Mathematics Advanced • Year 11 • Module 1 • Lesson 12
Combined Transformations
Build procedural fluency in reading the general form y = af(b(x − h)) + k and applying it to points.
1. Quick recall
Answer each part in the space provided. 1 mark each
Q1.1 In the general form y = af(b(x − h)) + k, name the role of each parameter:
a: ________________________________________________________
b: ________________________________________________________
h: ________________________________________________________
k: ________________________________________________________
Q1.2 Rewrite f(2x − 4) in the factored form f(b(x − h)): f(_____(x − _____))
Q1.3 If (x, y) is on y = f(x), the image on y = af(b(x − h)) + k is (__________, __________).
2. Worked example — describing y = −2f(x − 3) + 1
Follow each line. Every step matches the inside-out reading order from the lesson.
Problem. List the transformations that map y = f(x) onto y = −2f(x − 3) + 1.
Step 1 — Identify the inside change.
x − 3 → horizontal translation 3 units right (h = 3)
Reason: x − h shifts right by h units.
Step 2 — No coefficient on x inside, so b = 1.
No horizontal dilation; no y-axis reflection.
Step 3 — Read a from the outside coefficient.
a = −2 → vertical dilation by factor 2 and reflection in x-axis
Reason: negative a flips and stretches vertically.
Step 4 — Read k from the final + term.
k = +1 → vertical translation 1 unit up
Conclusion. Translation 3 right, vertical dilation factor 2, reflection in x-axis, translation 1 up.
3. Faded example — fill in the missing steps
The point (2, 4) lies on y = f(x). Find its image on y = 3f(2x − 4) + 5. Fill in each blank. 4 marks
Step 1 — Factor inside to standard form:
2x − 4 = 2(x − ____) so y = 3f(2(x − ____)) + 5
Read off: a = ____, b = ____, h = ____, k = ____.
Step 2 — Use the point-mapping formula:
x_new = x/b + h = ____/____ + ____ = ____
Step 3 — Transform y:
y_new = a·y + k = ____·____ + ____ = ____
Conclusion. The image of (2, 4) is (____, ____).
4. Graduated practice
Foundation — describe each single-step transformation (4 questions)
| Q | Equation | Transformation in one line |
|---|---|---|
| 4.1 1 | y = f(x) + 4 | |
| 4.2 1 | y = f(x − 2) | |
| 4.3 1 | y = −f(x) | |
| 4.4 1 | y = 3f(x) |
Standard — describe combined transformations (6 questions)
List every transformation with axis, direction and factor. Factorise inside first if the coefficient of x is not 1.
4.5 y = 2f(x − 3) + 1 2 marks
4.6 y = −f(2x + 4) 2 marks
4.7 y = 3f(−x/2) − 5 2 marks
4.8 y = ½ f(x + 1) − 2 2 marks
4.9 Point (3, 2) lies on y = f(x). Find its image on y = −f(x + 2) + 1. 2 marks
4.10 Point (4, 3) lies on y = f(x). Find its image on y = 3f(−2x + 4) − 1. 2 marks
Extension — synthesise (2 questions)
4.11 Write the equation of y = f(x) after the following transformations in order: reflection in the y-axis, horizontal dilation by factor 3, vertical dilation by factor 2, translation 1 right and 4 down. 3 marks
4.12 A student claims y = f(2x − 4) represents a horizontal dilation by factor ½ followed by a translation 4 units left. Identify the error and state the correct transformations. 3 marks
5. Self-check the easy 3
Tick once you've verified your method on the foundation row.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Roles of a, b, h, k
a: vertical dilation by factor |a| from the x-axis (and reflection in the x-axis if a < 0). b: horizontal dilation by factor 1/|b| from the y-axis (and reflection in the y-axis if b < 0). h: horizontal translation by h units (right if h > 0). k: vertical translation by k units (up if k > 0).
Q1.2 — Factor 2x − 4
2x − 4 = 2(x − 2), so f(2x − 4) = f(2(x − 2)).
Q1.3 — Image point formula
(x/b + h, ay + k).
Q3 — Faded example, image of (2, 4) on y = 3f(2x − 4) + 5
Step 1: 2x − 4 = 2(x − 2), so y = 3f(2(x − 2)) + 5. Read off a = 3, b = 2, h = 2, k = 5.
Step 2: x_new = x/b + h = 2/2 + 2 = 3.
Step 3: y_new = a·y + k = 3·4 + 5 = 17.
Image: (3, 17).
Q4.1 — y = f(x) + 4
Vertical translation 4 units up.
Q4.2 — y = f(x − 2)
Horizontal translation 2 units right.
Q4.3 — y = −f(x)
Reflection in the x-axis. (Each y-value is negated; the x-axis is the invariant line.)
Q4.4 — y = 3f(x)
Vertical dilation by factor 3 from the x-axis.
Q4.5 — y = 2f(x − 3) + 1
Vertical dilation factor 2 from x-axis; translation 3 right; translation 1 up.
Q4.6 — y = −f(2x + 4)
Factor first: 2x + 4 = 2(x + 2), so y = −f(2(x + 2)). Horizontal dilation by factor ½ from y-axis; translation 2 left; reflection in x-axis (from outside negative).
Q4.7 — y = 3f(−x/2) − 5
Rewrite inside: −x/2 = −½(x − 0). Horizontal dilation by factor 2 from y-axis; reflection in y-axis; vertical dilation factor 3 from x-axis; translation 5 down.
Q4.8 — y = ½ f(x + 1) − 2
Vertical dilation by factor ½ from x-axis; translation 1 left; translation 2 down.
Q4.9 — Image of (3, 2) on y = −f(x + 2) + 1
a = −1, b = 1, h = −2, k = 1. x_new = 3/1 + (−2) = 1. y_new = (−1)(2) + 1 = −1. Image: (1, −1).
Q4.10 — Image of (4, 3) on y = 3f(−2x + 4) − 1
Factor inside: −2x + 4 = −2(x − 2), so y = 3f(−2(x − 2)) − 1. a = 3, b = −2, h = 2, k = −1.
x_new = x/b + h = 4/(−2) + 2 = −2 + 2 = 0. y_new = 3(3) − 1 = 8. Image: (0, 8).
Q4.11 — Write the equation
Step-by-step build:
• Reflection in y-axis: f(−x).
• Horizontal dilation by 3: f(−x/3).
• Vertical dilation by 2: 2f(−x/3).
• Translation 1 right and 4 down: y = 2f(−(x − 1)/3) − 4, equivalently y = 2f(−⅓(x − 1)) − 4.
Q4.12 — Critique of "2x − 4 means dilation by ½ then left 4"
Error: the student did not factorise 2 out of 2x − 4 before reading the translation. The correct factorisation is 2x − 4 = 2(x − 2), so the translation is 2 units to the right, not 4 to the left.
Correct transformations: horizontal dilation by factor ½ from the y-axis, then translation 2 units right. (No reflection or vertical change because a = 1, k = 0.)