Graph Transformations – Grade 10 IGCSE Maths

Welcome to Graph Transformations!

Graph transformations let you take a known curve and predict the shape and position of related curves — without recalculating every point. The key insight is: changes outside the bracket affect y directly; changes inside the bracket affect x in the opposite/reciprocal way.

Outside bracket → direct effect on y | Inside bracket → opposite/reciprocal effect on x

Translations

y=f(x)+a and y=f(x+a)

Reflections & Stretches

y=-f(x), y=af(x), y=f(ax)

Combined & Trig

Order, amplitude, period, phase shift

Worked Examples

6 fully worked examples

Visualiser

4 sliders control y=a·sin(b(x+c))+d

Exercises

5 sets of 8q + Practice + Challenge

1. Translations

y = f(x) + a → shift UP by a | y = f(x) − a → shift DOWN by a

Vertical translations (outside bracket):
y = x² + 3: every point on y = x² moves UP 3 units. Vertex (0,0) → (0,3).
y = x² − 5: every point moves DOWN 5. Vertex (0,0) → (0,−5).
Key points: (p, q) → (p, q + a) for y = f(x) + a

y = f(x + a) → shift LEFT by a | y = f(x − a) → shift RIGHT by a (OPPOSITE direction)

Critical trap: y = f(x + 2) shifts LEFT 2, NOT right 2. The +2 is inside the bracket, so it acts on x in the OPPOSITE direction. Think: x + 2 = 0 when x = −2, so the graph passes through x = −2 where it used to pass through x = 0.

Horizontal translations (inside bracket):
y = (x − 3)²: vertex of y = x² moves from (0,0) to (3,0) — shifted RIGHT 3.
y = (x + 4)²: vertex moves from (0,0) to (−4,0) — shifted LEFT 4.
Key points: (p, q) → (p − a, q) for y = f(x + a)

Effect on key points:
y = f(x) + a: (p, q) → (p, q + a) — x-coord unchanged, y-coord +a
y = f(x + a): (p, q) → (p − a, q) — x-coord −a, y-coord unchanged

2. Reflections and Stretches

y = −f(x): reflect in x-axis (negate all y-coords)

Reflection in x-axis:
y = −x²: every y-value negated. Maximum at (0,0), opens downward.
Key points: (p, q) → (p, −q)

y = f(−x): reflect in y-axis (negate all x-coords)

Reflection in y-axis:
y = f(−x): replace x with −x in every formula.
Key points: (p, q) → (−p, q)

y = af(x): vertical stretch, scale factor a from x-axis (y-coords × a)

Vertical stretch by factor a:
y = 3sin(x): amplitude 3 — y-values multiplied by 3.
Key points: (p, q) → (p, aq)
If 0 < a < 1 this is a vertical compression (squash).

y = f(ax): horizontal stretch, scale factor 1/a from y-axis (x-coords ÷ a)

Horizontal stretch by factor 1/a:
y = sin(2x): x-coords halved, period halved (180° not 360°).
y = sin(x/2): x-coords doubled, period doubled.
Key points: (p, q) → (p/a, q)
Note: f(ax) halves the period if a=2 — the graph is compressed horizontally.

Memory aid — OUTSIDE vs INSIDE:
OUTSIDE bracket (affects y): a·f(x) → multiply y by a; −f(x) → negate y
INSIDE bracket (affects x — opposite/reciprocal): f(ax) → divide x by a; f(x+a) → subtract a from x; f(−x) → negate x

3. Combined Transformations and Trigonometric Functions

Order matters: apply INNER transformations first, then OUTER

y = 2f(x + 1):
Step 1 (inner): x+1 → shift LEFT 1
Step 2 (outer): ×2 → stretch vertically by 2
Combined effect on point (p, q): first → (p−1, q), then → (p−1, 2q)

Trig transformations — y = 3sin(2x):
Amplitude = 3 (vertical stretch by 3 — y-values range from −3 to +3)
Period = 360°/2 = 180° (horizontal compression by ½)
No phase shift, no vertical shift.

y = sin(x − 30°) + 1:
Phase shift: 30° to the RIGHT (inside: x−30 shifts right)
Vertical shift: +1 UP
Amplitude: 1 (unchanged), Period: 360° (unchanged)

Finding original point from image:
If y = f(x) has point (2, 5), find the corresponding point on y = f(2x) + 3.
f(2x) → x-coord halved: (2, 5) → (1, 5)
+3 → y-coord + 3: (1, 5) → (1, 8)
Image point: (1, 8)

To find original point when given image: REVERSE the transformation steps in reverse order.

Example 1 — Translate a Parabola

y = (x−3)² + 2 compared to y = x²:

f(x−3) → shift RIGHT 3. Then +2 → shift UP 2.

Vertex: (0,0) → (3, 2). All other points shift by (+3, +2).

Example 2 — Describe the Transformation

y = f(x) → y = −f(x+4):

f(x+4): shift LEFT 4. Then −: reflect in x-axis.

Description: Translation by vector (−4, 0) followed by reflection in the x-axis.

Example 3 — Apply to Specific Point

y = f(x) has point (6, 4). Find image on y = f(2x) − 3.

f(2x): x-coord ÷ 2 → (3, 4). Then −3: y-coord −3 → (3, 1).

Image point: (3, 1)

Example 4 — Sketch Transformed Trig

Sketch y = 2sin(3x) for 0° ≤ x ≤ 360°:

Amplitude = 2 (max=2, min=−2)

Period = 360°/3 = 120°. So 3 complete cycles in [0°, 360°].

Key points: (0,0), (40°,2), (60°,0), (80°,−2), (120°,0), then repeat.

Example 5 — Combined Transformation

y = 3f(x−2) + 1 from y = f(x) with point (5, 4):

f(x−2): x+2 → (5+2, 4) = (7, 4). Then ×3: (7, 12). Then +1: (7, 13).

Image: (7, 13)

Example 6 — Find Original Point from Image

y = f(3x) + 2 has image point (2, 7). Find corresponding point on y = f(x).

Reverse: y-coord − 2: 7−2 = 5. Reverse f(3x): x×3 = 2×3 = 6.

Original point: (6, 5)

Common Mistakes

Mistake 1: f(x+2) shifts RIGHT instead of LEFT
f(x+2) shifts LEFT by 2. The +2 is inside the bracket, so it acts oppositely on x. Many students confuse the direction. Always ask: "where does x+2=0? At x=−2." So the graph shifts left.
Mistake 2: f(2x) doubles the x-coordinates instead of halving
y=f(2x) means x-coordinates are HALVED (÷2). The period of sin(2x) is 180°, not 720°. The scale factor for horizontal stretch is 1/a, not a.
Mistake 3: Confusing −f(x) with f(−x)
y=−f(x) reflects in the x-axis (y-coords negate). y=f(−x) reflects in the y-axis (x-coords negate). These are different transformations with different effects.
Mistake 4: Confusing amplitude and period in trig
In y=3sin(2x): amplitude is 3 (NOT 2), period is 180° (NOT 3). The number OUTSIDE the bracket gives amplitude; the number INSIDE the bracket affects the period.
Mistake 5: Wrong order for combined transformations
For y=2f(x+1), the shift (inner) comes FIRST, then the stretch (outer). Getting the order wrong gives different coordinates.

Transformation Summary Table

Transformation	Equation	Effect on point (p, q)
Translate up a	y = f(x) + a	(p, q) → (p, q+a)
Translate down a	y = f(x) − a	(p, q) → (p, q−a)
Translate left a	y = f(x + a)	(p, q) → (p−a, q)
Translate right a	y = f(x − a)	(p, q) → (p+a, q)
Reflect in x-axis	y = −f(x)	(p, q) → (p, −q)
Reflect in y-axis	y = f(−x)	(p, q) → (−p, q)
Vertical stretch × a	y = af(x)	(p, q) → (p, aq)
Horizontal stretch × 1/a	y = f(ax)	(p, q) → (p/a, q)
Trig amplitude	y = a·sin(x)	Max = \|a\|, Min = −\|a\|
Trig period	y = sin(bx)	Period = 360°/b
Trig phase shift	y = sin(x − c)	Shift RIGHT by c
Trig vertical shift	y = sin(x) + d	Shift UP by d

Trig Transformation Visualiser

Control y = a·sin(b(x + c)) + d using the sliders. The original y = sin(x) is shown in grey for comparison.

a (amplitude) = 1
b (frequency) = 1
c (phase, °) = 0
d (v-shift) = 0

Equation: y = sin(x)

Exercise 1 — Identify the Transformation

Enter the numerical value of the transformation parameter described.

1. y=f(x)+5 — how many units does the graph shift UP?

2. y=f(x)−3 — how many units does the graph shift DOWN?

3. y=f(x+4) — how many units does the graph shift LEFT?

4. y=f(x−6) — how many units does the graph shift RIGHT?

5. y=3f(x) — vertical stretch scale factor?

6. y=f(4x) — horizontal stretch scale factor is 1/4. Enter denominator (4).

7. y=2sin(x): amplitude?

8. y=sin(3x): period in degrees? [360/3]

Exercise 2 — Apply Transformation to Points

y=f(x) has point (4, 6). Find the y-coordinate of the image point under each transformation.

1. y=f(x)+5. Image y-coord?

2. y=f(x)−2. Image y-coord?

3. y=−f(x). Image y-coord?

4. y=3f(x). Image y-coord?

5. y=f(x+4). Image y-coord? [x changes, y stays at 6]

6. y=2f(x)+1. Image y-coord? [2×6+1=13]

7. y=−f(x)+4. Image y-coord? [−6+4=−2]

8. y=4f(x)−2. Image y-coord? [4×6−2=22]

Exercise 3 — Find New x-Coordinates

y=f(x) has point (8, 3). Find the x-coordinate of the image point.

1. y=f(x+3). Image x-coord? [8−3=5]

2. y=f(x−5). Image x-coord? [8+5=13]

3. y=f(2x). Image x-coord? [8÷2=4]

4. y=f(x/4). Image x-coord? [8×4=32]

5. y=f(−x). Image x-coord? [−8]

6. y=f(4x). Image x-coord? [8÷4=2]

7. y=f(x+10). Image x-coord? [8−10=−2]

8. y=f(x/2). Image x-coord? [8×2=16]

Exercise 4 — Trig Transformations

Enter the requested property (amplitude, period in degrees, or shift).

1. y=4sin(x). Amplitude?

2. y=sin(2x). Period (degrees)?

3. y=3sin(4x). Amplitude?

4. y=3sin(4x). Period (degrees)?

5. y=sin(x−45°). Phase shift RIGHT? Enter 45.

6. y=2cos(3x). Period (degrees)?

7. y=−5sin(x). Amplitude? [amplitude = |−5| = 5]

8. y=sin(x/2). Period (degrees)?

Exercise 5 — Combined Transformations and Reverse

y=f(x) has point (4, 3). Find the y-coordinate of the combined image, or find the original y-coord from the image.

1. y=2f(x)+4. Image y-coord? [2×3+4=10]

2. y=−f(x)+7. Image y-coord? [−3+7=4]

3. y=3f(x)−5. Image y-coord? [9−5=4]

4. y=f(2x)−1. Point (4,3) → x÷2=2, y−1=2. Image y-coord?

5. y=2f(3x)+1. Image y-coord from (4,3): x÷3=4/3; y: 2×3+1=7. Enter 7.

6. REVERSE: y=f(x)+6 has image point (4, 9). Find original y-coord. [9−6=3]

7. REVERSE: y=3f(x) has image point (4, 15). Find original y-coord. [15÷3=5]

8. REVERSE: y=f(2x)+3 has image point (2, 11). Find original y-coord. [reverse +3: 8; original x=2×2=4 — we seek original y at (4,?). Image y=11 → y−3=8] Enter 8.

Practice — 25 Questions

Mixed transformation questions. y=f(x) has point (6, 4) unless stated.

1. y=f(x)+7. Image y-coord?

2. y=f(x)−3. Image y-coord?

3. y=−f(x). Image y-coord?

4. y=f(−x). Image x-coord?

5. y=5f(x). Image y-coord?

6. y=f(2x). Image x-coord?

7. y=f(x+5). Image x-coord?

8. y=f(x−8). Image x-coord?

9. y=2f(x)+3. Image y-coord? [2×4+3=11]

10. y=−f(x)+10. Image y-coord? [−4+10=6]

11. y=3sin(x). Amplitude?

12. y=sin(4x). Period (degrees)?

13. y=5sin(2x). Amplitude?

14. y=5sin(2x). Period (degrees)?

15. y=sin(x+30°). Phase shift LEFT? Enter 30.

16. y=cos(x−60°). Phase shift RIGHT? Enter 60.

17. y=f(3x). Image x-coord of (6,4)?

18. y=f(x/3). Image x-coord of (6,4)? [6×3=18]

19. REVERSE: y=f(x)+5 has image y=13. Find original y.

20. REVERSE: y=4f(x) has image y=20. Find original y.

21. y=2f(x−3). Image of (6,4): x+3=9, y×2=8. Image y-coord?

22. y=−f(x+2). Image of (6,4): x−2=4, y×−1=−4. Image y-coord?

23. y=−3sin(x). Amplitude?

24. y=sin(x/4). Period (degrees)?

25. REVERSE: y=2f(x)+1 has image y=9. Find original y. [(9−1)/2=4]

Challenge — 12 Questions

Harder combined transformations, reverse problems, and trig.

1. y=3f(2x)−5. Image of (4,3): x÷2=2, y×3−5=4. Image y-coord?

2. y=−2f(x+1)+6. Image of (5,3): x−1=4, y×−2+6=0. Image y-coord?

3. y=f(−2x). Image x-coord of (6,4)? [x÷(−2)=−3]

4. REVERSE: y=5f(x)−3 has image y=17. Find original y. [(17+3)/5=4]

5. REVERSE: y=−f(x)+8 has image y=2. Find original y. [8−original=2 → 6]

6. y=4cos(2x). Amplitude?

7. y=4cos(2x). Period (degrees)?

8. y=2sin(3(x−20°))+1. Amplitude?

9. y=2sin(3(x−20°))+1. Period (degrees)?

10. y=2sin(3(x−20°))+1. Phase shift RIGHT (degrees)?

11. y=2sin(3(x−20°))+1. Vertical shift UP?

12. y=f(x) has minimum at (3,−2). Under y=f(x+4)−1, what is the new minimum y-coord? [−2−1=−3]

Exam Style — 5 Questions

Cambridge IGCSE style. Show workings on paper.

[2 marks] The graph y=f(x) passes through (3, 5). The transformation y=2f(x)+1 is applied. Find the y-coordinate of the image. [2×5+1=11]

[2 marks] Describe the transformation from y=f(x) to y=f(x−4). The graph shifts RIGHT by how many units?

[3 marks] y=3sin(2x). State the amplitude.

[3 marks] y=3sin(2x). State the period in degrees. [360/2=180]

[4 marks] y=f(3x)−2 has image point (2, 7). Find the corresponding y-coordinate on y=f(x). [reverse: y+2=9; original x=2×3=6. So (6, 9) is on y=f(x). Original y-coord = 9]