Functions — Worked Solutions (Preliminary Maths Advanced)
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Worked examples for Preliminary Maths Advanced functions. 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 Preliminary Maths Advanced — Functions. Every question is worked through with the method and reasoning shown, so you can check how to get 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.
Example 1 — Domain and range
Question
Find the domain and range of the function $f(x) = \sqrt{x - 2} + 5$.
Solution
A square root only exists when what's inside is non-negative, so set $x - 2 \ge 0$.
That gives $x \ge 2$, so the domain is $x \ge 2$ (or $[2, \infty)$).
For the range, $\sqrt{x-2} \ge 0$ for every allowed $x$, with its smallest value $0$ at $x = 2$. Adding $5$ shifts everything up, so $f(x) \ge 5$.
Range: $y \ge 5$ (or $[5, \infty)$). Don't forget the $+5$ when stating the range — that's the mark most people drop.
Let's think about what could break this function. The only restriction is the square root: we can't take the root of a negative number, so we need $x - 2 \ge 0$.
Solving that, $x \ge 2$, so the domain is all real numbers from $2$ upwards, written $[2, \infty)$.
Now the range. The square root part, $\sqrt{x-2}$, is always zero or positive — its minimum is $0$, which happens right at $x = 2$. Because we then add $5$, the smallest output is $0 + 5 = 5$, and the function climbs from there.
So the range is $y \ge 5$, or $[5, \infty)$. The trick is to track what the square root can produce first, then apply the vertical shift.
Domain: square root needs a non-negative argument.
- $x - 2 \ge 0 \Rightarrow x \ge 2$
- Domain: $[2, \infty)$
Range: minimum of $\sqrt{x-2}$ is $0$, then shift up by $5$.
- $\sqrt{x-2} \ge 0$
- $f(x) \ge 5$
- Range: $[5, \infty)$
Where the marks go
- 1 mark: Sets the radicand non-negative: $x - 2 \ge 0$
- 1 mark: Correct domain $x \ge 2$
- 1 mark: Correct range $y \ge 5$
Key idea
For a square-root function the domain comes from making the inside $\ge 0$; the range comes from the root's minimum of $0$ adjusted by any vertical shift.
Example 2 — Features of a quadratic
Question
For the parabola $y = x^2 - 6x + 5$, find the coordinates of the vertex and the $x$-intercepts.
Solution
Complete the square to get the vertex fast: $y = x^2 - 6x + 5 = (x-3)^2 - 9 + 5 = (x-3)^2 - 4$.
So the vertex is $(3, -4)$.
For the $x$-intercepts set $y = 0$: $x^2 - 6x + 5 = (x-1)(x-5) = 0$, giving $x = 1$ and $x = 5$.
Intercepts $(1, 0)$ and $(5, 0)$. Completing the square hands you the vertex with no extra work — use it instead of memorising $x = -b/2a$.
Two things to find here, and each has a clean route. Start with the vertex by completing the square, because that rewrites the equation in a form where the turning point is obvious.
Take half of the $-6$ coefficient (that's $-3$) and square it: $y = (x-3)^2 - 9 + 5 = (x-3)^2 - 4$. In this form $(x-3)^2$ is smallest (zero) when $x = 3$, and then $y = -4$, so the vertex is $(3, -4)$.
For the $x$-intercepts we set $y = 0$, because intercepts on the $x$-axis are where the height is zero. Factoring, $x^2 - 6x + 5 = (x-1)(x-5)$, so $x = 1$ or $x = 5$.
The intercepts are $(1, 0)$ and $(5, 0)$. Notice the vertex sits exactly halfway between them at $x = 3$ — a handy check.
Vertex by completing the square.
- $y = (x-3)^2 - 9 + 5 = (x-3)^2 - 4$
- Vertex: $(3, -4)$
$x$-intercepts: set $y = 0$.
- $(x-1)(x-5) = 0$
- $x = 1,\ 5$
- Intercepts: $(1, 0)$, $(5, 0)$
Where the marks go
- 1 mark: Completes the square to $(x-3)^2 - 4$
- 1 mark: Correct vertex $(3, -4)$
- 1 mark: Factorises and solves $y = 0$
- 1 mark: Correct $x$-intercepts $(1, 0)$ and $(5, 0)$
Key idea
Completing the square gives the vertex directly; setting $y = 0$ and factorising gives the $x$-intercepts, which straddle the vertex symmetrically.
Frequently asked questions
Step-by-step solutions to Functions questions in Preliminary 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, rather than memorising answers.
Yes — they follow the NESA syllabus for Preliminary Maths Advanced, using the methods and notation expected in exams and assessments.