Architects do not draw every brick when they design a bridge. They start with a simple curve — usually a parabola — then stretch it, flip it, and move it until it fits the towers. In this lesson, you will learn to do the same thing: sketch any transformed function by tracking its key features, and use transformations to build mathematical models of real-world situations.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
The basic parabola $y = x^2$ has its vertex at $(0, 0)$ and passes through $(1, 1)$ and $(-1, 1)$. How would you quickly sketch the graph of $y = -2(x - 3)^2 + 4$ without plotting dozens of points? List the key features you would track.
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: f⁻¹(x) means 1/f(x).
Right: f⁻¹(x) denotes the inverse function, not the reciprocal. The reciprocal would be written [f(x)]⁻¹ or 1/f(x).
📚 Core Content
Sketching a transformed graph does not mean calculating 20 points and joining the dots. Examiners reward clear, accurate sketches that show the important structural features of the graph. Here is the method:
| Parent function | Key features | Anchor points |
|---|---|---|
| $y = x^2$ | Parabola, vertex at $(0, 0)$, opens up | $(0, 0), (1, 1), (-1, 1), (2, 4), (-2, 4)$ |
| $y = x^3$ | Cubic, point of inflection at $(0, 0)$ | $(0, 0), (1, 1), (-1, -1)$ |
| $y = |x|$ | V-shape, vertex at $(0, 0)$ | $(0, 0), (1, 1), (-1, 1)$ |
| $y = \dfrac{1}{x}$ | Hyperbola, asymptotes: $x = 0$ and $y = 0$ | $(1, 1), (-1, -1), (2, 0.5), (-2, -0.5)$ |
| $y = \sqrt{x}$ | Domain $x \geq 0$, starts at $(0, 0)$ | $(0, 0), (1, 1), (4, 2)$ |
| $y = e^x$ | Exponential growth, $y$-intercept $(0, 1)$, HA: $y = 0$ | $(0, 1), (1, e), (-1, \frac{1}{e})$ |
| $y = \ln x$ | Domain $x > 0$, $x$-intercept $(1, 0)$, VA: $x = 0$ | $(1, 0), (e, 1), (\frac{1}{e}, -1)$ |
If $(x_0, y_0)$ is a turning point on $y = f(x)$, then on $y = af(b(x - h)) + k$:
$$\text{Turning point} \rightarrow \left(\frac{x_0}{b} + h, ay_0 + k\right)$$
This is the single most useful formula for sketching transformed graphs quickly.
| Parent asymptote | After $y = af(b(x - h)) + k$ |
|---|---|
| Vertical: $x = c$ | $x = \dfrac{c}{b} + h$ |
| Horizontal: $y = d$ | $y = ad + k$ |
🧮 Worked Examples
🧪 Activities
1 $y = 2(x - 1)^2 + 3$
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Answer in your workbook.
2 $y = -\dfrac{1}{x + 2} + 3$
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3 $y = 3\sqrt{x - 2} + 1$
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4 $y = -2|x + 1| + 4$
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1 A small business finds that its daily profit $P$ (in dollars) is zero when 0 or 40 items are sold, and reaches a maximum of \$400 when 20 items are sold.
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Answer in your workbook.
2 The temperature $T$ (in °C) of a cooling cup of coffee approaches room temperature (20°C) over time. Initially the coffee is at 80°C, and after 1 hour it has cooled to 50°C.
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Answer in your workbook.
Earlier you were asked: How would you quickly sketch $y = -2(x - 3)^2 + 4$ without plotting dozens of points?
The key is to track the transformed features of the parent parabola $y = x^2$:
With just the vertex, intercepts, and the general shape, you have enough to sketch the graph accurately. No table of values needed.
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 graph of $y = f(x)$ has a turning point at $(2, -3)$ and $x$-intercepts at $x = -1$ and $x = 5$. Find the corresponding turning point and $x$-intercepts of $y = 2f(x - 1) + 4$. Show all working. 3 MARKS
Type your answer below:
Answer in your workbook.
9. Sketch the graph of $y = \dfrac{2}{x + 1} - 3$, clearly showing the asymptotes, the $x$-intercept, the $y$-intercept, and at least one other point. Label all features with their coordinates or equations. 4 MARKS
Type your answer below (describe the sketch):
Answer in your workbook.
10. A student is modelling the height of a ball thrown vertically upward. They propose the model $h = -5(t - 2)^2 + 20$, where $h$ is height in metres and $t$ is time in seconds. Evaluate the suitability of this model by finding the initial height, the maximum height, the time at which the maximum height occurs, and the time when the ball hits the ground. What are the domain and range of this model in the context of the problem? 4 MARKS
Type your answer below:
Answer in your workbook.
1. Parent: $y = x^2$. Transformations: vertical dilation by 2, right 1, up 3. Vertex: $(1, 3)$.
2. Parent: $y = \frac{1}{x}$. Transformations: reflection in $x$-axis, left 2, up 3. Asymptotes: $x = -2$, $y = 3$.
3. Parent: $y = \sqrt{x}$. Transformations: vertical dilation by 3, right 2, up 1. Domain: $x \geq 2$. Starting point: $(2, 1)$.
4. Parent: $y = |x|$. Transformations: vertical dilation by 2, reflection in $x$-axis, left 1, up 4. Vertex: $(-1, 4)$. $x$-intercepts: $-2|x + 1| + 4 = 0 \Rightarrow |x + 1| = 2 \Rightarrow x = 1$ or $x = -3$.
1. Parent: $y = x(x - 40)$ or vertex form $y = -(x - 20)^2 + 400$. Using vertex form: $P = -(x - 20)^2 + 400$. Vertex $(20, 400)$ means max profit \$400 at 20 items. $x$-intercepts at 0 and 40 confirm break-even points.
2. Parent: exponential decay $T = Ae^{-kt} + C$. As $t \to \infty$, $T \to 20$, so $C = 20$. At $t = 0$, $T = 80$, so $A = 60$. At $t = 1$, $50 = 60e^{-k} + 20 \Rightarrow k = \ln 2 \approx 0.693$. Equation: $T = 60e^{-0.693t} + 20$. The 60 is the initial temperature above room temperature, and the horizontal asymptote $T = 20$ is room temperature.
1. A — Right 1: $(3, 3)$; $\times 2$: $(3, 6)$; up 4: $(3, 10)$.
2. C — Reflection and vertical translation do not move points horizontally.
3. A — $f(2(x + 2))$: $x$-coordinate $\frac{1}{2} - 2 = -1.5$; $y$ stays $-2$.
4. A — Downward parabola with vertex $(3, 4)$ matches the basketball trajectory.
5. A — Reflection in $y$-axis negates $x$-intercepts.
Q8 (3 marks): Turning point: $(3, -2)$ [1]. $x$-intercepts: $2 - 1 = 1$ and $5 - 1 = 4$, but with vertical dilation by 2 and up 4, the $y$-values are no longer zero. Wait — actually, vertical dilation and translation change $y$, so $x$-intercepts need solving. Since we don't know $f(x)$, we can only state the transformed $x$-coordinates under the horizontal translation: $x = 0$ and $x = 4$ [1]. Full marks for correct turning point and stating that vertical transformations preserve $x$-coordinates of zeros if $a \neq 0$, but the $y$-intercept changes. Actually, simpler: horizontal translation shifts intercepts by $+1$, so $-1 \rightarrow 0$ and $5 \rightarrow 6$? No, right 1 means $x - 1$, so $f(x - 1) = 0$ when $x - 1 = -1$ or $5$, giving $x = 0$ and $x = 6$. Award 2 marks for correct turning point and intercepts.
Q9 (4 marks): VA: $x = -1$ [0.5]. HA: $y = -3$ [0.5]. $x$-intercept: $\frac{2}{x + 1} - 3 = 0 \Rightarrow x = -\frac{1}{3}$ [1]. $y$-intercept: $(0, -1)$ [1]. Other point: e.g. $(1, -2)$ [0.5]. Correct hyperbola shape approaching asymptotes [0.5].
Q10 (4 marks): Initial height at $t = 0$: $h = -5(4) + 20 = 0$ m [0.5]. Max height: 20 m at $t = 2$ s [1]. Hits ground when $-5(t - 2)^2 + 20 = 0 \Rightarrow (t - 2)^2 = 4 \Rightarrow t = 0$ or $t = 4$. So hits ground at $t = 4$ s [1]. Domain: $0 \leq t \leq 4$ [0.5]. Range: $0 \leq h \leq 20$ [0.5]. The model is suitable because it has realistic values and a single maximum.
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