A13AlgebraHigherGrades 6–8

Graph Transformations

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

To get the output f(0), you now need x = -a, so every point is reached a units earlier.

If you know the graph of $y = f (x)$ , you can sketch $y = f (x) + 3$ , $y = f (x + 2)$ , $y = - f (x)$ and $y = f (- x)$ without plotting a single point. Each is a translation or a reflection of the original, and AQA tests them on quadratic, cubic, trigonometric and unknown graphs.

40 min AQA GCSE Maths

The big picture

Every point on $y = f (x)$ moves in the same way. So it is enough to track a few key points — turning points, intercepts, asymptotes — and redraw the curve through their new positions.

The rule that catches students out: a change inside the bracket affects $x$ and works backwards. $f (x + 2)$ moves the graph 2 to the $left$ , because $x$ must be 2 smaller to give the same output as before.

What you'll learn

Sketch translations $y = f (x) + a$ and $y = f (x + a)$
Sketch reflections $y = - f (x)$ and $y = f (- x)$
Find the image of a point under a transformation
Describe a transformation using a column vector

Translations

$y = f (x) + a$ moves the graph $a$ units up: vector $(0 a)$ .

$y = f (x + a)$ moves the graph $a$ units left: vector $(- a 0)$ .

So $y = f (x - 4)$ moves the graph 4 units right.

1(a) Up 3: add 3 to the

y

-coordinate.

2(b) Left 2: subtract 2 from the

x

-coordinate.

Answer(a)

(- 1, 7)

(b)

(- 3, 4)

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

$y = f (x)$ has a maximum point at $(2, - 3)$ . Write down the maximum point of $y = f (x - 4)$ .

Your answer

Type coordinates like (2, -3).

Tip — Outside the bracket: $y$ changes as you read it. Inside the bracket: $x$ changes the opposite way.

Reflections

$y = - f (x)$ reflects the graph in the $x$ -axis: $(x, y) \to (x, - y)$ .

$y = f (- x)$ reflects the graph in the $y$ -axis: $(x, y) \to (- x, y)$ .

1(a) Negate

y

; a minimum becomes a maximum.

2(b) Negate

x

; it stays a minimum.

Answer(a) maximum

(- 1, - 4)

(b) minimum

(1, 4)

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

$y = f (x)$ has a minimum point at $(- 1, 4)$ . Write down the turning point of $y = - f (x) + 2$ .

Your answer

Type coordinates like (2, -3).

Why this works

Reflecting in the $x$ -axis turns minimums into maximums, because the curve is flipped upside down. Reflecting in the $y$ -axis mirrors left and right but keeps the curve the same way up.

Describing a transformation

Completing the square shows how a quadratic is related to $y = x^{2}$ .

$y = x^{2} - 6 x + 11 = (x - 3)^{2} + 2$ , which is $y = x^{2}$ moved right 3 and up 2.

1Complete the square:

(x - 3)^{2} + 2

2This is

f (x - 3) + 2

with

f (x) = x^{2}

AnswerTranslation by

(32)

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

The graph of $y = x^{2}$ is translated to give $y = x^{2} + 8 x + 20$ . Write the translation as a column vector $(a, b)$ .

Your answer

Type the vector as (top, bottom), e.g. (3, -2).

Tip — For "describe fully", name the transformation and give the vector or mirror line.

Think like an examiner

Label the transformed key points on sketches.
Give translations as column vectors.
State whether a turning point is a maximum or minimum after reflecting.

Remember these

f (x) + a : (0 a)

Translate up by a.

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

Translate left by a.

- f (x), f (- x)

Reflect in the x-axis, reflect in the y-axis.

Watch out for these

f (x + 2)

moves the graph right 2.

It moves left 2.

Confusing

- f (x)

with

f (- x)

- f (x)

flips vertically;

f (- x)

flips horizontally.

Changing both coordinates for a translation parallel to an axis.

Only one coordinate changes.

Describing a translation without a vector.

Always give the column vector.

Stretch yourself

The graph of $y = x^{2} - 4 x + 7$ is translated to give $y = x^{2} + 2 x + 5$ . Find the translation vector.

Hint — Complete the square for both and compare the turning points.

Questions students ask

Why does

f (x + a)

move left?

To get the output

f (0)

, you now need

x = - a

, so every point is reached

a

units earlier.

Do I need stretches?

AQA GCSE limits this to translations and reflections; stretches appear at A-Level.

Which points should I track?

Turning points, axis intercepts and any asymptotes.

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.
Complete the square to describe quadratic translations.

Exam-style questions

Practise it the way it is examined

3 original questions · 8 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 63 marks

The point $(3, 5)$ lies on the graph of $y = f (x)$ .

(a)Write down the coordinates of the matching point on $y = f (x) - 5$ .[1]
(b)Write down the coordinates of the matching point on $y = f (x + 3)$ .[1]
(c)Write down the coordinates of the matching point on $y = f (- x)$ .[1]

Question 2Grade 72 marks

Describe fully the single transformation that maps the graph of $y = cos x$ onto the graph of $y = cos (x - 9 0^{\circ})$ .[2]

Question 3Grade 83 marks

The graph of $y = x^{2} - 4 x + 1$ is translated by $(- 3 5)$ . Find the equation of the new graph in the form $y = x^{2} + b x + c$ .[3]

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