Teaching unit · UK Year 10 (Key Stage 4 / GCSE Foundation, ages 14 to 15)

Quadratic graphs: roots, intercepts and turning points

Reading a parabola's key points off its graph, and deducing roots algebraically from factorised form

About three lessons of 45 to 60 minutes

Start here · hook

Every thrown ball draws a quadratic graph in the air

Throw a ball to a friend and watch its path: it climbs, levels off for an instant at its highest point, then falls, tracing a smooth arch. That arch is a PARABOLA, the shape of every quadratic graph, and its three most useful features each answer a real question. Where the curve meets the ground level are the ROOTS (where the ball starts and lands). Where it crosses the starting axis is the INTERCEPT. Its highest (or lowest) point is the TURNING POINT.

GCSE Foundation asks two things: read those three features straight off a drawn graph, and deduce the roots ALGEBRAICALLY when the quadratic is handed to you in factorised form like (x - 4)(x + 2), using the fact that a product is zero only when one of its factors is zero.

A goal kick's flight paththe roots are where the ball leaves and lands; the turning point is its highest moment
A fountain jet's archa downward ('n'-shaped) parabola, because the water must come back down
The cable of a suspension bridgean upward ('u'-shaped) parabola whose turning point is its lowest sag
(x - 4)(x + 2) = 0a product is zero only when a factor is zero, so x = 4 or x = -2, no graph needed

Learning objective

What students will be able to do

Students will read the roots, y-intercept and turning point of a quadratic function off its drawn graph, state whether a parabola opens upward or downward from the sign of its x-squared term, and deduce roots algebraically from factorised form.

Success criteria

I can point to the roots of a quadratic graph: the x-values where the curve crosses the x-axis.
I can read off the y-intercept: the point where the curve crosses the y-axis, always at the constant term.
I can read off the coordinates of the turning point: the lowest point of a 'u' shape or the highest point of an 'n' shape.
I can tell from the sign of the x-squared term whether a parabola opens upward or downward.
I can solve a factorised quadratic like (x - 3)(x + 5) = 0 by setting each factor to zero.

Curriculum anchor

Standards this unit teaches

GCSE Algebra #11UK GCSE Mathematics (DfE, England)
Roots, intercepts and turning points of quadratic functions
Subject content statement (Department for Education, "GCSE mathematics: subject content and assessment objectives", published 1 November 2013, reference DFE-00233-2013, "Algebra" section, "Graphs", item 11, https://www.gov.uk/government/publications/gcse-mathematics-subject-content-and-assessment-objectives): students should "identify and interpret roots, intercepts, turning points of quadratic functions graphically; deduce roots algebraically and turning points by completing the square". The graphical clause and "deduce roots algebraically" are underlined type (Foundation, assessed for every GCSE student); the closing "turning points by completing the square" clause is bold type (Higher tier only) and is deliberately excluded from this unit.
GCSE Algebra #12UK GCSE Mathematics (DfE, England)
Recognise, sketch and interpret quadratic graphs
Subject content statement (same DfE document, "Algebra" section, "Graphs", item 12): students should "recognise, sketch and interpret graphs of linear functions, quadratic functions, simple cubic functions, the reciprocal function..." The quadratic-functions clause is standard type (all students); this unit is its recognition and interpretation companion.

Before you start

Prior knowledge

This unit builds on skills students should already have met. Revisit any that are shaky first.

Key vocabulary

Words to teach and display

Parabola: the symmetrical 'u' or 'n' shaped curve every quadratic function draws
Root: an x-value where the graph crosses the x-axis, i.e. where y = 0; also called a solution of the equation
y-intercept: the point where the graph crosses the y-axis, at x = 0
Turning point: the point where the curve stops falling and starts rising (or the reverse): the lowest point of a 'u', the highest of an 'n'
Factorised form: a quadratic written as a product of brackets, like (x - 4)(x + 2), which hands you its roots

Teaching sequence

Teach it: concrete, pictorial, abstract

The lesson moves from things students can hold, to pictures and diagrams, to the written maths. The diagrams below are drawn from data, so they are accurate and print cleanly. Teach straight from them.

1. The shape: 'u' up, 'n' down

Concrete

Every quadratic y = ax² + bx + c draws a parabola. The sign of the x-squared term decides the shape: positive opens upward (a 'u', with a lowest point), negative opens downward (an 'n', with a highest point). The curve is perfectly symmetrical about the vertical line through its turning point.

Have students trace the figure's curve with a finger: down through the left root, along the bottom, up through the right root. The symmetry means the two 'arms' are mirror images.

y =

x^{2}

- 2x - 8: an upward 'u' (positive

x^{2}

term). Roots at -2 and 4, y-intercept (0, -8), turning point (1, -9), all marked.

Check for understanding, ask

Before plotting anything: will y = - $x^{2}$ + 3x + 10 be a 'u' shape or an 'n' shape? How do you know?
Why must a 'u'-shaped parabola have a LOWEST point rather than a highest one?

2. Reading roots, intercept and turning point off the graph

Pictorial

The three key features are read straight off the picture: roots where the curve crosses the x-axis, y-intercept where it crosses the y-axis, turning point at the bottom (or top) of the curve. Roots are x-VALUES; the intercept and turning point are full COORDINATE PAIRS.

y =

x^{2}

+ 2x - 3. Crosses the x-axis at -3 and 1 (the roots), crosses the y-axis at (0, -3), and turns at (-1, -4).

Worked example

Using the figure above (y = $x^{2}$ + 2x - 3), write down (a) the roots, (b) the y-intercept, and (c) the turning point.

(a) The curve crosses the x-axis at x = -3 and x = 1: those are the roots.
(b) The curve crosses the y-axis at (0, -3). Notice this matches the constant term, because substituting x = 0 kills every x term.
(c) The lowest point of the 'u' is at (-1, -4): the turning point, read off the grid as a coordinate pair.

Answer: (a) x = -3 and x = 1 (b) (0, -3) (c) (-1, -4)

Check for understanding, ask

The roots are single numbers but the turning point is a coordinate pair. Why the difference?
Without any algebra, why must the y-intercept of y = $x^{2}$ + 2x - 3 sit at height -3?

3. Deducing roots algebraically from factorised form

Abstract

When a quadratic is written in factorised form, its roots can be DEDUCED without any graph: a product of two brackets equals zero exactly when one of the brackets equals zero. Each bracket hands you one root, with the sign flipped: (x - 4) = 0 gives x = 4, and (x + 2) = 0 gives x = -2.

The same downward logic works for 'n'-shaped curves: y = - $x^{2}$ + 4x - 3 factorises as -(x - 1)(x - 3), so its roots are x = 1 and x = 3, and the figure confirms the curve crosses exactly there.

y = -

x^{2}

+ 4x - 3, an 'n' shape (negative

x^{2}

term). Roots at 1 and 3, y-intercept (0, -3), turning point (2, 1) at the TOP of the curve.

Worked example

Solve (x - 4)(x + 2) = 0

A product is zero only when one of its factors is zero.
First factor: x - 4 = 0, so x = 4.
Second factor: x + 2 = 0, so x = -2.
Check by substituting each back in: (4 - 4)(4 + 2) = 0 x 6 = 0, and (-2 - 4)(-2 + 2) = -6 x 0 = 0.

Answer: x = 4 or x = -2

Check for understanding, ask

Why does (x - 4)(x + 2) = 0 force one of the two brackets to be zero, when (x - 4)(x + 2) = 12 forces neither?
What are the roots of (x + 7)(x - 7) = 0, and why do they have opposite signs?

Watch for

Common misconceptions and how to address them

MisconceptionReading the roots of (x - 4)(x + 2) = 0 as x = -4 and x = 2 (copying the signs inside the brackets).

Why it happens: The bracket (x - 4) LOOKS like it contains '-4', so students report -4 instead of solving the little equation x - 4 = 0.

How to address it: Always write the one-line equation for each bracket: x - 4 = 0 so x = 4. Then substitute both answers back into the original product; a sign-flipped 'root' fails the check immediately, because (-4 - 4)(-4 + 2) = 16, not 0.

MisconceptionGiving the turning point as a single number, or swapping its coordinates.

Why it happens: Roots are quoted as lone x-values, so students carry that habit to the turning point and drop half the information (or write (y, x) out of column-reading habit).

How to address it: Drill the sentence 'roots are x-values, intercepts and turning points are POINTS'. When reading a turning point, trace down to the x-axis first (x-coordinate), then across to the y-axis (y-coordinate), and write the pair in that order.

MisconceptionTrying to CALCULATE the turning point algebraically (completing the square or averaging the roots) on a Foundation paper.

Why it happens: Textbooks aimed at both tiers show the algebraic methods, so students assume they are required knowledge at Foundation.

How to address it: At Foundation the turning point is only ever READ OFF the graph; the algebraic derivation ('turning points by completing the square') is printed in bold, Higher-only type in the DfE subject content. Knowing the scope boundary saves exam time and anxiety.

Do it together

Guided practice (with answers)

1. Solve (x - 5)(x + 1) = 0
Answer: x = 5 or x = -1, because a product is zero only when one factor is zero.
2. Write down the coordinates of the point where y = (x - 2)(x - 6) crosses the y-axis.
Answer: (0, 12), because substituting x = 0 gives (-2) x (-6) = 12.
3. A quadratic graph crosses the x-axis at -4 and 2, and its lowest point is at (-1, -9). Write down its roots and the coordinates of its turning point.
Answer: Roots x = -4 and x = 2; turning point (-1, -9), quoted as a full coordinate pair.
4. Does the graph of y = - $x^{2}$ + 3x + 10 open upward or downward?
Answer: Downward (an 'n' shape), because the $x^{2}$ term has a negative coefficient.
5. Factorise $x^{2}$ - 5x + 6 and hence solve $x^{2}$ - 5x + 6 = 0
Answer: (x - 2)(x - 3) = 0, so x = 2 or x = 3 (two numbers multiplying to 6 and adding to -5 are -2 and -3).

On their own

Independent practice worksheets

Practise reading drawn quadratic graphs and deducing roots from factorised form, with computed, never-wrong answer keys and code-drawn graphs.

Quadratic Graphs: Roots, Intercepts & Turning Points (Year 10)read roots, y-intercepts and turning points off drawn graphs Quadratics from Factorised Form (Year 10)deduce roots algebraically, no graph needed Quadratic Graphs (Year 11)the same skill, freshly generated, for a Year 11 class or revision Year 9 quadratics worksheetextra expanding and factorising practice if the algebra is shaky

Reach every student

Differentiation

Support

Start every graph-reading question with the same three-step finger trace: along the x-axis for roots, up the y-axis for the intercept, then to the bottom (or top) of the curve for the turning point.
Keep the sign-flip rule visible: bracket (x - number) gives root +number; bracket (x + number) gives root -number, and ALWAYS check by substitution.
Offer graphs with the key points pre-marked as dots first, then fade to unmarked curves.

Extension

Given roots x = -3 and x = 5, write down a possible factorised quadratic, expand it, and check its y-intercept matches the constant term.
Sketch (unplotted) y = (x - 1)(x - 7): mark the roots, the y-intercept and the shape, and argue where the turning point must sit using only symmetry.
Investigate what happens to the graph when the factorised form is multiplied by -1: which features move and which stay fixed?

Check it stuck

Assessment: exit ticket

A three-question exit ticket: one graph read, one factorised solve, one y-intercept deduction.

1. From the section 1 figure (y = $x^{2}$ - 2x - 8), write down the roots and the coordinates of the turning point.
Answer: Roots x = -2 and x = 4; turning point (1, -9).
2. Solve (x + 3)(x - 7) = 0
Answer: x = -3 or x = 7, setting each factor to zero.
3. Write down the y-intercept of y = (x - 1)(x + 6).
Answer: (0, -6), because substituting x = 0 gives (-1) x 6 = -6.

For the teacher

Teacher notes and timings

Rough timing: Lesson 1 shape and symmetry (section 1), Lesson 2 reading graphs (section 2), Lesson 3 factorised-form roots plus the exit ticket (section 3).
Tier boundary, worth stating to the class: reading turning points OFF a graph is Foundation; DERIVING them (completing the square) is bold, Higher-only type in DfE Algebra item 11. This unit and its worksheets never ask for an algebraic turning point, and the midpoint-of-roots symmetry shortcut is left as extension talk only.
Every graph in this unit and its worksheets is drawn by the QuadraticGraph engine (components/StandardFigures.tsx), which derives the roots, intercept and turning point from the coefficients itself, so a figure can never disagree with its answer key.
The worksheet generators build every quadratic BACKWARD from integer roots, so all answers are exact whole numbers; the drawn graphs additionally keep the two roots the same parity so the turning point lands on a grid intersection students can actually read.

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