GCSE · Algebra

Graph Transformations Higher

If you know what y = f(x) looks like, you can sketch y = f(x) + 3, y = f(x − 2), y = −f(x) and y = f(−x) without plotting a single point. Each one moves or flips the whole graph in a predictable way. The one surprise: changes inside the bracket do the opposite of what you'd expect.

GCSE subject content (Higher): sketch translations and reflections of a given function; identify turning points and other key points after a transformation

Before you start. f(x) is function notation: f is a rule, and f(x) is what the rule gives when you put x in. If f(x) = x², then f(3) = 9 and f(x + 2) = (x + 2)². See Functions for more. You'll also need to know the basic graph shapes (Recognising graphs) and translations and reflections of shapes (Transformations). A translation is described by a column vector: ab means a right and b up.

1 · A graph is a set of points Higher

The graph of y = f(x) is every point (x, f(x)). To transform a graph, transform its points. Pick a starting graph and drag the plum point along it: each point is an input x with its output f(x) as the height. Every panel below starts from the graph you choose there.

y = f(x) = x =

2 · y = f(x) + a: up or down Higher

Adding a to the output raises every point by a. So y = f(x) + a is the graph of f translated by the vector 0a: up if a is positive, down if negative. Every point (x, y) moves to (x, y + a).

f(x) = a = point x =

3 · y = f(x + a): left or right, the opposite way Higher

Now the change is inside the bracket, to the input. In y = f(x + 2), the new graph at x = 0 shows f(2), the value the old graph had 2 units to the right. So everything the old graph did at x = 2, the new graph does at x = 0: the whole graph has moved 2 to the left.

y = f(x + a) is a translation by −a0

"Inside the bracket does the opposite": f(x + 2) moves left 2; f(x − 3) moves right 3. Every point (x, y) moves to (x − a, y).

f(x) = a = point x =

The classic exam trap. y = (x − 3)² is y = x² moved 3 to the right, not left: its turning point is (3, 0), because that's where the bracket is 0. Check any translation by asking "where does the bracket become what it used to be?"

4 · y = −f(x): reflect in the x-axis Higher

Making every output negative flips every point upside down: (x, y) goes to (x, −y). Points above the x-axis go below it by the same distance. The x-axis is the mirror line, and points on it (the roots) don't move.

f(x) = apply y = −f(x) point x =

5 · y = f(−x): reflect in the y-axis Higher

Now the minus sign is inside. The new graph at x = 3 shows f(−3), the old value from the other side of the y-axis. So (x, y) goes to (−x, y): a reflection in the y-axis. Points on the y-axis (the y-intercept) don't move. A symmetrical graph like y = x² doesn't change at all.

f(x) = apply y = f(−x) point x =

Equation	Transformation	Point (x, y) goes to
y = f(x) + a	translation by 0a	(x, y + a)
y = f(x + a)	translation by −a0	(x − a, y)
y = −f(x)	reflection in the x-axis	(x, −y)
y = f(−x)	reflection in the y-axis	(−x, y)

Memory hook: outside the f (changing y) does what it says; inside the bracket (changing x) does the opposite.

6 · Turning points after a transformation Higher

Exam questions often give you a turning point of y = f(x), like (2, 5), and ask for the turning point of a transformed graph. You don't need the equation: just move the point using the table above.

f(x) = transformation a =

Try it. Type the turning point of the new graph.

( , )

7 · Describe the transformation Higher

The teal dashed graph is y = f(x); the orange one is a transformation of it. Choose the full description. In an exam you must say which transformation (translation or reflection) and give the vector or the mirror line.

Worked examples

Describe how y = x² + 3 is related to y = x², and give its turning point Higher

Form: it's f(x) + 3 with f(x) = x², so 3 is added to every output.

Transformation: translation by the vector 03.

Turning point: (0, 0) moves to (0, 3).

y = f(x) has a turning point at (2, 5). Find the turning point of (a) y = f(x − 3), (b) y = f(x) − 4, (c) y = −f(x), (d) y = f(−x) Higher

(a) Inside the bracket, opposite way: −3 means 3 to the right. (2, 5) → (5, 5).

(b) Outside: down 4. (2, 5) → (2, 1).

(c) Reflect in the x-axis: change the sign of y. (2, −5) (and a maximum becomes a minimum).

(d) Reflect in the y-axis: change the sign of x. (−2, 5).

Sketch y = (x + 2)² Higher

Recognise: f(x + 2) with f(x) = x².

Inside the bracket, opposite way: translation by −20, 2 to the left.

Key points: turning point (0, 0) → (−2, 0); y-intercept: put x = 0, y = 4, so (0, 4). Sketch a U through those.

Describe the transformation from y = sin x to y = sin(x + 90°) Higher

Form: f(x + 90) with f(x) = sin x.

Transformation: translation by −900: 90° to the left.

Check a point: the maximum (90°, 1) moves to (0°, 1). The new graph starts at its maximum: it is the graph of y = cos x.

Show that reflecting y = x³ in the x-axis gives the same graph as reflecting it in the y-axis Higher

Reflect in the x-axis: y = −f(x) = −x³.

Reflect in the y-axis: y = f(−x) = (−x)³ = −x³.

Same equation, so the same graph. (This works because x³ has rotational symmetry about the origin.)

Exam-style: A(1, 4) lies on y = f(x). Find the image of A on y = f(x + 1) + 2 Higher

Inside: + 1 moves the graph 1 left: x-coordinate 1 → 0.

Outside: + 2 moves it 2 up: y-coordinate 4 → 6.

Image: (0, 6). Check: on the new graph at x = 0, y = f(0 + 1) + 2 = f(1) + 2 = 4 + 2 = 6 ✓

Quick check

What you've learned

y = f(x) + a: translation by 0a. Changes outside the f act on y and do what they say.
y = f(x + a): translation by −a0. Changes inside the bracket act on x and do the opposite.
y = −f(x): reflection in the x-axis, (x, y) → (x, −y). y = f(−x): reflection in the y-axis, (x, y) → (−x, y).
To find a turning point (or any point) on the new graph, move the old point with the same rule; you don't need the equation of f.
Describe fully: say "translation" with a vector, or "reflection" with the mirror line.