When you drag an app icon across your phone screen, nothing about the icon itself changes — it simply moves. The same thing happens with functions. A translation slides the entire graph up, down, left, or right without stretching or flipping it.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
The graph of $y = x^2$ has its vertex at $(0, 0)$. Imagine you wanted to move this parabola so its vertex is at $(2, 3)$. What changes would you need to make to the equation? Would you add or subtract numbers, and where would they go — inside or outside the squared term?
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Wrong: (a + b)² = a² + b².
Right: (a + b)² = a² + 2ab + b²; the middle term 2ab is essential and commonly forgotten.
📚 Core Content
A translation slides every point on a graph by the same distance in the same direction. The shape, size, and orientation of the graph do not change — only its position.
When you add or subtract a constant outside the function, the entire graph moves up or down:
Every point $(x, y)$ on the original graph moves to $(x, y + k)$. The $x$-coordinates stay the same; only the $y$-coordinates change.
When you add or subtract a constant inside the function, the graph moves left or right. This is where students often make mistakes, because the direction is opposite to what the sign suggests:
Think of it this way: if you want $f(x - 3)$ to produce the same output as $f(0)$, you need $x = 3$. So the feature that was at $x = 0$ has now moved to $x = 3$ — to the right.
When both transformations appear together, handle them one at a time:
$$y = f(x - h) + k$$
The order of these two shifts does not matter — horizontal and vertical translations commute with each other.
Translations affect different features in predictable ways:
🧮 Worked Examples
🧪 Activities
1 $y = f(x) + 5$
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2 $y = f(x - 4)$
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3 $y = f(x + 2) - 3$
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4 $y = f(x - 1) + 6$
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1 $y = f(x) + 2$
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2 $y = f(x - 3)$
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3 $y = f(x + 1) - 4$
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4 $y = f(x - 2) + 7$
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Earlier you were asked: How would you move $y = x^2$ so its vertex is at $(2, 3)$? What changes would you make to the equation?
To move the vertex from $(0, 0)$ to $(2, 3)$, you need to shift the graph 2 units to the right and 3 units up. The new equation is $y = (x - 2)^2 + 3$. The $-2$ inside the brackets causes the horizontal shift to the right, and the $+3$ outside causes the vertical shift up. This is the vertex form of a parabola, and it is one of the most useful equations in all of Year 11 mathematics because it lets you read the vertex directly from the equation.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
8. Explain in words why $y = f(x - 3)$ represents a shift 3 units to the right, even though the number $-3$ appears inside the function. Use the idea of inputs and outputs in your explanation. 2 MARKS
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Answer in your workbook.
9. The vertex of $y = f(x)$ is at $(-1, 2)$. Write the coordinates of the vertex after each transformation: (a) $y = f(x) + 4$ (b) $y = f(x - 3)$ (c) $y = f(x + 2) - 5$ 3 MARKS
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10. A parabola with equation $y = x^2$ is translated so that its vertex moves to $(4, -2)$. (a) Write the equation of the translated parabola. (b) State the domain and range of the translated parabola. (c) Explain why the domain changed or did not change, and why the range changed or did not change. 4 MARKS
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Answer in your workbook.
1. 5 units up
2. 4 units right
3. 2 units left and 3 units down
4. 1 unit right and 6 units up
1. $(2, 7)$ — $y$-coordinate increases by 2
2. $(5, 5)$ — $x$-coordinate increases by 3
3. $(1, 1)$ — $x$-coordinate decreases by 1, $y$-coordinate decreases by 4
4. $(4, 12)$ — $x$-coordinate increases by 2, $y$-coordinate increases by 7
1. A — $f(x) + 3$ shifts 3 units up.
2. D — $f(x - 2)$ shifts 2 units right.
3. B — Vertical shift up: $(1, 4) \to (1, 6)$.
4. B — Horizontal shift left: $(3, 5) \to (2, 5)$.
5. A — 3 right, 4 up gives vertex $(3, 4)$.
Q8 (2 marks): To get the same output from $f(x - 3)$ as you would from $f(0)$, you need $x - 3 = 0$, which means $x = 3$ [1]. So the point that was at $x = 0$ has moved to $x = 3$, which is 3 units to the right [1].
Q9 (3 marks): (a) $(-1, 6)$ [1] (b) $(2, 2)$ [1] (c) $(-3, -3)$ [1]
Q10 (4 marks): (a) $y = (x - 4)^2 - 2$ [1]. (b) Domain: $(-\infty, \infty)$; Range: $[-2, \infty)$ [1]. (c) The domain did not change because there was no horizontal restriction on the original parabola, and horizontal translations do not introduce restrictions [1]. The range changed because the vertical translation of 2 units down shifted the minimum $y$-value from $0$ to $-2$ [1].
Translations of Functions
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