + 3$ (b) $(0, 3)$
Examiner tip
Examiner tip: $3f(x)$ multiplies every term of $f(x)$ by $3$ . Use a bracket before you expand.

Common mistake: Writing $y = 3x^{2} + 1$ , stretching only the $x^{2}$ term. That graph has its turning point at $(0, 1)$ , which has not moved at all.
7Maximum points of $y = \sin(3x)^{\circ}$ [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• transformations, trigonometric graphs, stretch
▼
Question
Find the coordinates of all the maximum points of the graph of $y = \sin(3x)^{\circ}$ for $0 \leqslant x \leqslant 360$ . [3]
Step-by-step solution
1. Step 1
  M1 $\sin \theta^{\circ}$ has its maximum value $1$ when $\theta = 90, 450, 810, \ldots$ (every $360$ ). Here $\theta = 3x$ , and $x$ goes up to $360$ , so $3x$ goes up to $1080$ .
  $3x = 90,\ 450,\ 810$
2. Step 2
  A1 Divide by $3$ . The next value, $3x = 1170$ , gives $x = 390$ , which is outside the range.
  $x = 30,\ 150,\ 270$
3. Step 3
  A1 At each of these the value of $y$ is $1$ .
  $(30,\,1),\ (150,\,1),\ (270,\,1)$
Answer
$(30, 1)$ , $(150, 1)$ and $(270, 1)$
Examiner tip
Examiner tip: $y = \sin(3x)^{\circ}$ is $y = \sin x^{\circ}$ stretched parallel to the $x$ -axis with scale factor $\dfrac{1}{3}$ , so its period is $120^{\circ}$ and there are three full cycles in $0$ to $360$ : expect three maximum points.

Common mistake: Giving only $(30, 1)$ , the image of the one maximum of $y = \sin x^{\circ}$ , and not looking for the others in the range.
8Describing the transformation between two straight lines [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• transformations, linear graphs, stretch
▼
Question
Describe fully the single transformation that maps the graph of $y = 2x + 3$ onto the graph of $y = 6x + 3$ . [3]
Step-by-step solution
1. Step 1
  M1 Let $f(x) = 2x + 3$ . Write $6x + 3$ in terms of $f$ : replacing $x$ by $3x$ works.
  $f(3x) = 2(3x) + 3 = 6x + 3$
2. Step 2
  A1 $y = f(3x)$ is a stretch parallel to the $x$ -axis.
3. Step 3
  A1 The scale factor is the reciprocal of $3$ . Check: $y = 2x + 3$ meets the $x$ -axis at $-1.5$ and $y = 6x + 3$ meets it at $-0.5$ , which is $\dfrac{1}{3}$ of $-1.5$ .
  $\text{scale factor } \frac{1}{3}$
Answer
A stretch, scale factor $\dfrac{1}{3}$ , parallel to the $x$ -axis
Examiner tip
Examiner tip: The $y$ -intercept $(0, 3)$ has not moved. Points on the $y$ -axis stay fixed under a stretch parallel to the $x$ -axis, which is a quick sign that this is the right transformation.

Common mistake: Saying "stretch parallel to the $y$ -axis, scale factor $3$ ". That gives $y = 3(2x + 3) = 6x + 9$ , which moves the intercept to $(0, 9)$ .
9Finding a translation vector from two points on the image [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• transformations, translation, quadratic graphs
▼
Question
The graph of $y = x^{2}$ is translated by the vector $\begin{pmatrix} p \\ q \end{pmatrix}$ .

The new graph passes through the points $(1, 7)$ and $(5, 7)$ .

Find the value of $p$ and the value of $q$ . [4]
Step-by-step solution
1. Step 1
  M1 The two points have the same $y$ -coordinate, so they are symmetrical about the line through the turning point.
  $x = \frac{1 + 5}{2} = 3$
2. Step 2
  A1 The turning point moves from $(0, 0)$ to $x = 3$ .
  $p = 3$
3. Step 3
  M1 The new graph is $y = (x - 3)^{2} + q$ . Substitute $(1, 7)$ .
  $(1 - 3)^{2} + q = 7 \quad\Rightarrow\quad 4 + q = 7$
4. Step 4
  A1 Solve. Check with $(5, 7)$ : $(5 - 3)^{2} + 3 = 7$ .
  $q = 3$
Answer
$p = 3$ , $q = 3$
Examiner tip
Examiner tip: Every parabola $y = (x - p)^{2} + q$ is symmetrical about $x = p$ , so two points at the same height give $p$ straight away.

Common mistake: Writing the new graph as $y = (x + 3)^{2} + q$ because the vector has $+3$ on top. A move of $3$ to the right is $f(x - 3)$ .
10Translating a parabola so it touches the x-axis [5 marks]
Stretch• Paper 1H/2H style — 5 marks• transformations, translation, completing the square
▼
Question
$f(x) = x^{2} - 6x + 5$

The graph of $y = f(x) + k$ touches the $x$ -axis at exactly one point.

Find the value of $k$ and the coordinates of the point where the graph touches the $x$ -axis. [5]
Step-by-step solution
1. Step 1
  M1 Complete the square on $f(x)$ .
  $x^{2} - 6x + 5 = (x - 3)^{2} - 9 + 5$
2. Step 2
  A1 So the turning point of $y = f(x)$ is $(3, -4)$ .
  $f(x) = (x - 3)^{2} - 4$
3. Step 3
  M1 $y = f(x) + k$ moves the turning point up $k$ to $(3, -4 + k)$ . The graph touches the $x$ -axis at exactly one point only when its turning point is on the axis.
  $-4 + k = 0$
4. Step 4
  A1 Solve.
  $k = 4$
5. Step 5
  B1 The turning point is now on the $x$ -axis. Check: $f(x) + 4 = (x - 3)^{2}$ , which is $0$ only at $x = 3$ .
  $(3,\,0)$
Answer
$k = 4$ ; the graph touches the $x$ -axis at $(3, 0)$
Examiner tip
Examiner tip: "Touches" means the $x$ -axis is a tangent, so the turning point lies on it. Completing the square gives the turning point directly.

Common mistake: Giving $k = -4$ , moving the graph down instead of up, or taking $k = 5$ to cancel the constant term. $f(x) - 5 = x^{2} - 6x$ crosses the axis twice, at $0$ and $6$ .

Question 1
Paper 1H/2H style — 1 mark1 mark
The graph of $y = \sin x^{\circ}$ is reflected in the $x$ -axis.

Write down an equation of the new graph. [1]
Model answer
A reflection in the $x$ -axis changes the sign of every $y$ -coordinate: $y = -f(x)$ .

Answer: $y = -\sin x^{\circ}$
Why this scores
Mark scheme:
- B1 for $y = -\sin x^{\circ}$ oe
Why this scores full marks: The minus sign is placed outside the function, where it acts on $y$ . An equivalent form such as $y = \sin(-x)^{\circ}$ also scores B1, because $\sin(-x)^{\circ} = -\sin x^{\circ}$ for every $x$ : for sine, the reflection in the $y$ -axis gives the same graph as the reflection in the $x$ -axis. Writing $y = -\sin x^{\circ}$ shows the reflection directly.
Question 2
Paper 1H/2H style — 2 marks2 marks
(a) Write down the maximum value of $y = 4\sin x^{\circ}$ . [1]

(b) Write down the minimum value of $y = \sin x^{\circ} + 4$ . [1]
Model answer
(a) $4 \times 1 = 4$

Answer: $4$

(b) $-1 + 4 = 3$

Answer: $3$
Why this scores
Mark scheme:
- B1 for $4$ cao
- B1 for $3$ cao
Why this scores full marks: The candidate starts from the range of $y = \sin x^{\circ}$ , from $-1$ to $1$ . In (a) the stretch multiplies by $4$ ; in (b) the translation adds $4$ . Confusing the two gives $5$ in (a) or $-4$ in (b).
Question 3
Paper 1H/2H style — 2 marks2 marks
The graph of $y = 3x - 2$ is translated by the vector $\begin{pmatrix} 0 \\ a \end{pmatrix}$ .

The new graph passes through the point $(4, 7)$ .

Find the value of $a$ . [2]
Model answer
New graph: $y = 3x - 2 + a$

$3(4) - 2 + a = 7$

$10 + a = 7$

Answer: $a = -3$
Why this scores
Mark scheme:
- M1 for $3(4) - 2 + a = 7$ oe
- A1 for $a = -3$ cao
Why this scores full marks: The candidate writes the equation of the translated line before substituting. The original line has $y = 10$ at $x = 4$ , so it must move down $3$ ; a negative answer is expected.
Question 4
Paper 1H/2H style — 3 marks3 marks
The graph of $y = f(x)$ has a minimum point at $(-6, 3)$ .

Write down the coordinates of the image of this point on the graph of

(a) $y = f(x) - 3$ [1]

(b) $y = f\left(\dfrac{1}{2}x\right)$ [1]

(c) $y = -f(x)$ [1]
Model answer
(a) Down $3$ : $(-6, 0)$

(b) $x$ -coordinate multiplied by $2$ : $(-12, 3)$

(c) $y$ -coordinate changes sign: $(-6, -3)$
Why this scores
Mark scheme:
- B1 for $(-6, 0)$
- B1 for $(-12, 3)$
- B1 for $(-6, -3)$
Why this scores full marks: Each image is found by acting on only one coordinate. In (b) the candidate uses the reciprocal of $\dfrac{1}{2}$ , and in (c) notes that the minimum becomes a maximum point after the reflection in the $x$ -axis.
Question 5
Paper 1H/2H style — 3 marks3 marks
$f(x) = x^{2} - 1$

(a) Ali says, "The graph of $y = f(x + 3)$ is the graph of $y = f(x)$ translated $3$ units to the right."

Is Ali correct? Give a reason for your answer. [1]

(b) Find the coordinates of the points where the graph of $y = f(x + 3)$ crosses the $x$ -axis. [2]
Model answer
(a) No. $f(x + 3)$ is a translation $3$ units to the left, by $\begin{pmatrix} -3 \\ 0 \end{pmatrix}$ .

(b) $y = f(x)$ crosses the $x$ -axis where $x^{2} = 1$ , at $x = 1$ and $x = -1$ .

Move each $3$ to the left:

$1 - 3 = -2 \qquad -1 - 3 = -4$

Answer: $(-2, 0)$ and $(-4, 0)$
Why this scores
Mark scheme:
- B1 for No with a reason referring to a translation to the left, or by $\begin{pmatrix} -3 \\ 0 \end{pmatrix}$
- M1 for $(x + 3)^{2} - 1 = 0$ or for subtracting $3$ from $1$ or $-1$
- A1 for $(-2, 0)$ and $(-4, 0)$
Why this scores full marks: The reason in (a) states the correct direction, not just "No". In (b) both crossing points are given, and the check $(-2 + 3)^{2} - 1 = 0$ confirms the first.
Question 6
Paper 1H/2H style — 4 marks4 marks
The graph of $y = a\sin x^{\circ}$ passes through the point $(30, 3)$ .

(a) Find the value of $a$ . [2]

(b) Write down the coordinates of the maximum point and the minimum point of $y = a\sin x^{\circ}$ for $0 \leqslant x \leqslant 360$ . [2]
Model answer
(a) $a\sin 30^{\circ} = 3$

$0.5a = 3$

Answer: $a = 6$

(b) $y = \sin x^{\circ}$ has maximum $(90, 1)$ and minimum $(270, -1)$ . Multiply the $y$ -coordinates by $6$ .

Answer: maximum $(90, 6)$ , minimum $(270, -6)$
Why this scores
Mark scheme:
- M1 for $a\sin 30^{\circ} = 3$ or $0.5a = 3$
- A1 for $a = 6$ cao
- B1 for $(90, 6)$ ft their $a$
- B1 for $(270, -6)$ ft their $a$
Why this scores full marks: The candidate substitutes the given point rather than guessing, and knows the stretch parallel to the $y$ -axis leaves the $x$ -coordinates of the maximum and minimum unchanged.
Question 7
Paper 1H/2H style — 4 marks4 marks
$f(x) = x^{2} - 4x + 7$

Find the coordinates of the turning point of the graph of $y = f(2x)$ . [4]
Model answer
$f(x) = (x - 2)^{2} - 4 + 7 = (x - 2)^{2} + 3$

Turning point of $y = f(x)$ : $(2, 3)$

$f(2x)$ divides every $x$ -coordinate by $2$ :

$(2, 3) \rightarrow (1, 3)$

Check: $f(2x) = 4x^{2} - 8x + 7 = 4(x - 1)^{2} + 3$

Answer: $(1, 3)$
Why this scores
Mark scheme:
- M1 for $(x - 2)^{2} + c$ or for a method to find the turning point of $y = f(x)$
- A1 for $(2, 3)$
- M1 for halving their $x$ -coordinate, or for $4x^{2} - 8x + 7$
- A1 for $(1, 3)$ cao
Why this scores full marks: Finding the turning point of $y = f(x)$ first, then applying the stretch to that one point, keeps the working short. The check by expanding $f(2x)$ confirms that the $y$ -coordinate does not change.
Question 8
Paper 1H/2H style — 5 marks5 marks
The graph of $y = x^{2}$ is translated to give the graph of $y = (x + a)^{2} + b$ .

The turning point of the new graph lies on the line $y = 1$ , and the new graph passes through the point $(0, 5)$ .

Find the two possible equations of the new graph. [5]
Model answer
Turning point $(-a, b)$ is on $y = 1$ , so $b = 1$ .

Through $(0, 5)$ :

$(0 + a)^{2} + 1 = 5$

$a^{2} = 4$

$a = 2 \text{ or } a = -2$

Answer: $y = (x + 2)^{2} + 1$ or $y = (x - 2)^{2} + 1$
Why this scores
Mark scheme:
- B1 for $b = 1$
- M1 for $a^{2} + b = 5$ oe
- A1 for $a^{2} = 4$
- A1 for $a = 2$ and $a = -2$
- B1 for both equations correct
Why this scores full marks: The candidate takes both square roots of $4$ , which gives two different translations: by $\begin{pmatrix} -2 \\ 1 \end{pmatrix}$ and by $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ . Both curves pass through $(0, 5)$ , as the symmetry of the parabola suggests. Stopping at $a = 2$ loses the last two marks.
Question 9
Paper 1H/2H style — 5 marks5 marks
$f(x) = 2x + 1$

The graph of $y = f(x + a)$ passes through the point $(4, 3)$ .

(a) Find the value of $a$ . [2]

(b) The graph of $y = f(x + a)$ is also the graph of $y = f(x) + b$ .

Find the value of $b$ . [3]
Model answer
(a) $f(4 + a) = 2(4 + a) + 1 = 3$

$9 + 2a = 3$

Answer: $a = -3$

(b) $f(x - 3) = 2(x - 3) + 1 = 2x - 5$

$f(x) + b = 2x + 1 + b$

$1 + b = -5$

Answer: $b = -6$
Why this scores
Mark scheme:
- M1 for $2(4 + a) + 1 = 3$ oe
- A1 for $a = -3$ cao
- M1 for $2(x - 3) + 1$ or $2(x + a) + 1$ with their $a$
- A1 for $2x - 5$
- A1 for $b = -6$ ft their $a$
Why this scores full marks: For a straight line, a move sideways and a move up or down can give the same line. The candidate writes both equations in the form $y = 2x + c$ and compares intercepts. Check: moving $y = 2x + 1$ right $3$ lowers it by $2 \times 3 = 6$ .
Question 10
Paper 1H/2H style — 5 marks5 marks
The graph of $y = \cos(x + a)^{\circ}$ , where $0 < a < 180$ , passes through the point $(30, 0)$ .

(a) Find the value of $a$ . [2]

(b) For $0 \leqslant x \leqslant 360$ , find the coordinates of the maximum point and the minimum point of $y = \cos(x + a)^{\circ}$ . [2]

(c) Write down the coordinates of the point where the graph crosses the $y$ -axis. [1]
Model answer
(a) $\cos(30 + a)^{\circ} = 0$ , so $30 + a = 90$ (or $270$ , which gives $a = 240$ , too large)

Answer: $a = 60$

(b) Maximum when $x + 60 = 360$ , so $x = 300$

Minimum when $x + 60 = 180$ , so $x = 120$

Answer: maximum $(300, 1)$ , minimum $(120, -1)$

(c) $y = \cos 60^{\circ} = 0.5$

Answer: $(0, 0.5)$
Why this scores
Mark scheme:
- M1 for $\cos(30 + a)^{\circ} = 0$ leading to $30 + a = 90$
- A1 for $a = 60$ cao
- B1 for $(300, 1)$ ft their $a$
- B1 for $(120, -1)$ ft their $a$
- B1 for $(0, 0.5)$ ft their $a$
Why this scores full marks: The candidate uses the zero of $y = \cos x^{\circ}$ at $90$ and rejects the other value because of the condition $0 < a < 180$ . In (b) the maximum at $x + 60 = 0$ would be at $x = -60$ , outside the range, so the next one, at $x + 60 = 360$ , is used.

4MA1 Transformation of Graphs

At a glance

What you’ll learn

Translations

Reflections

Stretches

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Vertical translation $y = f(x) + a$

2Horizontal translation $y = f(x + a)$

3Reflections

4Stretches

5Image of a point under $y = f(-x)$ [1 mark]

6Equation and turning point after a stretch parallel to the y-axis [2 marks]

7Maximum points of $y = \sin(3x)^{\circ}$ [3 marks]

8Describing the transformation between two straight lines [3 marks]

9Finding a translation vector from two points on the image [4 marks]

10Translating a parabola so it touches the x-axis [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

4MA1 Transformation of Graphs

At a glance

What you’ll learn

Translations

Reflections

Stretches

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Vertical translation y=f(x)+ay = f(x) + ay=f(x)+a

2Horizontal translation y=f(x+a)y = f(x + a)y=f(x+a)

3Reflections

4Stretches

5Image of a point under y=f(−x)y = f(-x)y=f(−x) [1 mark]

6Equation and turning point after a stretch parallel to the y-axis [2 marks]

7Maximum points of y=sin⁡(3x)∘y = \sin(3x)^{\circ}y=sin(3x)∘ [3 marks]

8Describing the transformation between two straight lines [3 marks]

9Finding a translation vector from two points on the image [4 marks]

10Translating a parabola so it touches the x-axis [5 marks]

Model AnswersFull mark-scheme-style answers

1Vertical translation $y = f(x) + a$

2Horizontal translation $y = f(x + a)$

5Image of a point under $y = f(-x)$ [1 mark]

7Maximum points of $y = \sin(3x)^{\circ}$ [3 marks]