Maths · 3 Functions

3.6 Transforming graphs of functions

Before you start

Function notation, and that $f (x)$ is the output for input $x$ (3.3).
Writing a translation as a column vector $(\begin{matrix} a \\ b \end{matrix})$ .
The standard graph shapes from 3.1.

Syllabus objectives covered (1)

Describe and identify transformations to a graph of y = f(x) when y = f(x) + k or y = f(x + k).

Part 1 of 3

Two transformations, one of them counter-intuitive

Only two transformations are examined here, and both are translations — the graph slides without changing shape or size.
** $y = f (x) + k$ moves the graph UP by $k$ .** This is the one that behaves as you would expect: add $k$ to the output, and every point rises by $k$ .
As a column vector, the translation is $(\begin{matrix} 0 \\ k \end{matrix})$ . A negative $k$ moves it down.
** $y = f (x + k)$ moves the graph LEFT by $k$ .** This is the one that catches everyone, because $+ k$ moves it in the negative $x$ direction.
As a column vector, $(\begin{matrix} - k \\ 0 \end{matrix})$ . So $y = f (x + 3)$ is 3 to the left, and $y = f (x - 3)$ is 3 to the right.
The rule worth remembering: inside the bracket does the opposite, outside the bracket does what you expect. Changes to $x$ are inside; changes to $y$ are outside.

Part 2 of 3

Why the horizontal one goes backwards

This is worth understanding once rather than memorising forever, because a memorised rule with no reason behind it is exactly the thing that deserts you in an exam.
Take a specific case. Suppose $f$ has a maximum at $x = 5$ , and consider $g (x) = f (x + 3)$ .
Ask where $g$ has its maximum. It happens when the thing inside the bracket equals 5 — because that is where $f$ peaks.
So you need $x + 3 = 5$ , which gives $x = 2$ .
$x + 3 = 5 \Rightarrow x = 2$
The peak has moved from $x = 5$ to $x = 2$ — three units to the left. Adding inside the bracket shifted the graph backwards.
The reason is that $x + 3$ reaches any given value sooner than $x$ does, so the graph gets there earlier along the axis.
If you ever forget the direction, redo this one-line check with a point you can name. It takes fifteen seconds and it is always right.

Part 3 of 3

Describing and identifying

Questions come in two directions: given the transformation, find the new function; or given the two graphs, describe the transformation.
To describe a transformation, name it and give the vector. The full answer is 'a translation by $(\begin{matrix} 0 \\ - 4 \end{matrix})$ ', not just 'it moves down 4'.
Use the word translation, not 'shift', 'slide' or 'move'. It is the technical term and the mark scheme expects it.
One transformation, one description. If you find yourself writing 'a translation and then another translation', combine them into a single vector.
To find the new function, apply the rule to the expression. If $f (x) = x^{2}$ then $f (x) + 3 = x^{2} + 3$ , and $f (x + 3) = {(x + 3)}^{2}$ .
Notice how different those two are. In the first, the 3 is added after the squaring; in the second, before. Brackets are everything here.
The syllabus notes say ** $k$ will be an integer**, so you will never have to deal with fractional shifts.
A quick way to check any answer: pick one point on the original graph, apply the translation to it, and confirm it satisfies the new equation.

Worked examples

Example 1. The graph of $y = f (x)$ has a minimum at $(2, - 1)$ . Write down the coordinates of the minimum of (i) $y = f (x) + 5$ , (ii) $y = f (x - 4)$ .

Show the working

(i) The $+ 5$ is outside the bracket, so it behaves as expected: the graph moves up 5.
The $x$ -coordinate is unchanged and the $y$ -coordinate increases by 5, giving $(2, 4)$ .
(ii) The $- 4$ is inside the bracket, so the movement is horizontal and in the opposite direction to the sign — the graph moves 4 to the right.
Check it the reliable way: the minimum occurs when $x - 4 = 2$ , so $x = 6$ . The $y$ -coordinate is unchanged.
So the minimum is at $(6, - 1)$ .

Answer: (i) $(2, 4)$ (ii) $(6, - 1)$ .

Example 2. $f (x) = x^{2}$ . The graph of $y = f (x)$ is mapped onto the graph of $y = g (x)$ by a translation with vector $(\begin{matrix} - 3 \\ 0 \end{matrix})$ . Find $g (x)$ in terms of $x$ .

Show the working

Read the vector: the first entry is horizontal and the second vertical. Here the graph moves 3 to the left and not at all vertically.
A horizontal translation means the change goes inside the bracket, and because it is inside, the sign flips.
Moving left by 3 therefore corresponds to $f (x + 3)$ .
Substitute into $f$ .
$g (x) = f (x + 3) = {(x + 3)}^{2}$
Check with a point. The vertex of $y = x^{2}$ is at $(0,0)$ ; moving 3 left gives $(- 3, 0)$ . And $g (- 3) = {(- 3 + 3)}^{2} = 0$ . Correct.

Answer: $g (x) = {(x + 3)}^{2}$ .

Example 3. The graph of $y = \sin x$ is transformed to give the graph of $y = \sin x - 2$ . Describe the transformation fully, and state the new maximum and minimum values.

Show the working

The $- 2$ is outside the function, so this is a vertical translation, downwards by 2.
Describe it properly: a translation by the vector $(\begin{matrix} 0 \\ - 2 \end{matrix})$ .
The original $y = \sin x$ has a maximum of 1 and a minimum of $- 1$ .
Every point moves down 2, so the new maximum is $1 - 2 = - 1$ and the new minimum is $- 1 - 2 = - 3$ .
Note the shape is unchanged — same amplitude, same period. A translation never stretches a graph.

Answer: A translation by $(\begin{matrix} 0 \\ - 2 \end{matrix})$ . The new maximum is $- 1$ and the new minimum is $- 3$ .

Where people go wrong

Mistake: Thinking $y = f (x + 3)$ moves the graph 3 to the right.

Instead: It moves 3 to the left.

Inside the bracket, the effect is opposite to the sign. Check it by asking what value of $x$ makes the bracket equal a known point.

Mistake: Describing the transformation as a 'shift' or a 'move'.

Instead: Use the word translation, with a vector.

It is the technical term the mark scheme requires, and the vector carries the size and direction in one object.

Mistake: Changing both coordinates for a single translation.

Instead: $f (x) + k$ changes only the $y$ -coordinate; $f (x + k)$ changes only the $x$ -coordinate.

Each of these is a translation in one direction only, so the other coordinate is untouched.

Mistake: Writing $f (x + 3) = x^{2} + 3$ when $f (x) = x^{2}$ .

Instead: It is ${(x + 3)}^{2}$ — the whole of $x + 3$ gets squared.

The addition happens before the function acts, so it must sit inside the bracket.

Mistake: Assuming a translation changes the amplitude or period of a curve.

Instead: A translation slides the graph without altering its shape or size.

Only stretches change shape, and stretches are not in this subtopic.

Check yourself

Describe the transformation that maps $y = f (x)$ onto $y = f (x) - 6$ .
Answer
A translation by $(\begin{matrix} 0 \\ - 6 \end{matrix})$ , i.e. 6 units down.
Describe the transformation that maps $y = f (x)$ onto $y = f (x - 2)$ .
Answer
A translation by $(\begin{matrix} 2 \\ 0 \end{matrix})$ , i.e. 2 units to the right.
$f (x)$ has a zero at $x = 7$ . Where is the corresponding zero of $y = f (x + 4)$ ?
Answer
At $x = 3$ , since $x + 4 = 7$ gives $x = 3$ .
$f (x) = x^{3}$ . Write down $f (x) + 2$ and $f (x + 2)$ .
Answer
$f (x) + 2 = x^{3} + 2$ ; $f (x + 2) = {(x + 2)}^{3}$ .
Why does adding inside the bracket move a graph to the left?
Answer
Because $x + k$ reaches any given value at a smaller $x$ than $x$ does, so the graph reaches each feature earlier along the axis.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. The graph of $y = f (x)$ passes through the points $(0, 3)$ , $(2, 0)$ and $(5, - 4)$ .

(a) Write down the coordinates of the images of these three points on the graph of $y = f (x) + 4$ . 2 marks No calculator

Mark scheme

B1 any two correct of (0, 7), (2, 4), (5, 0)
B1 all three correct

(b) Write down the coordinates of the image of $(5, - 4)$ on the graph of $y = f (x + 1)$ . 1 mark No calculator

Mark scheme

B1 (4, -4)

(c) Describe fully the single transformation that maps $y = f (x)$ onto $y = f (x + 1)$ . 2 marks No calculator

Mark scheme

B1 translation
B1 by the vector (-1, 0), or 1 unit to the left

Question 2. $f (x) = x^{2} - 4$

(a) The graph of $y = f (x)$ is mapped onto the graph of $y = g (x)$ by a translation with vector $(\begin{matrix} 0 \\ 6 \end{matrix})$ . Find $g (x)$ in its simplest form. 2 marks No calculator

Mark scheme

M1 g(x) = f(x) + 6 seen oe
A1 x² + 2

(b) The graph of $y = f (x)$ is mapped onto the graph of $y = h (x)$ by a translation with vector $(\begin{matrix} - 5 \\ 0 \end{matrix})$ . Find $h (x)$ . 2 marks No calculator

Mark scheme

M1 h(x) = f(x + 5) seen
A1 (x + 5)² - 4 oe

(c) Write down the vertex of $y = h (x)$ . 1 mark No calculator

Mark scheme

B1 (-5, -4); FT their (b)

Question 3. A student says: 'The graph of $y = f (x + 2)$ is the graph of $y = f (x)$ moved 2 units to the right.'

(a) Comment on the student's statement. 1 mark No calculator

Mark scheme

B1 it is incorrect; the graph moves 2 units to the left

(b) Explain, using a point on the graph, why the statement is wrong. 2 marks No calculator

Mark scheme

B1 chooses a feature of f at a named value, e.g. a maximum at x = a
B1 shows the corresponding feature of f(x + 2) occurs at x = a - 2, which is to the left oe

Words to know

Translation: A transformation that slides a graph without changing its shape or size, described by a column vector.
$y = f (x) + k$: The graph of $f$ translated by $(\begin{matrix} 0 \\ k \end{matrix})$ — vertically, in the direction of the sign.
$y = f (x + k)$: The graph of $f$ translated by $(\begin{matrix} - k \\ 0 \end{matrix})$ — horizontally, opposite to the sign.
Image: The point or graph produced by applying a transformation.

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