Worked Solutions
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.
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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.