Are quadratic graphs on the Foundation paper as well?

In part. Spec point A11 asks both tiers to identify roots, intercepts and turning points from a quadratic graph, and A12 puts linear, quadratic, simple cubic and reciprocal shapes on both papers. Higher adds the algebra behind them, deducing roots by factorising and the turning point from the completed square form, along with exponential graphs y = k^x and the trigonometric curves.

How many marks is a quadratic graph question worth on Edexcel Higher?

A full plot-and-read question runs to roughly 5-6 marks across its parts, with the table and the plotting worth 2-3 and each read-off part 1-2. A single 'write down the coordinates of the turning point' part on its own is 1 mark. Shape-matching and 'which curve is this' parts are usually 1-2 marks.

What does 'by drawing a suitable line' actually mean?

It means the equation you have to solve is not the one already drawn on the grid. Rearrange it until one side is exactly the plotted expression, and whatever is left on the other side is the line you draw. To solve x^2 - 4x + 1 = 0 from a drawn y = x^2 - 4x - 3, the rearrangement gives x^2 - 4x - 3 = -4, so you draw y = -4.

Do I have to learn the cubic, reciprocal and exponential shapes?

Yes. The Exam Aid formulae sheet issued with every 1MA1 paper carries the quadratic formula, the circle and prism results and the trigonometric rules, and says nothing about graph shapes. A12 is recall: an x^3 term gives an S-shaped cubic, y = (a)/(x) gives two branches that never touch either axis, and y = k^x passes through (0, 1) and flattens towards the x-axis.

Quadratic Graphs Questions - Edexcel GCSE Higher Maths

Quadratic Graphs - Edexcel GCSE Higher Maths

Typically 3-6 marks, usually as one multi-part graph question · Spec 1MA1

Every quadratic graph is a parabola: one vertical line of symmetry, one turning point, and a U shape when the coefficient of $x^2$ is positive or an n shape when it is negative. Edexcel splits the work across two spec points. A11 is about reading a parabola, its roots, its intercepts and its turning point; its second clause, deducing the roots algebraically and locating the turning point from a completed square, carries the bold type that marks content as Higher tier only. A12 is the shape list: linear, quadratic, simple cubic and the reciprocal $y = \frac{1}{x}$ sit on both tiers, while exponential curves $y = k^x$ and the sine, cosine and tangent graphs are added for Higher candidates.

The classic version arrives in parts. You are given a grid and a half-filled table of values, you plot the curve, and then the short parts follow: the $y$ -intercept for a mark, the turning point for a mark, and 'use your graph to find estimates for the solutions of $x^2 - 3x - 1 = 0$ ' for two. The word estimate is doing real work, because the mark scheme accepts a range either side of each root. Papers 2 and 3 let you fill the table with the table function on a calculator rather than retyping the expression for every value of $x$ , while Paper 1 sets the same question with numbers small enough to square in your head. The harder variant asks you to solve an equation that is not the one drawn, which means working out which straight line to add.

Three things decide the mark. Solutions are the $x$ -coordinates of the crossing points, so a candidate who writes down the height of an intersection scores nothing even with a perfectly drawn curve. A line of symmetry is an equation, $x = 3$ , not the bare number 3 and not $y = 3$ . And when a stem says 'use the graph', an algebraic solution earns zero however accurate it is, because the marks are attached to the graphical method. Practise the read-off parts on curves you have not plotted: handed only $y = (x - 2)(x + 8)$ , you should be able to state the roots, the $y$ -intercept, the line of symmetry and the turning point in under a minute.

Worked example

The curve

C

has equation

y = x^2 - 6x + 5

. (a) Write down the coordinates of the point where

C

crosses the

y

-axis. (b) Find the coordinates of the two points where

C

crosses the

x

-axis. (c) Write down the equation of the line of symmetry of

C

. (d) Hence find the coordinates of the turning point of

C

[5 marks]

Put $x = 0$ : $y = 5$ . So $C$ crosses the $y$ -axis at $(0, 5)$ .
B1: the constant term is the $y$ -intercept, no working needed
The curve meets the $x$ -axis where $y = 0$ , so factorise: $x^2 - 6x + 5 = (x - 1)(x - 5)$ .
M1: a correct factorisation (or the formula) as the route to the roots
$(x - 1)(x - 5) = 0$ gives $x = 1$ and $x = 5$ , so the crossing points are $(1, 0)$ and $(5, 0)$ .
A1: both roots, written as coordinates because the question asked for points
A parabola is symmetrical about the vertical line halfway between its roots: $\frac{1 + 5}{2} = 3$ , so the line of symmetry is $x = 3$ .
B1: the answer must be the equation $x = 3$ ; a bare 3 is not a line
The turning point sits on that line, so substitute $x = 3$ : $y = 9 - 18 + 5 = -4$ . The turning point is $(3, -4)$ .
B1 follow through: $y$ found by substituting their line of symmetry

Practice questions

These are original questions in Edexcel 1MA1 Higher style. There are no grids on this page, so every curve arrives as an equation and every sketch is described in words, which is how the write-down and 'hence' parts of a real paper work once the plotting is done. Coordinates take two boxes with $x$ first, and the shape questions are multiple choice because on a paper they would be a matching exercise.

No account needed: the questions below are always free to check, plus 8 extra questions today. Your score so far updates as you go.

Question 1 Warm-up [2 marks]

The curve $y = (x - 4)^2 - 9$ has one turning point. Write down its coordinates.

x = y =
Question 2 Warm-up [1 mark]

What is the equation of the line of symmetry of the curve $y = (x - 6)^2 + 2$ ?

Answer options for question 2
$x = 6$ $x = -6$ $y = 2$ $x = 2$
Question 3 Warm-up [1 mark]

Which of these is the equation of an exponential graph?

Answer options for question 3
$y = x^5$ $y = 5^x$ $y = \frac{5}{x}$ $y = 5x^2$
Question 4 Exam pace [2 marks]

The curve $y = 2(x - 3)^2 - 7$ crosses the $y$ -axis at one point. Write down the $y$ -coordinate of that point.

Your answer (fractions like 1/2 are fine)
Question 5 Exam pace [1 mark]

A curve crosses the $x$ -axis at exactly three points and has two turning points, one a local maximum and one a local minimum. Which equation matches this curve?

Answer options for question 5
$y = x^2 - 4$ $y = x^3 - 4x$ $y = \frac{4}{x}$ $y = 4^x$
Question 6 Exam pace [3 marks]

For the same ball, the height in metres after $t$ seconds is $h = 20t - 5t^2$ . The maximum height is reached at the turning point of the parabola. Find the time in seconds at which the maximum height is reached, and the maximum height in metres.

time (s) = height (m) =

Common mistakes examiners see

Reading the height of an intersection rather than the value underneath it, so a curve meeting the line $y = 4$ at two points produces the answer 4 twice.

Drop a vertical line from each intersection down to the $x$ -axis and read the scale there. Solutions of an equation are values of $x$ ; a $y$ value is wanted only when the question names the intercept or the turning point.
Answering a 'use the graph' part with the quadratic formula. The two decimals are right and the part scores zero.

Treat 'use the graph' the way you treat 'hence': it names the method. Mark the intersections, drop the vertical lines, and quote the readings to 1 decimal place. Exact-looking answers on a graphical part are what tell an examiner you ignored the instruction.
Writing the line of symmetry as $3$ , or as $y = 3$ , for a curve that is symmetrical about the vertical line through its turning point.

Decide what kind of line it is before you write anything: it is vertical, so its equation starts $x =$ . Find it as the midpoint of the two roots, or as $-\frac{b}{2a}$ when the curve has no roots to average.
Picking the line to draw by matching the constant. With $y = x^2 + 2x - 5$ on the grid, solving $x^2 + 2x - 8 = 0$ tempts the line $y = -8$ , which cuts the curve in the wrong places.

Rearrange the target equation until its left side is exactly the plotted expression: $x^2 + 2x - 8 = 0$ becomes $x^2 + 2x - 5 = 3$ , so the line is $y = 3$ . Write that rearrangement on the page, because it carries the method mark.
Assuming every parabola opens upwards, so the turning point of $y = 12 + 4x - x^2$ is labelled a minimum.

Check the sign of the $x^2$ term before anything else: negative gives an n shape whose turning point is a maximum. When you are plotting, join the points freehand in one sweep with a rounded vertex, since a curve built from ruled segments or finished with a point loses the accuracy mark.

Frequently asked questions

Are quadratic graphs on the Foundation paper as well?: In part. Spec point A11 asks both tiers to identify roots, intercepts and turning points from a quadratic graph, and A12 puts linear, quadratic, simple cubic and reciprocal shapes on both papers. Higher adds the algebra behind them, deducing roots by factorising and the turning point from the completed square form, along with exponential graphs $y = k^x$ and the trigonometric curves.
How many marks is a quadratic graph question worth on Edexcel Higher?: A full plot-and-read question runs to roughly 5-6 marks across its parts, with the table and the plotting worth 2-3 and each read-off part 1-2. A single 'write down the coordinates of the turning point' part on its own is 1 mark. Shape-matching and 'which curve is this' parts are usually 1-2 marks.
What does 'by drawing a suitable line' actually mean?: It means the equation you have to solve is not the one already drawn on the grid. Rearrange it until one side is exactly the plotted expression, and whatever is left on the other side is the line you draw. To solve $x^2 - 4x + 1 = 0$ from a drawn $y = x^2 - 4x - 3$ , the rearrangement gives $x^2 - 4x - 3 = -4$ , so you draw $y = -4$ .
Do I have to learn the cubic, reciprocal and exponential shapes?: Yes. The Exam Aid formulae sheet issued with every 1MA1 paper carries the quadratic formula, the circle and prism results and the trigonometric rules, and says nothing about graph shapes. A12 is recall: an $x^3$ term gives an S-shaped cubic, $y = \frac{a}{x}$ gives two branches that never touch either axis, and $y = k^x$ passes through $(0, 1)$ and flattens towards the $x$ -axis.

Quadratic Graphs - Edexcel GCSE Higher Maths

Worked example

Practice questions

Common mistakes examiners see

Frequently asked questions

Related topics