J560 · 7.01–7.04GraphsHigherGrades 7–8

Graph Transformations

Q: Why does f(x + a) move left?

To get the same output as before, x + a must equal the old x, so each new x is a smaller than the original — the whole graph shifts left.

If you know what $y = \mathrm{f}(x)$ looks like, you can sketch $y = \mathrm{f}(x) + 3$ , $y = \mathrm{f}(x - 2)$ or $y = -\mathrm{f}(x)$ without plotting a single new point. Each is a $\textbf{transformation}$ — a translation or a reflection — of the original graph. The challenge is remembering which way each change moves the graph.

45 min Graphs

The big picture

Changes $\textit{outside}$ the function affect $y$ directly and behave as you would expect: $\mathrm{f}(x) + 3$ moves every point up 3. Changes $\textit{inside}$ the bracket affect $x$ and behave the opposite way: $\mathrm{f}(x - 2)$ moves the graph 2 units to the $\textit{right}$ , because $x$ must now be 2 bigger to produce the same output.

OCR questions give a graph or a point such as a turning point, and ask for its image under a transformation, or ask you to identify the transformation from two graphs. Tracking one key point is usually the fastest route.

Transformations of Graphs - CorbettmathsCurated by Kepler, not produced by us. Watch it, then read the notes below.

Open on YouTube

Before you start

Try a short calculation from the preceding topic without the notes.
Check the tier and grade band shown in this lesson.

Watch, pause, try

Video by corbettmaths. The video’s own board label is retained; follow the specification reference and scope notes on this lesson for assessed content.

Viewing focus: Translate graphs using $y = \mathrm{f}(x) + a$ and $y = \mathrm{f}(x + a)$

After watching, attempt these questions before opening the answers. You can also use the written lesson below.

After watching 1: $y = \mathrm{f}(x) + 5$ is a translation by:

Show answer and reasoning

$\begin{pmatrix} 0 \\ 5 \end{pmatrix}$ . Up 5.

After watching 2: $y = -\mathrm{f}(x)$ is a reflection in:

Show answer and reasoning

The $x$ -axis. Every $y$ -value changes sign.

What you'll learn

Translate graphs using $y = \mathrm{f}(x) + a$ and $y = \mathrm{f}(x + a)$
Reflect graphs using $y = -\mathrm{f}(x)$ and $y = \mathrm{f}(-x)$
Find the image of a point under a transformation
Describe a transformation, including as a column vector

Vertical translations: f(x) + a

$y = \mathrm{f}(x) + a$ adds $a$ to every $y$ -value, translating the graph $a$ units up (or down if $a$ is negative).

As a column vector, this is a translation by $\begin{pmatrix} 0 \\ a \end{pmatrix}$ .

A point $(p, q)$ moves to $(p, q + a)$ .

y = \mathrm{f}(x) + a: \ \begin{pmatrix} 0 \\ a \end{pmatrix}

Up a units.

Horizontal translations: f(x + a)

$y = \mathrm{f}(x + a)$ translates the graph $a$ units to the $\textbf{left}$ — the vector $\begin{pmatrix} -a \\ 0 \end{pmatrix}$ .

So $y = \mathrm{f}(x - 2)$ moves the graph 2 units right.

A point $(p, q)$ moves to $(p - a, q)$ .

y = \mathrm{f}(x + a): \ \begin{pmatrix} -a \\ 0 \end{pmatrix}

Left a units.

Your turnSame method, new numbers — try it before moving on.

$y = f(x)$ has a minimum at $(-1, 4)$ . Write down the minimum point of $y = f(x + 3)$ .

Your answer

Type coordinates like (2, -3).

Why this works

For $\mathrm{f}(x) = x^{2}$ , the minimum is where the inside of the function is 0. In $\mathrm{f}(x - 2) = (x - 2)^{2}$ , the inside is 0 when $x = 2$ — so the minimum has moved to $x = 2$ , to the right.

Reflections

$y = -\mathrm{f}(x)$ changes the sign of every $y$ -value: a $\textbf{reflection in the x-axis}$ . A point $(p, q)$ moves to $(p, -q)$ .

$y = \mathrm{f}(-x)$ changes the sign of every $x$ -value: a $\textbf{reflection in the y-axis}$ . A point $(p, q)$ moves to $(-p, q)$ .

Points on the mirror line do not move: roots stay put under $-\mathrm{f}(x)$ , and the $y$ -intercept stays put under $\mathrm{f}(-x)$ .

y = -\mathrm{f}(x): (p, q) \to (p, -q)

Reflection in the x-axis.

y = \mathrm{f}(-x): (p, q) \to (-p, q)

Reflection in the y-axis.

Your turnSame method, new numbers — try it before moving on.

$y = f(x)$ has a maximum at $(3, 2)$ . Write down the turning point of $y = -f(x)$ .

Your answer

Type coordinates like (2, -3).

Images of points and trig graphs

To find where a key point goes, apply the rule to its coordinates rather than redrawing the graph.

Transformations of trigonometric graphs use the same rules: $y = \sin(x + 90^{\circ})$ is $y = \sin x$ translated $90^{\circ}$ left, which is exactly the graph of $y = \cos x$ .

Example 1The curve

y = \mathrm{f}(x)

has a maximum at

(2, 5)

. Find the maximum or minimum point of (a)

y = \mathrm{f}(x - 3)

(b)

y = \mathrm{f}(x) - 4

(c)

y = -\mathrm{f}(x)

(d)

y = \mathrm{f}(-x)

1(a) Right 3:

(5, 5)

, still a maximum.

2(b) Down 4:

(2, 1)

, still a maximum.

3(c) Reflect in the

x

-axis:

(2, -5)

, now a minimum.

4(d) Reflect in the

y

-axis:

(-2, 5)

, still a maximum.

Answer(a) max

(5, 5)

(b) max

(2, 1)

(2, -5)

(d) max

(-2, 5)

Your turnSame method, new numbers — try it before moving on.

$y = f(x)$ has a maximum at $(2, 5)$ . Write down the maximum point of $y = f(x) - 7$ .

Your answer

Type coordinates like (2, -3).

Tip — Say whether a turning point is a maximum or minimum after reflecting in the $x$ -axis — the reflection turns one into the other.

Think like an examiner

Describe translations precisely, preferably with a column vector.
Name the mirror line for reflections.
Label transformed key points with coordinates.

Transformations

\mathrm{f}(x) + a

Translate up a.

\mathrm{f}(x + a)

Translate left a.

-\mathrm{f}(x)

Reflect in the x-axis.

\mathrm{f}(-x)

Reflect in the y-axis.

Watch out for these

Moving

y = \mathrm{f}(x + 4)

four units right.

Inside the bracket acts the opposite way: four units left.

Reflecting in the

y

-axis for

-\mathrm{f}(x)

-\mathrm{f}(x)

reflects in the

x

-axis.

Changing the

x

-coordinate for

\mathrm{f}(x) + a

Only

y

changes:

(p, q) \to (p, q + a)

Keeping a maximum as a maximum after

-\mathrm{f}(x)

It becomes a minimum.

Stretch yourself

The graph of $y = x^{2}$ is translated by the vector $\begin{pmatrix} 3 \\ -2 \end{pmatrix}$ and then reflected in the $x$ -axis. Find the equation of the resulting graph and its turning point.

Hint — Translate first: right 3 means replacing $x$ with $x - 3$ , and down 2 means subtracting 2. Then negate the whole expression.

Questions students ask

Why does f(x + a) move left?

To get the same output as before,

x + a

must equal the old

x

, so each new

x

a

smaller than the original — the whole graph shifts left.

Does the order of transformations matter?

Sometimes. A translation followed by a reflection can give a different result from the reflection first, so follow the order given.

Do I need stretches for GCSE?

No. The J560 specification covers translations and reflections; stretches such as

\mathrm{f}(2x)

are A-Level.

Key takeaways

$\mathrm{f}(x) + a$ : up $a$ . $\mathrm{f}(x + a)$ : left $a$ .
$-\mathrm{f}(x)$ : reflect in the $x$ -axis. $\mathrm{f}(-x)$ : reflect in the $y$ -axis.
Changes inside the bracket act on $x$ in the opposite direction.
Track key points by applying the rule to their coordinates.
A reflection in the $x$ -axis swaps maxima and minima.

Exam-style questions

Practise it the way it is examined

3 original questions · 6 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 72 marks

Describe fully the single transformation that maps $y = x^2$ onto $y = (x - 3)^2 + 1$ .[2]

Question 2Grade 72 marks

The point $(-2, 6)$ lies on the curve $y = f(x)$ .

(a)Write down the matching point on $y = f(-x)$ .[1]
(b)Write down the matching point on $y = -f(x)$ .[1]

Question 3Grade 82 marks

Write down the coordinates of the maximum point of $y = \sin x + 1$ for $0^\circ \le x \le 360^\circ$ .[2]

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

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Explore further

OCR J560 official specification (opens in a new tab)
Check the assessed scope and practical or mathematical requirements.
Read or explore this topic (opens in a new tab)
An external explanation or interactive activity to use alongside the lesson. It may include extension material.

Return to this tomorrow

Close the notes and answer from memory. Revisit any explanation you cannot yet reconstruct.

The point $(4, 1)$ is on $y = \mathrm{f}(x)$ . Which point is on $y = \mathrm{f}(x + 2)$ ?
The point $(-3, 7)$ is on $y = \mathrm{f}(x)$ . Which point is on $y = \mathrm{f}(-x)$ ?
Which equation is $y = x^{2}$ translated 4 units right?
$y = \sin(x - 90^{\circ})$ is the graph of $y = \sin x$ :

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

Related lessons

All 5 lessons

Graph Transformations

If you know what

y = \mathrm{f}(x)

looks like, you can sketch

y = \mathrm{f}(x) + 3

y = \mathrm{f}(x - 2)

y = -\mathrm{f}(x)

without plotting a single new point. Each is a

\textbf{transformation}

— a translation or a reflection — of the original graph. The challenge is remembering which way each change moves the graph.

45 min Graphs

Watch, pause, try

Video by corbettmaths. The video’s own board label is retained; follow the specification reference and scope notes on this lesson for assessed content.

Viewing focus: Translate graphs using

y = \mathrm{f}(x) + a

and

y = \mathrm{f}(x + a)

After watching, attempt these questions before opening the answers. You can also use the written lesson below.

After watching 1: $y = \mathrm{f}(x) + 5$ is a translation by:

Show answer and reasoning

$\begin{pmatrix} 0 \\ 5 \end{pmatrix}$ . Up 5.

After watching 2: $y = -\mathrm{f}(x)$ is a reflection in:

Show answer and reasoning

The $x$ -axis. Every $y$ -value changes sign.

Practise it the way it is examined

Question 1Grade 72 marks

Describe fully the single transformation that maps $y = x^2$ onto $y = (x - 3)^2 + 1$ .[2]

Question 2Grade 72 marks

The point $(-2, 6)$ lies on the curve $y = f(x)$ .

(a)Write down the matching point on $y = f(-x)$ .[1]
(b)Write down the matching point on $y = -f(x)$ .[1]

Question 3Grade 82 marks

Write down the coordinates of the maximum point of $y = \sin x + 1$ for $0^\circ \le x \le 360^\circ$ .[2]

Return to this tomorrow

Close the notes and answer from memory. Revisit any explanation you cannot yet reconstruct.

The point

(4, 1)

is on

y = \mathrm{f}(x)

. Which point is on

y = \mathrm{f}(x + 2)

The point

(-3, 7)

is on

y = \mathrm{f}(x)

. Which point is on

y = \mathrm{f}(-x)

Which equation is

y = x^{2}

translated 4 units right?

y = \sin(x - 90^{\circ})

is the graph of

y = \sin x