What is A11: Quadratic graphs?

A root of a quadratic is an x-coordinate where its graph meets the x-axis, so y=0. The y-intercept is found by setting x=0.

How many marks are A11 questions worth?

In verified past-paper appearances, A11 questions on Edexcel GCSE Maths 1MA1 have been worth 2–7 marks.

Which past exam papers has A11 appeared in?

A11 has appeared in 15 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

A11

Identify and interpret roots, intercepts, turning points of quadratic functions graphically; deduce roots algebraically and turning points by completing the square

Quadratic graphs

Worked answers, methods and verified real exam appearances for A11 on Edexcel GCSE Maths 1MA1.

Explanation

A root of a quadratic is an $x$ -coordinate where its graph meets the $x$ -axis, so $y=0$ . The $y$ -intercept is found by setting $x=0$ .
A turning point is the maximum or minimum point, and its vertical line is the axis of symmetry.
Factorising can reveal roots algebraically and the graph confirms their positions.
Higher tier: completing the square into $a(x-h)^2+k$ reveals turning point $(h,k)$ .
Examiners expect coordinates for points, equations for axes, and algebraic working when roots are to be deduced.

A quadratic's roots are its

x

-axis intersections, and its turning point lies on the axis of symmetry.

Worked example

Find the roots and $y$ -intercept of $y=x^2-5x+6$ .

1.Factorise: $x^2-5x+6=(x-2)(x-3)$ .
2.Set $y=0$ : each factor can be zero, giving $x=2$ or $x=3$ .
3.Set $x=0$ : $y=6$ , so the $y$ -intercept is $(0,6)$ .

Answer: Roots $x=2$ and $x=3$ ; $y$ -intercept $(0,6)$ .

Common mistakes

Don't report roots as $y$ -values instead of $x$ -coordinates.
Don't find the $y$ -intercept by setting $y=0$ .
Don't read $(x-h)^2+k$ as having turning point $(-h,k)$ (Higher tier).

Exam tip

State roots as $x$ -values, intercepts as coordinates, and the symmetry line as an equation.

Worked practice

Tier 2 · Standard

For

y=(x-4)^2-9

, state the turning point and axis of symmetry, and find the roots.

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	Turning point $(4,-9)$ Axis $x=4$ Roots $x=1$ and $x=7$	3	The completed-square form gives turning point $(4,-9)$ and axis $x=4$ . For the roots, set $y=0$ : $(x-4)^2=9$ , so $x-4=\pm3$ and $x=1$ or $x=7$ .

Tier 3 · Hard

Complete the square for

y=2x^2+8x-10

. Hence state the turning point and find the roots.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$y=2(x+2)^2-18$ Turning point $(-2,-18)$ Roots $x=-5$ and $x=1$	4	Factor $2$ from the quadratic and linear terms: $y=2(x^2+4x)-10=2((x+2)^2-4)-10=2(x+2)^2-18$ . The turning point is therefore $(-2,-18)$ . Setting $y=0$ gives $(x+2)^2=9$ , so $x+2=\pm3$ and the roots are $-5$ and $1$ .

Tier 1 · Easy

One root of a quadratic is

x=-6

. Write down the coordinates of the corresponding

x

-axis intercept.

(1)

(Total for Question 3 is 1 mark)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$(-6,0)$	1	At an $x$ -axis intercept, the $y$ -coordinate is $0$ . The root gives the $x$ -coordinate, so the point is $(-6,0)$ .

Tier 2 · Standard

For

y=9-x^2

, find the roots and the turning point. Is the turning point a maximum or a minimum?

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	Roots $x=-3$ and $x=3$ Turning point $(0,9)$ , a maximum	3	For the roots, set $y=0$ : $9-x^2=0$ , so $x^2=9$ and $x=-3$ or $x=3$ . The graph is symmetric about $x=0$ , where $y=9$ , and the negative coefficient of $x^2$ makes $(0,9)$ a maximum.

Tier 3 · Hard

By first factorising, find the roots and the turning point of

y=2x^2-6x-56

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	Roots $x=-4$ and $x=7$ Turning point $\left(\dfrac{3}{2},-\dfrac{121}{2}\right)$ (or $(1.5,-60.5)$ )	4	Factorise: $y=2(x-7)(x+4)$ , so the roots are $x=7$ and $x=-4$ . The axis of symmetry is midway between the roots at $x=\dfrac{3}{2}$ . Substitution gives $y=2\left(\dfrac{3}{2}\right)^2-6\left(\dfrac{3}{2}\right)-56=-\dfrac{121}{2}$ , so the turning point is $\left(\dfrac{3}{2},-\dfrac{121}{2}\right)$ .

Tier 2 · Standard

A quadratic graph has roots

x=-4

and

x=2

and passes through

(0,-16)

. Find its equation in the form

y=a(x+4)(x-2)

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$y=2(x+4)(x-2)$	3	The roots give the factors $(x+4)$ and $(x-2)$ , so write $y=a(x+4)(x-2)$ . Substituting $(0,-16)$ gives $-16=a(4)(-2)=-8a$ , so $a=2$ . Therefore $y=2(x+4)(x-2)$ .

Tier 3 · Hard

Ravi says that the graph of

y=-(x+5)(x-1)

has roots

x=5

and

x=-1

and has a minimum turning point. Explain Ravi's two errors.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	The roots are $x=-5$ and $x=1$ ; the graph has a maximum turning point because the coefficient of $x^2$ is negative	3	For a root, set each factor equal to zero: $x+5=0$ gives $x=-5$ , and $x-1=0$ gives $x=1$ . The negative sign makes the coefficient of $x^2$ negative, so the parabola opens downwards and its turning point is a maximum, not a minimum.

Tier 3 · Hard

Higher only: Write

y=x^2+6x+13

in completed-square form. Hence state the turning point and explain why the graph has no roots.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$y=(x+3)^2+4$ Turning point $(-3,4)$ Since $(x+3)^2\geq0$ , $y\geq4$ , so the graph never reaches the $x$ -axis and has no roots	4	$x^2+6x+13=(x+3)^2-9+13=(x+3)^2+4$ . The turning point is $(-3,4)$ . Since $(x+3)^2\geq0$ , every value of $y$ is at least $4$ , so the graph never reaches the $x$ -axis and has no roots.

Tier 3 · Hard

Higher only: The quadratic graph

y=2x^2+kx+16

has axis of symmetry

x=3

. Work out

k

. Hence find the roots and the turning point of the graph.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$k=-12$ Roots $x=2$ and $x=4$ Turning point $(3,-2)$	5	For $y=2x^2+kx+16$ , the axis is $x=-\dfrac{k}{2\times2}$ . Hence $-\dfrac{k}{4}=3$ , so $k=-12$ . Then $y=2x^2-12x+16=2(x-2)(x-4)$ , giving roots $2$ and $4$ . Substituting $x=3$ gives $y=18-36+16=-2$ , so the turning point is $(3,-2)$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2019-06	1H	Q19	3	Non-calculator	Higher	QP MS
2019-06	1F	Q29	3	Non-calculator	Foundation	QP MS
2019-11	1H	Q21	5	Non-calculator	Higher	QP MS
2022-11	3H	Q21	3	Allowed	Higher	QP MS
2024-06	2H	Q20	3	Allowed	Higher	QP MS
2021-11	1H	Q22	4	Non-calculator	Higher	QP MS
2021-11	2H	Q4	6	Allowed	Higher	QP MS
2022-11	2F	Q24	4	Allowed	Foundation	QP MS
2022-06	1H	Q6	6	Non-calculator	Higher	QP MS
2022-06	1F	Q28	6	Non-calculator	Foundation	QP MS
2023-06	3F	Q26	2	Allowed	Foundation	QP MS
2022-11	2H	Q7	4	Allowed	Higher	QP MS
2024-06	2F	Q24	6	Allowed	Foundation	QP MS
2024-11	3H	Q23	5	Allowed	Higher	QP MS
2023-11	3H	Q17	7	Allowed	Higher	QP MS

Other points in A Algebra · equations and reasoning

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