What are wrong order of vertical operations?

a\,f(x) + d stretches first, then translates; reversing them gives a different graph.

What are not moving the asymptotes?

Asymptotes transform with the curve; a translated curve has translated asymptotes.

Describe the transformation taking y = f(x) to y = f(x) - 4. [1 mark]

The point (1, 6) lies on y = f(x). Find its image on y = f(2x). [2 marks]

SingaporeMathsSyllabus dot point

How do translations, stretches and reflections change the graph and the equation of a function?

Relate the graph of y equals a f(b(x + c)) + d to the graph of y equals f(x) through translations, stretches and reflections, and apply combined transformations in the correct order

A focused answer to the H2 Mathematics outcome on graph transformations. Translations, stretches and reflections, the effect of each parameter, the order of combined transformations, and the effect on asymptotes and key points.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to connect the graph of $y = a\,\mathrm{f}(b(x + c)) + d$ to the graph of $y = \mathrm{f}(x)$ by describing the translations, stretches and reflections involved, applying them in the correct order, and tracking what happens to key points and asymptotes.

The answer

The four basic transformations

Working from $y = \mathrm{f}(x)$ :

Vertical translation: $y = \mathrm{f}(x) + d$ shifts the graph up by $d$ (down if $d < 0$ ).
Horizontal translation: $y = \mathrm{f}(x + c)$ shifts the graph left by $c$ (right if $c < 0$ ). The sign is counter-intuitive: $x + c$ moves in the negative direction.
Vertical stretch: $y = a\,\mathrm{f}(x)$ stretches by scale factor $a$ parallel to the $y$ -axis. If $a < 0$ it also reflects in the $x$ -axis.
Horizontal stretch: $y = \mathrm{f}(bx)$ stretches by scale factor $\dfrac{1}{b}$ parallel to the $x$ -axis. If $b < 0$ it also reflects in the $y$ -axis.

Reflections

$y = -\mathrm{f}(x)$ reflects in the $x$ -axis.
$y = \mathrm{f}(-x)$ reflects in the $y$ -axis.

These are the special cases of vertical or horizontal stretch with a negative scale factor.

Order of combined transformations

For $y = a\,\mathrm{f}(b(x + c)) + d$ :

Changes inside the function affect $x$ and are applied to the input (horizontal), in the reverse of the usual order.
Changes outside the function affect $y$ (vertical), applied in the natural order.

A safe routine for the vertical part: stretch by $a$ , then translate by $d$ . For the horizontal part: factor out $b$ to read the stretch by $\frac{1}{b}$ and the translation by $c$ .

Effect on features

Translations move asymptotes and key points by the same shift. A vertical stretch by $a$ multiplies $y$ -coordinates (and the height of horizontal asymptotes) by $a$ ; a horizontal stretch by $\frac{1}{b}$ scales $x$ -coordinates and vertical asymptote positions.

Worked example

The curve $y = \mathrm{f}(x)$ has a vertical asymptote $x = 1$ and passes through $(3, 4)$ . Find the asymptote and the image of $(3, 4)$ on $y = -2\mathrm{f}(x - 2) + 1$ .

Step 1: Handle the horizontal change

The replacement $x \to x - 2$ translates everything $2$ units right. The asymptote $x = 1$ moves to $x = 3$ . The point's $x$ -coordinate $3$ moves to $5$ .

Step 2: Apply the vertical stretch and reflection

The factor $-2$ stretches $y$ by $2$ and reflects in the $x$ -axis, so the $y$ -coordinate $4$ becomes $-2 \times 4 = -8$ .

Step 3: Apply the vertical translation

Adding $1$ shifts up by $1$ , so $-8$ becomes $-7$ .

Step 4: State the results

The vertical asymptote is $x = 3$ , and the image of $(3, 4)$ is $(5, -7)$ .

Examples in context

Example 1. Building a curve from a parent. The graph $y = \dfrac{1}{x - 2} + 3$ is the basic reciprocal $y = \dfrac{1}{x}$ translated $2$ right and $3$ up, so its asymptotes shift from $x = 0$ , $y = 0$ to $x = 2$ , $y = 3$ . Reading the transformation gives the sketch instantly.

Example 2. A sinusoidal model. A tide height $h = 2\sin\left(\dfrac{\pi}{6}t\right) + 5$ takes the base $\sin t$ , stretches it horizontally (period $12$ hours), stretches it vertically by $2$ (amplitude $2$ m), and translates it up by $5$ m (mean level), so transformations describe the whole model.

Try this

Q1. Describe the transformation taking $y = \mathrm{f}(x)$ to $y = \mathrm{f}(x) - 4$ . [1 mark]

Cue. Translation $4$ units in the negative $y$ -direction (down).

Q2. The point $(1, 6)$ lies on $y = \mathrm{f}(x)$ . Find its image on $y = \mathrm{f}(2x)$ . [2 marks]

Cue. Horizontal compression by $\frac{1}{2}$ : $x$ -coordinate halves to $\frac{1}{2}$ , giving $\left(\frac{1}{2}, 6\right)$ .

Q3. Describe fully the transformations mapping $y = \cos x$ to $y = \cos(x - \frac{\pi}{2})$ . [2 marks]

Cue. Translation $\frac{\pi}{2}$ in the positive $x$ -direction (right), which gives $\sin x$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksThe graph of

y = \mathrm{f}(x)

has a maximum point at

(2, 5)

. State the coordinates of the corresponding point on the graph of

y = 3\mathrm{f}(x - 1) - 2

Show worked answer →

Work outwards. The replacement $x \to x - 1$ translates the graph $1$ unit in the positive $x$ -direction, so the $x$ -coordinate becomes $2 + 1 = 3$ .

The factor $3$ stretches vertically by scale factor $3$ about the $x$ -axis, so the $y$ -coordinate $5$ becomes $15$ .

The $-2$ translates down by $2$ , so the $y$ -coordinate becomes $15 - 2 = 13$ .

The corresponding point is $(3, 13)$ .

Markers reward the horizontal translation in the correct direction, the vertical stretch, the downward translation, and the final coordinates.

Original4 marksDescribe fully a sequence of transformations that maps the graph of

y = x^2

onto the graph of

y = 2(x + 3)^2 - 1

Show worked answer →

Starting from $y = x^2$ :

Replace $x$ by $x + 3$ : translate $3$ units in the negative $x$ -direction, giving $y = (x + 3)^2$ .

Multiply by $2$ : stretch vertically by scale factor $2$ (parallel to the $y$ -axis), giving $y = 2(x + 3)^2$ .

Subtract $1$ : translate $1$ unit in the negative $y$ -direction, giving $y = 2(x + 3)^2 - 1$ .

Markers reward all three transformations with correct directions and scale factor, and a sensible order (horizontal translation, then stretch, then vertical translation).