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 $y = \mathrm{f}(x)$ 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.

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

Show worked solution

Solution

1 $f(x + 3)$ moves the graph 3 to the left.
2 $(-1 - 3, 4) = (-4, 4)$ .

Answer: $(-4, 4)$

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).

Show worked solution

Solution

1 $-f(x)$ reflects in the $x$ -axis.
2The maximum becomes a minimum at $(3, -2)$ .

Answer: $(3, -2)$

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).

Show worked solution

Solution

1 $-7$ outside the function moves the graph down 7.
2 $(2, 5 - 7) = (2, -2)$ .

Answer: $(2, -2)$

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.

Reveal solution

SolutionTranslating

y = x^{2}

right 3 and down 2 gives

y = (x - 3)^{2} - 2

, with minimum

(3, -2)

. Reflecting in the

x

-axis negates every

y

-value:

y = -\big((x - 3)^{2} - 2\big) = -(x - 3)^{2} + 2

. The turning point is

(3, 2)

, now a maximum. Expanded, the equation is

y = -x^{2} + 6x - 7

. Check: at

x = 3

-9 + 18 - 7 = 2

✓.

\blacksquare

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

\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]
Show mark scheme
- B1Translation
- B1By $\begin{pmatrix} 3 \\ 1 \end{pmatrix}$

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]
Show mark scheme (a)
- B1 $(2, 6)$
(b)Write down the matching point on $y = -f(x)$ .[1]
Show mark scheme (b)
- B1 $(-2, -6)$

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]
Show mark scheme
- B1 $x = 90^\circ$
- B1 $(90, 2)$
Answer: $(90^\circ, 2)$

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