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Module 1 · L13 of 15 ~45 min ⚡ +50 XP in Learn · +25 to complete

Sketching & Modelling Transformed Functions

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.

Today's hook, 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? What key features would you track?
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Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.

Worksheets

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Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

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Recall, your gut answer first
+5 XP warm-up

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.

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Formula reference · this lesson
core notation

Key features to track when sketching

$x$-intercepts (set $y = 0$)
$y$-intercept (set $x = 0$)
Turning points / vertices
Asymptotes (horizontal, vertical, oblique)
Domain and range
End behaviour

Key insight: You only need 3–5 precise points and the correct general shape to produce a sketch that earns full marks in an exam.

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What you'll master
Know

Key facts

  • The standard features that define the shape of a function's graph
  • How each transformation affects intercepts, turning points, and asymptotes
  • Common parent functions used in modelling
Understand

Concepts

  • Why tracking key features is more efficient than plotting every point
  • How transformations preserve some properties and change others
  • How to select an appropriate parent function for a real-world model
Can do

Skills

  • Sketch transformed graphs by tracking key features
  • Find the equation of a transformed graph given its key features
  • Model real-world situations using transformed functions
  • Interpret the meaning of parameters in applied contexts
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Key terms
Base graph
The simple graph you start from before any change is applied. Like this: for $y = 2(x - 1)^2 + 3$ the base graph is $y = x^2$.
Key feature
The points a marker looks for on a sketch. Like this: for a parabola, the vertex and both intercepts, each labelled with coordinates.
Asymptote
A line the graph gets closer and closer to but never touches. Like this: $y = \tfrac{1}{x} + 2$ approaches the line $y = 2$ as $x$ gets large.
Domain and range after a change
Horizontal changes move the domain, vertical changes move the range. Like this: $y = f(x) + 4$ leaves the domain alone and lifts the range by 4.
Modelling with a transformation
Choosing a transformed graph that matches a real situation. Like this: a tank starting at 50 litres and losing 2 litres a minute fits $y = 50 - 2x$.
Checking a sketch
Substituting one easy value to confirm the picture. Like this: put $x = 0$ into the rule and check the $y$-intercept you drew.
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The key-feature method

Work through the core explanation before applying it.

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The key-feature method
core concept · +3 XP at end

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:

  1. Identify the parent function. Is it a parabola? A hyperbola? A cubic? A square root? An exponential?
  2. Read the transformations. Write the equation in the form $y = af(b(x - h)) + k$.
  3. Track the anchor points. Identify 3–5 key points on the parent graph, then transform each one.
  4. Draw asymptotes first. If the parent has asymptotes, transform them and draw them as dashed lines.
  5. Sketch the curve. Draw a smooth curve through the transformed points, approaching asymptotes correctly and respecting the general shape.
Why bridge designers use parabolas. The cables of a suspension bridge naturally form a parabola (or a catenary, which is very similar) under uniform load. Engineers start with $y = x^2$, stretch it horizontally to match the span between towers, stretch it vertically to match the sag, and translate it so the vertex sits at the lowest point of the cable. One parent function + three transformations = the entire bridge design.

Common parent functions

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: $x = 0$ $(1, 0), (e, 1), (\frac{1}{e}, -1)$

Sketching checklist: parent function → read transformations → transform anchor points → draw asymptotes → smooth curve; Turning point formula: $(x_0, y_0) \to \left(\frac{x_0}{b} + h,\; ay_0 + k\right)$

Pause, copy the five-step sketching checklist (parent → read params → transform anchor points → draw asymptotes → smooth curve) and the turning-point mapping formula into your book.

Quick check: The vertex of $y = x^2$ is at $(0, 0)$. After applying $y = 3(x - 2)^2 + 1$, where does the vertex move?

Did you get this? True or false: when sketching, you should always calculate 20+ points to ensure accuracy.

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How transformations affect features
core concept

We just saw a five-step checklist for sketching: identify the parent, read the parameters, map anchor points, place asymptotes, then draw. That raises a question: can we go further and write exact formulas for how turning points and asymptotes move under any transformation? This card answers it → each type of feature has a specific transformation rule derived directly from the general form.

Intercepts

  • Vertical dilation: multiplies $y$-coordinates; the $y$-intercept becomes $(0, af(0) + k)$
  • Horizontal dilation: multiplies $x$-coordinates by $\frac{1}{b}$; the $x$-intercepts become $\frac{x_0}{b} + h$
  • Reflections: can change the sign of intercepts or eliminate them entirely

Turning points

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.

Asymptotes

Parent asymptote After $y = af(b(x - h)) + k$
Vertical: $x = c$ $x = \dfrac{c}{b} + h$
Horizontal: $y = d$ $y = ad + k$

Domain and range

  • Horizontal transformations (dilation, translation, reflection) affect the domain
  • Vertical transformations affect the range

Turning point: $(x_0, y_0) \to \left(\frac{x_0}{b} + h, ay_0 + k\right)$; Vertical asymptote $x = c \to x = \frac{c}{b} + h$; Horizontal asymptote $y = d \to y = ad + k$

Pause, copy the three feature-transformation rules: turning point, vertical asymptote ($x = c \to x = c/b + h$), and horizontal asymptote ($y = d \to y = ad + k$) into your book.

Fill the blanks: drag each token into place.

domain range horizontal vertical

___ transformations (translation, dilation in $x$) affect the ___. ___ transformations affect the ___.

Worked examples · reveal as you go

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Work examples end to end

Follow the reasoning through complete worked solutions.

Worked example 1 · sketching a transformed parabola +5 XP on full reveal

Sketch the graph of $y = -2(x - 3)^2 + 4$, showing the vertex and $x$-intercepts.

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Parent function: $y = x^2$
Parabola opening upward with vertex at $(0, 0)$.
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Transformations: $a = -2$, $h = 3$, $k = 4$
Reflection in $x$-axis, vertical dilation by 2, translation 3 right and 4 up.
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Vertex: $(0, 0) \to (3, 4)$
$x_{\text{new}} = \frac{0}{1} + 3 = 3$; $y_{\text{new}} = -2(0) + 4 = 4$.
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$x$-intercepts: $-2(x - 3)^2 + 4 = 0 \Rightarrow (x-3)^2 = 2 \Rightarrow x = 3 \pm \sqrt{2}$
Set $y = 0$ and solve.
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Sketch: downward parabola, vertex $(3, 4)$, $x$-intercepts at $x = 3 \pm \sqrt{2} \approx 1.59$ and $4.41$, $y$-intercept $(0, -14)$. ✓
Draw, label all features.
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Worked example 2 · sketching a transformed hyperbola +5 XP on full reveal

Sketch the graph of $y = \dfrac{3}{x - 2} + 1$, showing asymptotes and key points.

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Parent function: $y = \frac{1}{x}$
Hyperbola with asymptotes $x = 0$ and $y = 0$, passing through $(1, 1)$ and $(-1, -1)$.
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Write as $y = 3f(x - 2) + 1$: $a = 3$, $b = 1$, $h = 2$, $k = 1$
Vertical dilation by 3, right 2, up 1.
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Asymptotes: VA $x = 0 \to x = 2$; HA $y = 0 \to y = 1$
Draw as dashed lines first.
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Key points: $(1, 1) \to (3, 4)$; $(-1, -1) \to (1, -2)$
Apply point transformation formula.
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Sketch: hyperbola centred at $(2, 1)$ passing through $(3, 4)$ and $(1, -2)$, approaching asymptotes $x = 2$ and $y = 1$. ✓
Draw and label all features.
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Worked example 3 · modelling with transformations +5 XP on full reveal

The profit $P$ (in thousands of dollars) from selling $x$ hundred items is modelled by a parabola with a maximum profit of $\$50{,}000$ when 300 items are sold, and breaking even ($P = 0$) at 100 and 500 items. Find the equation of the model.

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Choose parent function: downward parabola $y = -x^2$
Profit has a maximum, so we use a downward parabola.
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Vertex form: $P = -a(x - 3)^2 + 50$
Vertex at $(3, 50)$ since $x$ is in hundreds and $P$ is in thousands.
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Find $a$ using $x$-intercept at $x = 1$: $0 = -a(1 - 3)^2 + 50 \Rightarrow 4a = 50 \Rightarrow a = 12.5$
Substitute the known intercept to find the dilation factor.
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Answer: $P = -12.5(x - 3)^2 + 50$ or $P = -12.5x^2 + 75x - 62.5$ ✓
State in vertex form and expanded form.
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Common mistakes · the 4 traps that cost marks

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Dodge the traps, then apply

Meet the mistakes that cost marks, then do it yourself.

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Forgetting to transform asymptotes

Students often draw the transformed graph with the original asymptotes. For $y = \frac{3}{x - 2} + 1$, the vertical asymptote moves to $x = 2$ and the horizontal asymptote moves to $y = 1$.

✓ Fix: Always write down the transformed asymptotes before sketching the curve.

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Not labelling key points on the sketch

A sketch without labelled intercepts or turning points will lose marks in exams, even if the shape is correct. Examiners cannot read your mind.

✓ Fix: Label at least the vertex/turning point and all intercepts. Asymptotes should be drawn as dashed lines with their equations noted.

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Using the wrong parent function for a model

If a quantity has a maximum or minimum, a parabola is usually appropriate. If it approaches a limit over time, an exponential or hyperbola is better. Choosing the wrong parent function means the model will never fit the data.

✓ Fix: Ask: does it have a max/min? (Parabola) Does it approach a limit? (Exponential/hyperbola) Does it start at zero and grow with decreasing rate? (Square root/logarithm)

4

Confusing the units in applied questions

In modelling questions, $x$ might represent "thousands of items" or "hours after 9am". If you treat $x = 3$ as "3 items" when it really means "3000 items", your intercepts and vertex will be completely wrong.

✓ Fix: Read the question carefully and write a note defining your variables with units before you start calculating.

Activity 1, Track the features

For each function, identify the parent function, describe the transformations, and find the coordinates of the transformed vertex (or turning point). Do not sketch yet, just calculate.

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$y = 2(x - 1)^2 + 3$

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$y = -\dfrac{1}{x + 2} + 3$

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$y = 3\sqrt{x - 2} + 1$

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$y = -2|x + 1| + 4$

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Odd one out: Which function does NOT have asymptotes?

Quick-fire practice · 5 reps +2 XP per reveal

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Drill it, then lock it in

Run the quick drill and copy the summary into your book.

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What are the asymptotes of $y = \dfrac{2}{x + 1} - 3$?

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The graph of $y = f(x)$ has a turning point at $(1, 5)$. Find the turning point on $y = 2f(x - 3) + 1$.

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Which parent function would you choose to model a quantity that grows with a decreasing rate from zero?

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The domain of $y = \sqrt{x}$ is $x \geq 0$. What is the domain of $y = \sqrt{x - 3} + 2$?

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A ball is thrown and its height follows a downward parabola with maximum height 20 m at $t = 3$ s. Write the vertex form of the equation.

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Revisit your thinking

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 $y = x^2$:

  • Vertex: The basic vertex $(0, 0)$ moves to $(3, 4)$ because of the right 3 and up 4.
  • Reflection: The negative sign means the parabola opens downward.
  • Dilation: The factor of 2 makes it narrower (steeper) than $y = x^2$.
  • Intercepts: Set $y = 0$ to find $x$-intercepts at $x = 3 \pm \sqrt{2}$. Set $x = 0$ to find the $y$-intercept at $(0, -14)$.

With just the vertex, intercepts, and the general shape, you have enough to sketch the graph accurately. No table of values needed.

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Multiple choice

Answer the drill bank and rate your confidence.

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Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

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Short answer

Write full responses, then check them against the model answers.

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Short answer
ApplyBand 43 marks

Q8. 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)

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ApplyBand 44 marks

Q9. 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)

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EvaluateBand 54 marks

Q10. 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)

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📖 Comprehensive answers (click to reveal)

Multiple choice, drill bank

MC answers and feedback are shown inline as you complete each question. Use the retry button to attempt a fresh set.

Drill bank answers:

Order of combined transformations. Right 1: $(3, 3)$; $\times 2$: $(3, 6)$; up 4: $(3, 10)$.

Which transformations move a point horizontally. Reflection and vertical translation do not move points horizontally.

Factorised horizontal form. $f(2(x + 2))$: $x$-coordinate $\frac{1}{2} - 2 = -1.5$; $y$ stays $-2$.

Modelling a trajectory. Downward parabola with vertex $(3, 4)$ matches the basketball trajectory.

Reflection and $x$-intercepts. Reflection in $y$-axis negates $x$-intercepts.

Activity 1, Track the features model answers

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 $x = -3$.

Short answer model answers

Q8 (3 marks): Turning point: $(3, -2)$, $x_{\text{new}} = 2 + 1 = 3$; $y_{\text{new}} = 2(-3) + 4 = -2$ [1]. $x$-intercepts: horizontal translation right 1 shifts intercepts: $-1 \to 0$ and $5 \to 6$. So $x$-intercepts at $x = 0$ and $x = 6$ [2]. Note: vertical transformations preserve $x$-coordinates where $f(x) = 0$.

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 $-5(t - 2)^2 + 20 = 0 \Rightarrow (t - 2)^2 = 4 \Rightarrow t = 4$ 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 [0.5].

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