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

Quadratic functions and graphs — Paper 2 Challenge MC

Eighteen harder multiple-choice questions. In most of them one option is the answer a student gets by doing something reasonable and slightly wrong — a vertex outside the given interval, a transformation applied in the wrong order. Check the second thing before you commit. Detailed solutions can be requested by email.

18 questions · about 35 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 — the vertex lying outside the given interval
Question 2 — a minimum value of zero
Question 3 — the vertex of y = f(2x)
Question 4 — a vertical stretch combined with a translation
Question 5 — reflection in the line x = 2
Question 6 — counting integers satisfying a discriminant condition
Question 7 — signs of products of coefficients read from a sketch
Question 8 — an 'above for all x' condition on the difference of two functions
Question 9 — locating the vertex in a named quadrant
Question 10 — eliminating the parameter from the vertex coordinates
Question 11 — a symmetry condition given as a functional equation
Question 12 — what a positive-for-all-x condition guarantees
Question 13 — transformations do not commute
Question 14 — distance between the x-intercepts
Question 15 — tangency between a line and a parabola
Question 16 — describing the set of possible output values
Question 17 — a vertex constrained to lie on a given line
Question 18 — the number of common points of two quadratic graphs

Question 1 · New sub-type — an extreme value on a restricted interval

The least value of

x^{2} - 6 x + 11

, where

1 \leq x \leq 2

, is

$2$ .
$3$ .
$6$ .
$11$ .

Show answer

B. The vertex is at $x = 3$ , outside the interval, so $2$ is the right number to the wrong question. On $1 \leq x \leq 2$ the function decreases, so the least value is at $x = 2$ .

Question 2 · Modelled on 2014 P2 Q35, with the condition stated instead of the method

It is given that the least value of

x^{2} + k x + k

0

. The greatest possible value of

k

$0$ .
$2$ .
$4$ .
$8$ .

Show answer

C. The least value is $k - \frac{k ^{2}}{4}$ , so $k^{2} = 4 k$ and $k = 0$ or $4$ . Both work, and the question asks for the greater.

Question 3 · New sub-type — a horizontal scaling rather than a translation

Let

f (x) = (x - 6)^{2} + 1

. The coordinates of the vertex of the graph of

y = f (2 x)

are

$(3, 1)$ .
$(12, 1)$ .
$(3, 2)$ .
$(6, 2)$ .

Show answer

A. $f (2 x) = (2 x - 6)^{2} + 1 = 4 (x - 3)^{2} + 1$ . Halving the $x$ -coordinate is correct; doubling it gives option B.

Question 4 · Modelled on 2024 P1 Q19(b)

The vertex of the graph of

y = f (x)

(a, b)

, where

f

is a quadratic function. The coordinates of the vertex of the graph of

y = 2 f (x) - 3

are

$(2 a - 3, b)$ .
$(2 a, 2 b - 3)$ .
$(2 a - 3, 2 b - 3)$ .
$(a, 2 b - 3)$ .

Show answer

D. Multiplying the output and subtracting $3$ touch $y$ only. The $x$ -coordinate of the turning point cannot move if $x$ itself is untouched.

Question 5 · New sub-type — reflection in a vertical line

The graph of

y = x^{2} - 10 x + 28

is reflected with respect to the straight line

x = 2

. The coordinates of the vertex of the resulting graph are

$(- 5, 3)$ .
$(- 1, - 3)$ .
$(5, - 3)$ .
$(- 1, 3)$ .

Show answer

D. The vertex is $(5, 3)$ , and reflecting in $x = 2$ sends $5$ to $2 - (5 - 2) = - 1$ . Reflecting in the $y$ -axis instead would give $- 5$ .

Question 6 · Modelled on 2017 P1 Q18, then counted

The number of integers

k

for which the graph of

y = x^{2} + k x + 9

has no

x

-intercept is

$5$ .
$10$ .
$11$ .
infinitely many.

Show answer

C. $Δ = k^{2} - 36 < 0$ gives $- 6 < k < 6$ , so $k$ runs from $- 5$ to $5$ . That is $11$ integers, since $0$ counts.

Question 7 · Modelled on 2012 P2 Q38

The figure shows the graph of

y = a x^{2} + b x + c

. Which of the following must be true?
I.

ab > 0

II.

a c < 0

III.

b c < 0

I only
II only
I and II only
II and III only

Show answer

B. The graph opens upwards, so $a > 0$ ; the axis of symmetry is to the right of the $y$ -axis, so $- \frac{b}{2 a} > 0$ and $b < 0$ ; the graph cuts the $y$ -axis below the origin, so $c < 0$ . That makes $ab < 0$ and $b c > 0$ , leaving II alone.

Question 8 · New sub-type — one graph lying above a line for every x

If the graph of

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

lies above the straight line

y = 2 x + 1

for all real values of

x

, then

$k > 1$ .
$k > 3$ .
$k > 5$ .
$k > 9$ .

Show answer

C. Subtract first: $x^{2} + 4 x + (k - 1) > 0$ for all $x$ , so $16 - 4 (k - 1) < 0$ . Comparing the two constant terms alone gives $k > 1$ , which is the trap.

Question 9 · Modelled on 2020 P1 Q17

The vertex of the graph of

y = x^{2} - 2 k x + 3 k

lies in the fourth quadrant. The range of values of

k

$k < 0$ .
$k > 0$ .
$0 < k < 3$ .
$k > 3$ .

Show answer

D. The vertex is $(k, 3 k - k^{2})$ . The fourth quadrant needs $k > 0$ and $3 k - k^{2} < 0$ , and both hold only when $k > 3$ .

Question 10 · New sub-type — a locus of vertices as a constant varies

As the real constant

t

varies, the vertex of the graph of

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

always lies on

the straight line $y = x$ .
the straight line $y = 2 x$ .
the parabola $y = x^{2}$ .
the parabola $y = - x^{2}$ .

Show answer

A. The vertex is $(t, t)$ , so its two coordinates are always equal.

Question 11 · Modelled on 2023 P2 Q35

A quadratic function

f

satisfies

f (2 + x) = f (2 - x)

for all real

x

, and

f (0) = 0

. Then

f (4) =

$- 4$ .
$0$ .
$4$ .
It cannot be determined.

Show answer

B. The condition says the graph is symmetric about $x = 2$ , and $0$ and $4$ are mirror images of each other, so $f (4) = f (0)$ . No coefficient is needed.

Question 12 · Modelled on 2017 P1 Q18(b), inverted

It is given that

x^{2} + p x + q > 0

for all real values of

x

. Which of the following must be true?

$p > 0$
$p > q$
$p + q > 0$
$q > 0$

Show answer

D. Putting $x = 0$ gives $q > 0$ at once. The sign of $p$ is free: both $x^{2} + 4 x + 5$ and $x^{2} - 4 x + 5$ are always positive.

Question 13 · New sub-type — the order of two transformations

The graph of

y = f (x)

, where

f (x) = (x - 1)^{2}

, is first translated rightwards by

3

units and the resulting graph is then reflected with respect to the

y

-axis. The coordinates of the vertex of the final graph are

$(4, 0)$ .
$(2, 0)$ .
$(- 2, 0)$ .
$(- 4, 0)$ .

Show answer

D. The vertex moves $(1, 0) \to (4, 0) \to (- 4, 0)$ . Reflecting first and translating afterwards would give $(2, 0)$ instead, which is why the order has to be read carefully.

Question 14 · Modelled on 2015 P1 Q18

The graph of

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

cuts the

x

-axis at two points that are

4

units apart. Then

c =

$1$ .
$3$ .
$5$ .
$9$ .

Show answer

C. The roots are $3 \pm 9 - c$ , so $2 9 - c = 4$ and $9 - c = 4$ .

Question 15 · Modelled on 2021 P1 Q19(c)

The straight line

y = 4 x + k

is a tangent to the graph of

y = x^{2} + 6 x + 13

. Then

k =

$12$ .
$13$ .
$14$ .
$15$ .

Show answer

A. $x^{2} + 2 x + (13 - k) = 0$ must have a double root, so $4 - 4 (13 - k) = 0$ .

Question 16 · New sub-type — the range of a quadratic function

Let

f (x) = - 2 x^{2} + 4 x + 1

, where

x

is any real number. The range of values of

f (x)

$f (x) \geq 1$ .
$f (x) \leq 1$ .
$f (x) \leq 3$ .
$f (x) \geq 3$ .

Show answer

C. $f (x) = - 2 (x - 1)^{2} + 3$ . The graph opens downwards, so $3$ is a ceiling, not a floor; $1$ is merely the $y$ -intercept.

Question 17 · Modelled on 2026 P1 Q19, condensed

The vertex of the graph of

y = x^{2} - 4 k x + 6 k^{2} + 4 k

lies on the straight line

y = 3 x

, where

k

is a positive constant. Then

k =

$1$ .
$2$ .
$3$ .
$4$ .

Show answer

A. Completing the square gives $(x - 2 k)^{2} + 2 k^{2} + 4 k$ , so the vertex is $(2 k, 2 k^{2} + 4 k)$ . The condition $2 k^{2} + 4 k = 3 (2 k)$ gives $k = 0$ or $k = 1$ , and $k = 0$ is rejected.

Question 18 · New sub-type — two parabolas meeting exactly once

The graphs of

y = x^{2} + 2 x + 5

and

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

have exactly one point in common. Then

k =

$1$ .
$3$ .
$5$ .
$7$ .

Show answer

B. Setting them equal gives $2 x^{2} - 4 x + (5 - k) = 0$ , and a double root needs $16 - 8 (5 - k) = 0$ .

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

Where this sits in the syllabus

Nothing here goes outside the Compulsory Part syllabus. The difficulty is in the reading: a restricted interval, an order of transformations, or a condition stated about a line rather than about the graph.

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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