Why is the extreme value always at the vertex?

After completing the square the squared term cannot pass zero, so the extreme value sits at the vertex.

How do I set up an applied question?

Let the changing quantity be x, write the quadratic from the condition, then complete the square.

S4 Quadratic Functions, Paper 2: completing the square and applications

2026-11-02

Tutor Prep Guide · Primary English

Vertex form exposes the maximum or minimum at once and answers applied questions directly.

Focus　Fix the opening direction, complete the square for the vertex, then use that vertex for greatest area or revenue.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10
Suggested time	25
Difficulty mix	Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1
Chapter	Chapter: S4 Maths: Quadratic Functions and Graphs
Level	S4

02　Questions (10 items)

FoundationState the vertex, axis of symmetry and minimum of y = (x + 3)² − 5.
FoundationState the vertex and minimum of y = 3(x − 1)² + 2.
FoundationFind the y-intercept and the x-intercepts of y = x² − 8x + 12.
FoundationFor y = −(x − 4)² + 9, find the maximum value and the x-intercepts.
IntermediateComplete the square for y = x² + 10x + 18 and state the vertex.
IntermediateComplete the square for y = 2x² − 12x + 13 and state the vertex.
IntermediateHow many times does y = x² − 2x + 5 meet the x-axis? Give the vertex too.
ChallengeA rectangle has perimeter 24 m. Find the greatest possible area.
ChallengeAn item sells at $(60 − x) with (200 + 10x) units sold. Find the price that maximises revenue.
Error redoError: a student says the minimum of y = x² + 4x + 1 is 1. What is correct?

03　Answers and one-line notes

#	Answer	One-line note
1	Vertex (−3, −5), axis x = −3, minimum −5	The bracket carries a plus sign, so the vertex sits at minus 3.
2	Vertex (1, 2), minimum 2	The coefficient 3 is positive so the curve opens upwards and the least value is 2.
3	y-intercept 12, x-intercepts 2 and 6	Putting x to zero gives 12, and factorising gives (x minus 2)(x minus 6).
4	Maximum 9 at x = 4, x-intercepts 1 and 7	The minus sign makes 9 the largest value, and solving the square equal to 9 gives 1 and 7.
5	y = (x + 5)² − 7, vertex (−5, −7)	Adding and subtracting 25 turns the first two terms into a perfect square, leaving minus 7.
6	y = 2(x − 3)² − 5, vertex (3, −5)	Take out 2 first to get 2(x squared minus 6x) plus 13, then add and subtract 9 inside.
7	No crossings, vertex (1, 4)	The discriminant is 4 minus 20, which is negative, and completing the square gives (x minus 1) squared plus 4.
8	36 m², as a 6 m by 6 m square	Let the width be x so the length is 12 minus x; the area peaks at 36 when x is 6.
9	$40, revenue $16 000	Revenue is 16 000 minus 10(x minus 20) squared, which peaks when x is 20.
10	Minimum −3 at x = −2	The constant term is not the minimum; completing the square gives (x plus 2) squared minus 3.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

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