Why do maximum and minimum problems use quadratics?

Quantities such as area or profit often take the form x(times a constant minus x), and completing the square reveals the extreme value.

What is the tangency condition?

After solving the two equations together, a zero discriminant means exactly one shared point.

How is an inverse function found?

Rearrange y = f(x) to make x the subject, then swap x and y.

S4 Maths functions and graphs paper 3: optimisation, tangency and inverses

2026-11-10

Tutor Prep Guide · S4 Maths Functions and Graphs

Ten applied items: the largest area for a fixed perimeter, a parameter from a tangency condition, the order of transformations, and evaluating an inverse function.

Direct answers　Items 01 to 03 cover an area maximum, a minimum and x-intercepts, 04 to 06 tangency for a parameter, an inverse value and the order of transformations, 07 to 09 a perimeter maximum, a completed-square minimum and a composite domain, and 10 a sign error in the vertex formula.

01　Paper details (check answers afterwards)

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

02　Questions (10 items)

FoundationA rectangle has perimeter 20 cm and one side x cm. Express the area A in x and find its maximum.
FoundationFind the minimum of y = 2x^2 – 8x + 3 and the x where it occurs.
FoundationFind the x-intercepts of y = -(x – 1)^2 + 9.
IntermediateThe line y = 3x + k touches y = x^2 + 2x + 5. Find k.
IntermediateGiven f(x) = 3x – 4, find f-inverse of 8.
IntermediateDescribe the steps taking y = x^2 to y = -2(x – 1)^2 + 6.
ChallengeWhat is the largest area of a rectangle with perimeter 36 cm?
ChallengeFind the minimum value of y = x^2 – 6x + 13.
ChallengeGiven f(x) = sqrt(x) and g(x) = 5 – x, state the domain of f(g(x)).
Error redoError drill: a student thinks the vertex x-coordinate of y = x^2 + bx + c is b / 2, so for y = x^2 – 6x + 13 he answered -3. What is correct?

03　Answers and one-line notes

#	Answer	One-line note
1	A = x(10 – x), maximum 25 cm^2	The other side is 10 – x; completing the square gives 25 – (x – 5)^2, largest when x = 5.
2	Minimum -5 at x = 2	Completing the square gives 2(x – 2)^2 – 5, least when the squared term vanishes.
3	4 and -2	Setting y = 0 gives (x – 1)^2 = 9, so x = 4 or -2.
4	k = 19/4 = 4.75	Subtracting gives x^2 – x + (5 – k) = 0; setting 1 – 4(5 – k) = 0 yields k = 19/4.
5	4	Rearranging y = 3x – 4 gives x = (y + 4) / 3, and y = 8 gives 4.
6	Reflect in the x-axis, stretch vertically by 2, shift right 1, shift up 6	The minus sign turns the curve downward, the factor 2 stretches it vertically, the -1 inside shifts it right and the +6 outside shifts it up.
7	81 cm^2 (a 9 cm by 9 cm square)	With one side x the area is x(18 – x) = 81 – (x – 9)^2, largest at x = 9.
8	4	Completing the square gives (x – 3)^2 + 4, and the squared term is never negative.
9	x <= 5	The root needs 5 – x >= 0, so x <= 5.
10	x = 3	The formula is x = -b / (2a); here a = 1 and b = -6, giving x = 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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