Worked Solutions
Functions & Graphing — Worked Solutions (HSC Maths Advanced)
Created with Intu AI Reviewed by Intuition's expert tutors
Studying this? See our HSC Maths Advanced course →
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.
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.