Algebra

Graph transformations

Q: Why does f(x + 2) move the graph left?

The new graph reaches each y value 2 earlier: at x = -2 it gives f(0), which the old graph gave at x = 0. So every point shifts 2 to the left.

Q: What is the difference between -f(x) and f(-x)?

-f(x) reflects the graph in the x-axis. f(-x) reflects it in the y-axis.

13 past-paper questions, easiest first. Try each one, then reveal the answer.

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Graph transformations: what you need to know

Graph transformations move or flip a curve by changing its equation. You need to know which change to $y = f(x)$ gives which movement, and apply it to a sketch or a single point.

Key facts

$y = f(x) + a$ : translate up by $a$ (vector $\binom{0}{a}$ ).
$y = f(x + a)$ : translate LEFT by $a$ (vector $\binom{-a}{0}$ ). $y = f(x - a)$ moves right.
$y = -f(x)$ : reflect in the $x$ -axis, so $(x, y) \to (x, -y)$ .
$y = f(-x)$ : reflect in the $y$ -axis, so $(x, y) \to (-x, y)$ .
Combined changes: do the inside change (to $x$ ) and the outside change (to $y$ ) separately. On $y = f(-x) + 2$ , the point $(x, y)$ goes to $(-x, y + 2)$ .
The same rules work for trig graphs, in degrees: $y = \sin(x + 90)$ is $y = \sin x$ moved $90$ to the left.

Transforming a point or a sketch

Decide if the change is outside the bracket (affects $y$ , vertical) or inside (affects $x$ , horizontal).
Inside changes do the opposite of what they look like: $+3$ inside means left 3.
Move a few key points (turning points, crossings) and redraw the same shape through them.

Worked examples

Example 1 Higher · grade 7

The point $(3, 5)$ lies on $y = f(x)$ . Write down its image on (a) $y = f(x - 2)$ , (b) $y = -f(x)$ , (c) $y = f(-x)$ .

Show the solution

(a) Inside the bracket, $-2$ means right 2: $(5, 5)$ .
(b) Reflect in the $x$ -axis, so $y$ changes sign: $(3, -5)$ .
(c) Reflect in the $y$ -axis, so $x$ changes sign: $(-3, 5)$ .

Example 2 Higher · grade 8

The curve $y = x^2$ is translated by the vector $\binom{3}{-4}$ . Find the equation of the new curve in the form $y = x^2 + bx + c$ and its turning point.

Show the solution

Right 3 means replace $x$ with $x - 3$ . Down 4 means subtract 4.
$y = (x - 3)^2 - 4$
$= x^2 - 6x + 9 - 4$
$= x^2 - 6x + 5$
The turning point moves from $(0, 0)$ to $(3, -4)$ .

Example 3 Higher · grade 8

The point $(0, 1)$ is a maximum point on the graph of $y = \cos x$ . Write down the coordinates of its image on the graph of $y = \cos(x - 90) - 2$ .

Show the solution

Inside the bracket, $-90$ means right 90: $x = 0 + 90 = 90$ .
Outside, $-2$ means down 2: $y = 1 - 2 = -1$ .
Image: $(90, -1)$

Common mistakes

Moving right for $f(x + a)$ . A plus inside the bracket moves the graph left.
Mixing up $-f(x)$ and $f(-x)$ . The minus outside flips top to bottom; inside flips left to right.
Moving a graph sideways for $f(x) + a$ . A number added outside moves it up or down only.

Exam tips

When sketching, mark where the key points end up, such as the new turning point or axis crossings. Those points usually carry the marks.
If you are asked for the equation of a transformed graph, write it in full, e.g. $y = f(x - 3)$ or $y = -\sin x$ , not just $f(x - 3)$ .

Questions people ask

Why does f(x + 2) move the graph left?

The new graph reaches each $y$ value 2 earlier: at $x = -2$ it gives $f(0)$ , which the old graph gave at $x = 0$ . So every point shifts 2 to the left.

What is the difference between -f(x) and f(-x)?

$-f(x)$ reflects the graph in the $x$ -axis. $f(-x)$ reflects it in the $y$ -axis.

Exam questions: Graph transformations (13, easiest first)

Graph transformations4 marksGrade 7CalculatorHigher

November 2024 · Paper 3 Higher · Q21

Reflect a graph in the x-axis, then find where a turning point of y = sin x moves under a translation.

Show answer

(a)	Graph of $y = -f(x)$ : the given graph reflected in the x-axis	1 mark
(b)	$(90,\ 3)$	3 marks

Worked solution Checked against the mark scheme

(a) Each point $(x, y)$ moves to $(x, -y)$ : reflect the graph in the x-axis.

(b) P is the minimum of $\sin x$ at $(270, -1)$ .

$\sin(x + 180) + 4$ moves the graph 180 left and 4 up.

$R = (270 - 180,\ -1 + 4) = (90, 3)$

Watch out: Moving the graph 180 to the right instead of the left.

Graph transformations2 marksGrade 7CalculatorHigher

June 2023 · Paper 2 Higher · Q21

Draw a translated graph and a reflected graph of a given piecewise function.

Show answer

(a)	Shape moved down 4: through $(-3,-4)$ , $(-2,-3)$ , $(-1,-3)$ , $(0,-4)$	1 mark
(b)	Shape reflected in the $y$ -axis: through $(0,0)$ , $(1,1)$ , $(2,1)$ , $(3,0)$	1 mark

Worked solution Checked against the mark scheme

(a) $f(x) - 4$ moves every point down 4: $(-3,0) \to (-3,-4)$ , $(-2,1) \to (-2,-3)$ , $(-1,1) \to (-1,-3)$ , $(0,0) \to (0,-4)$ .

(b) $f(-x)$ reflects in the $y$ -axis: change the sign of each $x$ : $(3,0)$ , $(2,1)$ , $(1,1)$ , $(0,0)$ .

Watch out: Reflecting in the x-axis for f(−x), which is the rule for −f(x).

Graph transformations4 marksGrade 7CalculatorHigher

November 2021 · Paper 2 Higher · Q18

Use a sine graph to solve an equation over two cycles, write the equation of its reflection and translate another graph.

Show answer

(a)	$37°, 143°, 397°, 503°$	2 marks
(b)	$y = -\sin x°$	1 mark
(c)	The curve moved 2 units to the right	1 mark

Worked solution Checked against the mark scheme

(a) Draw $y = 0.6$ across the graph: it meets the curve at about $37°$ and $180 - 37 = 143°$ .

The graph repeats every 360°, so also $37 + 360 = 397°$ and $143 + 360 = 503°$ .

(b) Reflecting in the $x$ -axis makes every $y$ negative: $y = -\sin x°$ .

Watch out: Moving the graph 2 to the left for f(x − 2).

Graph transformations2 marksGrade 7CalculatorHigher

June 2019 · Paper 3 Higher · Q15

Find where a point moves under the graph transformation y = f(−x) − 3.

Show answer

Answer	$(-7, -1)$	2 marks

Worked solution Checked against the mark scheme

$f(-x)$ reflects in the $y$ -axis: $(7, 2) \to (-7, 2)$

$-3$ moves it 3 down: $(-7, 2) \to (-7, -1)$

Watch out: Reflecting in the x-axis instead, giving (7, −5).

Graph transformations2 marksGrade 7No calculatorHigher

November 2018 · Paper 1 Higher · Q18

Sketch a horizontal translation of a graph and write the equation of a graph reflected in the y-axis.

Show answer

(a)	The curve moved 2 units right: crosses the x-axis at (2, 0) and (4, 0), minimum at (3, −1)	1 mark
(b)	$y = g(-x)$	1 mark

Worked solution Checked against the mark scheme

(a) $f(x - 2)$ moves the graph 2 units to the right. The roots 0 and 2 become 2 and 4, and the minimum $(1, -1)$ becomes $(3, -1)$ .

(b) Graph B is graph A reflected in the y-axis, so its equation is $y = g(-x)$ .

Watch out: Moving the graph 2 units left for $f(x - 2)$ .

Graph transformations2 marksGrade 8CalculatorHigher

June 2024 · Paper 2 Higher · Q21

Draw the graph of g(-x) + 2 from a given curve y = g(x).

Show answer

Answer	The curve reflected in the $y$ -axis and moved up 2: through $(1,3)$ , $(-1,0)$ and $(-3,3)$	2 marks

Worked solution Checked against the mark scheme

$g(-x)$ reflects the curve in the $y$ -axis; $+2$ then moves it up 2 units.

Key points: $(-1,1) \to (1,3)$ , $(1,-2) \to (-1,0)$ , $(3,1) \to (-3,3)$ .

Draw the U-shaped curve through these points, minimum at $(-1, 0)$ .

Watch out: Reflecting in the $x$ -axis instead of the $y$ -axis, or moving 2 units horizontally.

Graph transformations3 marksGrade 8CalculatorHigher

November 2023 · Paper 3 Higher · Q20

Give the turning point after a horizontal translation of a graph, and sketch a reflected and shifted version of a curve.

Show answer

(a)	$(4, 6)$	1 mark
(b)	Curve with maximum $(2, 6)$ and end points $(-2, -4)$ and $(6, -4)$	2 marks

Worked solution Checked against the mark scheme

(a) $g(x - 7)$ moves the graph 7 to the right: $(-3 + 7, 6) = (4, 6)$ .

(b) $f(-x)$ reflects the curve in the $y$ -axis: maximum $(-2, 3) \to (2, 3)$ , ends $(-6, -7) \to (6, -7)$ and $(2, -7) \to (-2, -7)$ .

$+3$ moves it up 3: maximum $(2, 6)$ , ends $(6, -4)$ and $(-2, -4)$ .

Watch out: In (a), moving the turning point 7 to the left, giving (−10, 6).

Graph transformations3 marksGrade 8CalculatorHigher

June 2022 · Paper 2 Higher · Q21

Sketch a reflection of a curve in the y-axis, then find the equation of a translated tan graph from where one point moves.

Show answer

(a)	The curve reflected in the $y$ -axis	1 mark
(b)	$g(x) = \tan(x + 270)^\circ - 5$	2 marks

Worked solution Checked against the mark scheme

(a) $f(-x)$ swaps left and right: reflect every point of the curve in the $y$ -axis.

(b) Q is at $(360, 0)$ and R is at $(90, -5)$ .

The graph moves 270 left and 5 down, i.e. by $\begin{pmatrix}-270\\-5\end{pmatrix}$

Moving left by 270 means replacing $x$ with $x + 270$ ; moving down 5 means subtracting 5:

$g(x) = \tan(x + 270)^\circ - 5$

Watch out: Writing tan(x − 270) because the graph moves to the left.

Graph transformations2 marksGrade 8CalculatorHigher

November 2022 · Paper 3 Higher · Q23

Sketch a horizontal translation of a cubic graph and write the equation of a reflected graph.

Show answer

(a)	The curve moved 2 units left: crossing the x-axis at $-3$ , $-1$ and $0$ , with a maximum at $(-2, 2)$	1 mark
(b)	$y = -g(x)$	1 mark

Worked solution Checked against the mark scheme

(a) $f(x + 2)$ moves the graph 2 units to the left. The roots $-1, 1, 2$ become $-3, -1, 0$ and the peak at $(0, 2)$ moves to $(-2, 2)$ .

(b) Graph B is graph A flipped in the x-axis, so each y-value changes sign: $y = -g(x)$

Watch out: In (a), moving the graph 2 units to the right.

Graph transformations4 marksGrade 8CalculatorHigher

November 2020 · Paper 2 Higher · Q21

Sketch the reflection of a curve in the y-axis, then find the equation of a translated quadratic curve.

Show answer

(a)	The curve reflected in the $y$ -axis, through $(0, 0)$ , $(1, 2)$ , $(2, 1)$ , $(3, 0)$ , $(4, 2)$	2 marks
(b)	$y = 5 + 2(x - 3) - (x - 3)^2$	2 marks

Worked solution Checked against the mark scheme

(a) $f(-x)$ reflects the curve in the $y$ -axis: each point $(a, b)$ moves to $(-a, b)$ . So $(-4, 2)$ , $(-3, 0)$ , $(-1, 2)$ , $(0, 0)$ become $(4, 2)$ , $(3, 0)$ , $(1, 2)$ , $(0, 0)$ .

(b) The curve moves 3 to the right, so replace $x$ with $x - 3$ :

$y = 5 + 2(x - 3) - (x - 3)^2$

which simplifies to $y = -x^2 + 8x - 10$ .

Watch out: Replacing $x$ with $x + 3$ for a shift to the right.

Graph transformations3 marksGrade 8No calculatorHigher

November 2019 · Paper 1 Higher · Q20

Draw a translated graph and find where a point moves under a reflection in the y-axis.

Show answer

(a)	The graph moved 1 left and 3 down	2 marks
(b)	$(2, 1)$	1 mark

Worked solution Checked against the mark scheme

(a) $f(x+1)$ moves the graph 1 unit left; $-3$ moves it 3 units down.

So each corner moves by $\begin{pmatrix}-1\\-3\end{pmatrix}$ , for example $(-2, 1) \to (-3, -2)$ and $(1, 4) \to (0, 1)$ .

(b) $f(-x)$ reflects in the $y$ -axis, so $x$ changes sign: $B = (2, 1)$

Watch out: Moving the graph 1 unit right for f(x + 1).

Graph transformations2 marksGrade 8No calculatorHigher

June 2018 · Paper 1 Higher · Q18

Sketch a sine graph translated down by 2 on the given axes.

Show answer

Answer	The sine curve moved down 2, through $(-180,-2)$ , $(-90,-3)$ , $(0,-2)$ , $(90,-1)$ , $(180,-2)$	2 marks

Worked solution Checked against the mark scheme

Subtracting 2 moves every point of $y = \sin x$ down by 2.

Key points: $(-180, -2)$ , $(-90, -3)$ , $(0, -2)$ , $(90, -1)$ , $(180, -2)$

Join them with a smooth wave.

Watch out: Moving the graph 2 to the left or right instead of down.

Graph transformations2 marksGrade 8CalculatorHigher

Specimen · Paper 3 Higher · Q16

Find where a point moves under a combined graph transformation (a horizontal translation and a reflection in the x-axis).

Show answer

Answer	$(6, -1)$	2 marks

Worked solution Checked against the mark scheme

$f(x + 3)$ shifts the graph 3 units left: $(9, 1) \to (6, 1)$

$-f(\ldots)$ reflects in the $x$ -axis: $(6, 1) \to (6, -1)$

So $P = (6, -1)$

Watch out: Moving the point 3 units right instead of left, giving (12, −1).