1MA1 · A13AlgebraHigherGrades 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 $y = f (x)$ looks like, you can sketch $y = f (x) + 3$ , $y = f (x - 2)$ or $y = - f (x)$ without plotting a single new point. Each is a $transformation$ — 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 $outside$ the function affect $y$ directly and behave as you would expect: $f (x) + 3$ moves every point up 3. Changes $inside$ the bracket affect $x$ and behave the opposite way: $f (x - 2)$ moves the graph 2 units to the $right$ , because $x$ must now be 2 bigger to produce the same output.

Edexcel 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.

What you'll learn

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

Vertical translations: f(x) + a

$y = f (x) + a$ 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 $(0 a)$ .

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

y = f (x) + a : (0 a)

Up a units.

Horizontal translations: f(x + a)

$y = f (x + a)$ translates the graph $a$ units to the $left$ — the vector $(- a 0)$ .

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

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

y = f (x + a) : (- a 0)

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

Reflections

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

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

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

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

Reflection in the x-axis.

y = 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: $y = sin (x + 9 0^{\circ})$ is $y = sin x$ translated $9 0^{\circ}$ left, which is exactly the graph of $y = cos x$ .

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

f (x) + a

Translate up a.

f (x + a)

Translate left a.

- f (x)

Reflect in the x-axis.

f (- x)

Reflect in the y-axis.

Watch out for these

Moving

y = f (x + 4)

four units right.

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

Reflecting in the

y

-axis for

- f (x)

- f (x)

reflects in the

x

-axis.

Changing the

x

-coordinate for

f (x) + a

Only

y

changes:

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

Keeping a maximum as a maximum after

- f (x)

It becomes a minimum.

Stretch yourself

The graph of $y = x^{2}$ is translated by the vector $(3 - 2)$ 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 1MA1 specification covers translations and reflections; stretches such as

f (2 x)

are A-Level.

Key takeaways

$f (x) + a$ : up $a$ . $f (x + a)$ : left $a$ .
$- f (x)$ : reflect in the $x$ -axis. $f (- x)$ : 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 $0^{\circ} \leq x \leq 36 0^{\circ}$ .[2]

Unlimited practice

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Real past-paper questions on Graph Transformations, marked mark-by-mark. How you do feeds straight into your weak-topic list, so your revision keeps targeting what actually needs work.

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Graph Transformations

Vertical translations: f(x) + a

Horizontal translations: f(x + a)

Reflections

Images of points and trig graphs

Practise it the way it is examined

Skill drill

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Related topics

Graph Transformations

Vertical translations: f(x) + a

Horizontal translations: f(x + a)

Reflections

Images of points and trig graphs

Practise it the way it is examined

Skill drill

Practise this topic

Related topics