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).