Practice › Number and Algebra › Functions and Graphs of Quadratic Functions

Quadratic functions and graphs — Paper 1 Challenge

Ten harder questions. Each one states a condition rather than naming a method, so the first job is deciding what the condition means. Work the standard Paper 1 set first — this tier assumes completing the square is already automatic. Detailed solutions can be requested by email.

10 questions · about 90 minutes · answers below each question

All questions are original, written for Math and AI Academy and modelled on past-paper style. They are not reproductions of HKEAA examination questions.

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Question 1 — least and greatest value on a closed interval when the vertex lies outside it
Question 2 — translating 'lies entirely above' into a discriminant condition, then counting integers
Question 3 — treating the vertex coordinates as a parametric pair and eliminating the parameter
Question 4 — recovering a quadratic from a symmetry condition and one value
Question 5 — relating the distance between the x-intercepts to the vertex
Question 6 — reflection in the line y = 3 as the mapping y ↦ 6 − y
Question 7 — disproving a universal claim about x-intercepts with a counter-example
Question 8 — two parabolas meeting at exactly one point
Question 9 — an 'above for all x' condition reduced to a discriminant on the difference
Question 10 — locating a vertex from a stated condition; a translation described by its effect; testing an equilateral claim

Question 1 · 5 marks · New sub-type — the standard set never asks for a value on a restricted interval

Let

f (x) = x^{2} - 6 x + 11

, where

0 \leq x \leq 2

(a) Find the least value of $f (x)$ . (3 marks)

(b) Find the greatest value of $f (x)$ . (2 marks)

Show answer

(a) $3$ , at $x = 2$ . (b) $11$ , at $x = 0$ .
The vertex is at $x = 3$ , which is outside the interval, so $2$ is not the answer to (a). On $0 \leq x \leq 2$ the function is decreasing, so both extreme values sit at the endpoints.

Question 2 · 5 marks · Modelled on 2017 P1 Q18, condition stated rather than method named

The graph of

y = x^{2} - 2 a x + a + 6

has no

x

-intercept, where

a

is a constant.

(a) Find the range of values of $a$ . (3 marks)

(b) How many integers satisfy the condition in (a)? (2 marks)

Show answer

(a) $- 2 < a < 3$ . (b) $4$ integers: $- 1, 0, 1, 2$ .
$Δ < 0$ gives $a^{2} - a - 6 < 0$ , i.e. $(a - 3) (a + 2) < 0$ . The endpoints are excluded, so $- 2$ and $3$ are not counted.

Question 3 · 6 marks · New sub-type — a locus of vertices, in the spirit of the fixed-point questions in recent Paper 1 Section B

Let

f (x) = x^{2} + p x + 5

, where

p

is a real constant. As

p

varies, the vertex of the graph of

y = f (x)

traces a curve

C

(a) Express the coordinates of the vertex in terms of $p$ . (2 marks)

(b) Prove that $C$ is the graph of $y = 5 - x^{2}$ . (4 marks)

Show answer

(a) $(- \frac{p}{2}, 5 - \frac{p ^{2}}{4})$ . (b) Writing $X = - \frac{p}{2}$ gives $\frac{p ^{2}}{4} = X^{2}$ , so $Y = 5 - X^{2}$ .
Every point of $y = 5 - x^{2}$ is also reached, since $p = - 2 X$ is defined for every real $X$ — worth saying, or the proof is only half done.

Question 4 · 6 marks · Modelled on 2014 P2 Q35, run backwards

Let

f

be a quadratic function such that

f (1) = f (5)

f (3) = - 8

and

f (0) = 10

(a) Write down the equation of the axis of symmetry of the graph of $y = f (x)$ . (2 marks)

(b) Express $f (x)$ in the form $a x^{2} + b x + c$ . (4 marks)

Show answer

(a) $x = 3$ . (b) $f (x) = 2 x^{2} - 12 x + 10$ .
$f (1) = f (5)$ says the two points are mirror images, so the axis is the midpoint $x = 3$ ; then $f (3) = - 8$ is the vertex value and $f (x) = a (x - 3)^{2} - 8$ .

Question 5 · 6 marks · New sub-type — the gap between the roots, expressed through the completed square

The graph of

y = x^{2} - 2 k x + k^{2} + k - 2

cuts the

x

-axis at two distinct points

A

and

B

, where

k

is a constant and

A B = 4

(a) Find $k$ . (4 marks)

(b) Find the coordinates of $A$ and $B$ . (2 marks)

Show answer

(a) $k = - 2$ . (b) $(- 4, 0)$ and $(0, 0)$ .
Completing the square gives $(x - k)^{2} + (k - 2)$ , so the roots are $k \pm 2 - k$ and $A B = 2 2 - k$ .

Question 6 · 6 marks · New sub-type — reflection in a horizontal line, which the standard set never touches

Let

f (x) = - x^{2} + 6 x + 7

. The graph of

y = g (x)

is obtained by reflecting the graph of

y = f (x)

with respect to the straight line

y = 3

(a) Express $g (x)$ in the form $a x^{2} + b x + c$ . (4 marks)

(b) Find the coordinates of the vertex of the graph of $y = g (x)$ . (2 marks)

Show answer

(a) $g (x) = x^{2} - 6 x - 1$ . (b) $(3, - 10)$ .
A point $(x, y)$ reflects to $(x, 6 - y)$ , so $g (x) = 6 - f (x)$ . The vertex $(3, 16)$ maps to $(3, - 10)$ , and $16$ and $- 10$ are indeed equidistant from $3$ .

Question 7 · 7 marks · Modelled on 2017 P1 Q18(b)(ii), as a claim to test

Let

f (x) = x^{2} + 2 (k - 1) x + k^{2} - 3

, where

k

is a real constant. Someone claims that however

k

is chosen, the graph of

y = f (x)

has at least one

x

-intercept.

(a) Find the range of values of $k$ for which the graph of $y = f (x)$ has at least one $x$ -intercept. (4 marks)

(b) Is the claim correct? Explain your answer. (3 marks)

Show answer

(a) $k \leq 2$ . (b) The claim is not correct.
$Δ = 4 (k - 1)^{2} - 4 (k^{2} - 3) = - 8 k + 16$ , so $Δ \geq 0$ only when $k \leq 2$ . Taking $k = 3$ gives $f (x) = x^{2} + 4 x + 6$ , whose graph lies entirely above the $x$ -axis.

Question 8 · 8 marks · Modelled on 2021 P1 Q19(c), with the tangency condition unnamed

Let

f (x) = x^{2} + 4 x + 7

and

g (x) = - x^{2} + 8 x + k

, where

k

is a constant. The graphs of

y = f (x)

and

y = g (x)

have exactly one point in common.

(a) Find $k$ . (5 marks)

(b) Find the coordinates of that common point. (3 marks)

Show answer

(a) $k = 5$ . (b) $(1, 12)$ .
Setting $f (x) = g (x)$ gives $2 x^{2} - 4 x + 7 - k = 0$ ; exactly one common point means this has a double root, so $Δ = 16 - 8 (7 - k) = 0$ .

Question 9 · 9 marks · New sub-type — one graph lying above another for every x

It is given that the graph of

y = x^{2} + 6 x + k

lies above the straight line

y = 2 x + 1

for every real value of

x

, where

k

is a constant.

(a) Find the range of values of $k$ . (5 marks)

(b) When $k$ takes its least integral value, find the shortest vertical distance between the graph and the straight line. (4 marks)

Show answer

(a) $k > 5$ . (b) $k = 6$ , shortest vertical distance $1$ .
The condition is that $x^{2} + 4 x + (k - 1) > 0$ for all $x$ , so $Δ < 0$ . In (b) that difference is $(x + 2)^{2} + 1$ , whose least value is $1$ at $x = - 2$ .

Question 10 · 12 marks · Modelled on 2026 P1 Q19, with every sub-step unscaffolded

Let

f (x) = x^{2} - 2 k x + 5 k^{2} - 6 k

, where

k

is a positive constant. The vertex of the graph of

y = f (x)

V

, and

V

lies on the straight line

y = 2 x

. Denote the origin by

O

(a) Find $k$ and the coordinates of $V$ . (5 marks)

(b) The graph of $y = f (x)$ is translated so that its vertex becomes the point $W (- 2, 4)$ . Express the new function in the form $a x^{2} + b x + c$ . (3 marks)

(c) Is $△ O V W$ an equilateral triangle? Explain your answer. (4 marks)

Show answer

(a) $k = 2$ , $V = (2, 4)$ . (b) $y = x^{2} + 4 x + 8$ . (c) No.
$V = (k, 4 k^{2} - 6 k)$ , and $4 k^{2} - 6 k = 2 k$ gives $k = 2$ once $k = 0$ is rejected. In (c), $V W = 4$ but $O V = O W = 20$ , so the triangle is isosceles, not equilateral.

Continue practising this topic: Try the Paper 1 long questions for harder variants, or move to Paper 2 multiple-choice — also available as Paper 2 Challenge MC.

Where this sits in the syllabus

The techniques here stay inside the Compulsory Part syllabus: discriminants, completing the square and coordinate geometry. What changes is the scaffolding. A question that says the vertex lies on a given line has not told you to complete the square — you have to see that for yourself.

Full coverage of this topic, and the rest of the course, is on the Mathematics page. More sets are listed on the practice index.

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