In a video game, every character on screen is just a basic image that has been stretched, flipped, rotated, and moved into position. Game developers do not redraw the character for every frame — they apply combined transformations. In this lesson, you will learn to do the same thing with functions: stack multiple transformations together and read the result like a pro.
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 $(0, 0)$. How would you transform this graph so that it opens downward, is twice as steep, and has its vertex at $(3, -2)$? Try to write an equation that achieves all three changes at once.
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: The domain of a function is always all real numbers.
Right: The domain depends on the function; rational functions exclude values that make the denominator zero.
📚 Core Content
All the transformations you have learned so far can be combined into a single, powerful equation:
$$y = af(b(x - h)) + k$$
Each letter in this equation controls a specific transformation:
The safest way to read combined transformations is from the inside out:
Some transformations can be applied in any order without changing the final result:
However, translations and dilations along the same axis generally do not commute. That is why we write the equation in the standard form above — the parentheses fix the correct order.
If you know a point $(x, y)$ on the original graph $y = f(x)$, you can find its image on the transformed graph $y = af(b(x - h)) + k$ using this process:
So the final image point is:
$$\left(\frac{x + h}{b}, ay + k\right)$$
Wait — actually, the more natural way to think about it is: if $(x, y)$ is on the original, then for the transformed graph, the point that has the same "relative position" is found by reversing the transformations. Let's be more careful.
If $(x, y)$ is on $y = f(x)$, then to find the corresponding point on $y = af(b(x - h)) + k$:
So the correct transformed point is $\left(\frac{x}{b} + h, ay + k\right)$.
🧮 Worked Examples
🧪 Activities
1 $y = 2f(x - 3) + 1$
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2 $y = -f(2x + 4)$
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3 $y = 3f\left(-\frac{x}{2}\right) - 5$
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4 $y = \frac{1}{2}f(x + 1) - 2$
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1 $y = 2f(x - 1) + 4$
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2 $y = -f(2x) + 1$
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3 $y = 3f\left(\frac{x}{2} - 1\right) - 2$
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4 $y = -2f(-x + 3) + 1$
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Earlier you were asked: How would you transform $y = x^2$ so it opens downward, is twice as steep, and has its vertex at $(3, -2)$?
The original parabola $y = x^2$ opens upward with vertex $(0, 0)$. To make it open downward, we need a reflection in the $x$-axis: $-x^2$. To make it twice as steep vertically, we multiply by 2: $-2x^2$. Finally, to move the vertex to $(3, -2)$, we replace $x$ with $(x - 3)$ and subtract 2: $y = -2(x - 3)^2 - 2$. This single equation combines three distinct transformations: reflection, vertical dilation, and translation. That is the power of the general transformation form.
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. The point $(1, 2)$ lies on the graph of $y = f(x)$. Find the corresponding point on the graph of $y = -2f(3x - 6) + 4$. Show all working, including rewriting the function in the form $af(b(x - h)) + k$. 3 MARKS
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Answer in your workbook.
9. (a) Write the equation of $y = f(x)$ after a reflection in the $x$-axis, a horizontal dilation by factor $\frac{1}{2}$, and a translation 3 units left and 2 units up. (b) If $f(x) = x^2$, simplify your equation from part (a) into expanded polynomial form. 4 MARKS
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10. A student claims that $y = f(2x - 4)$ represents a horizontal dilation by factor $\frac{1}{2}$ followed by a translation 4 units to the left. Evaluate this claim. If it is incorrect, explain the error and state the correct transformations. 3 MARKS
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Answer in your workbook.
1. Vertical dilation by factor 2 from the $x$-axis, translation 3 units right, translation 1 unit up.
2. $2x + 4 = 2(x + 2)$: horizontal dilation by factor $\frac{1}{2}$ from the $y$-axis, translation 2 units left, reflection in the $x$-axis.
3. Horizontal dilation by factor 2 from the $y$-axis, reflection in the $y$-axis, vertical dilation by factor 3 from the $x$-axis, translation 5 units down.
4. Vertical dilation by factor $\frac{1}{2}$ from the $x$-axis, translation 1 unit left, translation 2 units down.
1. $(3, 10)$ — $x$: $2 + 1 = 3$; $y$: $2(3) + 4 = 10$
2. $(1, -2)$ — $x$: $\frac{2}{2} = 1$; $y$: $-(3) + 1 = -2$
3. $(6, 7)$ — Rewrite: $\frac{x}{2} - 1 = \frac{1}{2}(x - 2)$. $x$: $\frac{2}{\frac{1}{2}} + 2 = 4 + 2 = 6$; wait, using formula: $x_{\text{new}} = \frac{x}{b} + h = \frac{2}{\frac{1}{2}} + 2 = 6$. $y$: $3(3) - 2 = 7$.
4. $(1, -5)$ — Rewrite: $-x + 3 = -(x - 3)$. $x$: $\frac{2}{-1} + 3 = 1$. $y$: $-2(3) + 1 = -5$.
1. A — Vertical dilation 2, right 3, up 1.
2. B — $-f(x)$ = $x$-axis reflection; $x + 2$ = left 2.
3. A — Right 1: $(3, 4)$; $\times(-2)$: $(3, -8)$; up 3: $(3, -5)$.
4. B — $y$-axis reflection: $f(-x)$; horizontal dilation 2: $f(-\frac{x}{2})$; up 1: $+1$.
5. A — Right 2, up 5, with $x$-axis reflection. Vertex moves to $(2, 5)$.
Q8 (3 marks): $3x - 6 = 3(x - 2)$, so $y = -2f(3(x - 2)) + 4$ [1]. $x_{\text{new}} = \frac{1}{3} + 2 = \frac{7}{3}$ [0.5]. $y_{\text{new}} = -2(2) + 4 = 0$ [1]. New point: $(\frac{7}{3}, 0)$ [0.5].
Q9 (4 marks):
(a) y = -(2x)^2 \text{ with left 3 and up 2} = -4(x + 3)^2 + 2 (b) y = -4(x^2 + 6x + 9) + 2 = -4x^2 - 24x - 36 + 2 = -4x^2 - 24x - 34Award 2 marks for (a) and 2 marks for correct expansion in (b).
Q10 (3 marks): The student's claim is incorrect [1]. The error is not factoring out the 2: $2x - 4 = 2(x - 2)$, so the translation is 2 units to the right, not 4 units to the left [1]. The correct transformations are: horizontal dilation by factor $\frac{1}{2}$ from the $y$-axis, followed by a translation 2 units to the right [1].
Sprint through questions on sketching graphs with combined transformations. Pool: lessons 1–12.
Tick when you've finished all activities and checked your answers.