Mathematics Advanced • Year 11 • Module 1 • Lesson 9
Translations of Functions
Build procedural fluency in reading horizontal and vertical translations from an equation, and applying them to key features.
1. Quick recall — the translation rules
Answer each question in the space provided. 1 mark each
Q1.1 Complete the translation rules:
y = f(x) + k shifts the graph ________ if k > 0 and ________ if k < 0.
y = f(x − h) shifts the graph ________ if h > 0 and ________ if h < 0.
Q1.2 The "opposite-direction" trap. Circle the correct direction for each:
(a) y = f(x + 5) shifts LEFT / RIGHT by 5.
(b) y = f(x − 2) shifts LEFT / RIGHT by 2.
(c) y = f(x) − 4 shifts UP / DOWN by 4.
Q1.3 For y = a(x − h)² + k the vertex is at (____, ____). Horizontal shifts affect (domain / range / both); vertical shifts affect (domain / range / both). Circle one for each.
2. Worked example — describe y = f(x + 4) − 1
Follow each line. The reason is given on the right.
Problem. Describe the transformation that maps y = f(x) to y = f(x + 4) − 1.
Step 1 — Identify the horizontal shift.
Write x + 4 in the form x − h: x + 4 = x − (−4) ⇒ h = −4
Reason: shift trap — the value of h is the opposite sign of what appears with x inside the brackets.
Step 2 — Translate h to a direction.
h = −4 ⇒ shift 4 units LEFT
Reason: h < 0 means shift in the negative-x direction.
Step 3 — Identify the vertical shift.
Outside the function: −1 ⇒ k = −1 ⇒ shift 1 unit DOWN
Conclusion. The graph is translated 4 units left and 1 unit down.
3. Faded example — find the new vertex
The vertex of y = f(x) is at (2, −3). Find the vertex of y = f(x − 5) + 2. 4 marks
Step 1 — Identify the translation vector (h, k):
h = ____ (this is the horizontal shift: ____ units ________)
k = ____ (this is the vertical shift: ____ units ________)
Step 2 — Apply the shift to the original vertex (2, −3):
New x-coordinate = original x + h = 2 + ____ = ________
New y-coordinate = original y + k = (−3) + ____ = ________
Conclusion. The new vertex is at ( ________, ________ ).
4. Graduated practice — describe, predict, and write
Foundation — describe the translation (4 questions)
| Q | Equation | Translation (direction + units) |
|---|---|---|
| 4.1 1 | y = f(x) + 5 | |
| 4.2 1 | y = f(x − 4) | |
| 4.3 1 | y = f(x + 2) − 3 | |
| 4.4 1 | y = f(x − 1) + 6 |
Standard — predict the new coordinates (6 questions)
Take y = f(x) to have key point (3, 5). Find the image of this point under each translation.
4.5 y = f(x) + 2 → ( ____, ____ ) 1 mark
4.6 y = f(x − 3) → ( ____, ____ ) 1 mark
4.7 y = f(x + 1) − 4 → ( ____, ____ ) 1 mark
4.8 The graph y = x² is translated 2 units right and 4 units down. Write its equation in vertex form. 2 marks
4.9 Write the equation of y = √x after a translation 1 unit left and 5 units up. State the new domain. 2 marks
4.10 The graph y = 1/x is translated 3 right and 2 down. Write its equation, and state the equations of the new vertical and horizontal asymptotes. 2 marks
Extension — read the translation from the equation (2 questions)
4.11 A parabola has equation y = (x + 1)² − 4. (a) State its vertex. (b) State its range. (c) Find its x-intercepts. 3 marks
4.12 The graph of y = x² is translated so that the point (0, 0) moves to (a, b). (a) Write the equation of the translated graph in vertex form, in terms of a and b. (b) Hence write the equation when (a, b) = (−2, 7), and state the y-intercept of this translated graph. 3 marks
5. Self-check the easy 3
Tick the first three once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Translation rules
y = f(x) + k: up if k > 0, down if k < 0. y = f(x − h): right if h > 0, left if h < 0.
Q1.2 — Direction trap
(a) f(x + 5): LEFT 5 (inside + is opposite direction). (b) f(x − 2): RIGHT 2. (c) f(x) − 4: DOWN 4 (outside − is intuitive).
Q1.3 — Vertex and ranges
Vertex at (h, k). Horizontal shifts affect domain; vertical shifts affect range.
Q3 — Faded example: vertex of y = f(x − 5) + 2
Step 1: h = 5 (5 units right). k = 2 (2 units up).
Step 2: New x = 2 + 5 = 7. New y = (−3) + 2 = −1.
Conclusion: new vertex at (7, −1).
Q4.1–4.4 — Describe the translation
4.1: 5 units up. 4.2: 4 units right. 4.3: 2 units left, 3 units down. 4.4: 1 unit right, 6 units up.
Q4.5 — Image of (3, 5) under y = f(x) + 2
x unchanged, y + 2: (3, 7).
Q4.6 — Image under y = f(x − 3)
x + 3, y unchanged: (6, 5).
Q4.7 — Image under y = f(x + 1) − 4
x − 1 (left 1), y − 4: (2, 1).
Q4.8 — y = x² translated 2 right, 4 down
y = (x − 2)² − 4 (vertex at (2, −4)). y = (x − 2)² − 4.
Q4.9 — y = √x translated 1 left, 5 up
y = √(x + 1) + 5. Need x + 1 ≥ 0 ⇒ x ≥ −1, so domain is [−1, ∞).
Q4.10 — y = 1/x translated 3 right, 2 down
y = 1/(x − 3) − 2. New vertical asymptote: x = 3. New horizontal asymptote: y = −2.
Q4.11 — y = (x + 1)² − 4
(a) Vertex: read off h = −1, k = −4, so vertex at (−1, −4).
(b) Parabola opens upward and vertex is the minimum, so range = [−4, ∞).
(c) (x + 1)² − 4 = 0 ⇒ (x + 1)² = 4 ⇒ x + 1 = ±2 ⇒ x = 1 or x = −3. x-intercepts: (1, 0) and (−3, 0).
Q4.12 — Vertex moves from (0, 0) to (a, b)
(a) Shift a right (or |a| left if a < 0) and b up (or down): y = (x − a)² + b.
(b) (a, b) = (−2, 7): y = (x − (−2))² + 7 = (x + 2)² + 7. y-intercept: substitute x = 0: y = (2)² + 7 = 11, so y-intercept at (0, 11).