Functions & Graphing — Worked Solutions (HSC Maths Advanced)
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Worked examples for HSC Maths Advanced graphing techniques. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for HSC Maths Advanced — Functions & Graphing. Every question is worked through with the method and reasoning shown, in NESA exam style, so you can check how to reach the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
These examples focus on graphing techniques — describing transformations of a base function and sketching graphs built from simpler ones.
Example 1 — Transformations of a function
Question
The graph of $y = f(x)$ has a single maximum turning point at $(2, 5)$.
Describe the sequence of transformations that maps $y = f(x)$ onto $y = -2f(x - 1) + 3$, and state the coordinates of the turning point on the transformed graph.
Solution
Read the changes off the equation in the right order: inside the function affects $x$, outside affects $y$.
$f(x-1)$ shifts the graph right 1. The factor $-2$ does two things to the $y$-values: vertical stretch by factor 2 and reflection in the $x$-axis. The $+3$ shifts up 3.
Track the point $(2, 5)$. Shift right 1: $x$ becomes $3$. Apply $-2$ to the $y$-value: $5 \to -10$. Add 3: $-10 \to -7$.
New turning point: $(3, -7)$. A reflected maximum becomes a minimum, so check your nature claim if the next part asks for it.
Transformations come from how the equation is built around $f$. Anything inside the brackets acts on the $x$-values (and works in reverse to what you'd expect); anything outside acts on the $y$-values.
Working from the inside out: $f(x - 1)$ means we replace $x$ with $x - 1$, which slides the whole graph 1 unit right. The $-2$ multiplying the function stretches every $y$-value to twice its size (vertical stretch, factor 2) and the minus sign flips it over the $x$-axis (reflection in the $x$-axis). Finally $+3$ lifts the graph up 3 units.
Now follow the turning point $(2, 5)$ through each step. Right 1 takes the $x$ from $2$ to $3$. Multiplying the $y$-value by $-2$ takes $5$ to $-10$. Adding 3 takes $-10$ to $-7$.
So the turning point lands at $(3, -7)$. It helps to transform the key point rather than the whole curve — one point carries all the information you need.
Decompose $y = -2f(x-1) + 3$:
- $x - 1$ inside → shift right 1
- $\times (-2)$ → vertical stretch factor 2, reflect in $x$-axis
- $+3$ → shift up 3
Map the point $(2, 5)$:
- $x: 2 \to 3$
- $y: 5 \to 5 \times (-2) = -10 \to -10 + 3 = -7$
Turning point: $(3, -7)$.
Where the marks go
- 1 mark: Identifies the horizontal shift right 1 from $f(x-1)$
- 1 mark: Identifies the vertical stretch factor 2 and reflection in the $x$-axis from $-2$, and the shift up 3
- 1 mark: Correct transformed turning point $(3, -7)$
Key idea
Changes inside the function act on $x$ (horizontally, in reverse); changes outside act on $y$ — apply them to a known point in order.
Example 2 — Reciprocal and absolute-value graphs
Question
Consider the function $f(x) = \dfrac{1}{x - 2} + 1$.
State the equations of the asymptotes, find the $x$- and $y$-intercepts, and hence describe the shape of the graph.
Solution
This is the reciprocal $\frac{1}{x}$ shifted right 2 and up 1, so read the asymptotes straight off.
Vertical asymptote where the denominator is zero: $x - 2 = 0$, so $x = 2$. Horizontal asymptote is the constant added on: $y = 1$.
$y$-intercept ($x = 0$): $f(0) = \frac{1}{-2} + 1 = \frac{1}{2}$, so $\left(0, \tfrac{1}{2}\right)$.
$x$-intercept ($y = 0$): $\frac{1}{x-2} = -1 \Rightarrow x - 2 = -1 \Rightarrow x = 1$, so $(1, 0)$.
It's a standard hyperbola with two branches about the asymptotes $x = 2$ and $y = 1$ — top-right and bottom-left branches. Don't forget the horizontal asymptote moves with the $+1$, not just the vertical one.
Think of $f(x) = \frac{1}{x-2} + 1$ as the familiar curve $y = \frac{1}{x}$ that's been shifted. The $-2$ inside moves it right 2, and the $+1$ outside moves it up 1.
Asymptotes are the lines the curve never crosses. The vertical one sits where we'd divide by zero: $x - 2 = 0$, so $x = 2$. The horizontal one is the value $\frac{1}{x-2}$ approaches as $x$ gets very large — that fraction shrinks to $0$, leaving $y = 1$.
For the $y$-intercept set $x = 0$: $f(0) = \frac{1}{0-2} + 1 = -\frac{1}{2} + 1 = \frac{1}{2}$, giving $\left(0, \tfrac{1}{2}\right)$. For the $x$-intercept set $y = 0$: $0 = \frac{1}{x-2} + 1$, so $\frac{1}{x-2} = -1$, meaning $x - 2 = -1$ and $x = 1$, giving $(1, 0)$.
Putting it together, the graph is a hyperbola in two pieces, hugging the lines $x = 2$ and $y = 1$: one branch in the upper-right region and one in the lower-left. Knowing the asymptotes and a couple of intercepts is enough to place both branches confidently.
$f(x) = \frac{1}{x-2} + 1$ is $\frac{1}{x}$ shifted right 2, up 1.
Asymptotes:
- Vertical: $x - 2 = 0 \Rightarrow x = 2$
- Horizontal: $y = 1$
Intercepts:
- $y$-int ($x=0$): $\frac{1}{-2} + 1 = \frac{1}{2} \Rightarrow \left(0, \tfrac{1}{2}\right)$
- $x$-int ($y=0$): $\frac{1}{x-2} = -1 \Rightarrow x = 1 \Rightarrow (1, 0)$
Shape: rectangular hyperbola, two branches about $x = 2$ and $y = 1$ (upper-right and lower-left).
Where the marks go
- 1 mark: Correct vertical asymptote $x = 2$
- 1 mark: Correct horizontal asymptote $y = 1$
- 1 mark: Correct intercepts $\left(0, \tfrac{1}{2}\right)$ and $(1, 0)$
- 1 mark: Correct description of the two-branch hyperbola relative to its asymptotes
Key idea
A reciprocal graph $\frac{1}{x-h} + k$ has asymptotes $x = h$ and $y = k$; intercepts then fix the two branches.
Frequently asked questions
Step-by-step solutions to exam-style questions on Functions & Graphing in HSC Maths Advanced, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions in the exam, rather than memorising answers.
Yes — they follow the NESA HSC Maths Advanced syllabus for Functions & Graphing, using the methods and notation expected in the exam.