J560 · 6.03QuadraticsFoundation & HigherGrades 4–7

Quadratic Graphs

Q: Why is a quadratic graph symmetric?

Values of x equally far either side of the turning point give the same y, because of the squared term.

Q: How do I find the turning point without completing the square?

Find the line of symmetry halfway between the roots, then substitute that x value into the equation.

Q: What if the graph doesn’t cross the x-axis?

The equation ax^2 + bx + c = 0 has no real solutions.

The graph of any quadratic $y = ax^{2} + bx + c$ is a smooth U-shaped curve called a $\textbf{parabola}$ . Three features describe it: where it crosses the axes, and where it turns. Each feature connects to algebra you already know — roots from factorising, the turning point from completing the square — so the graph and the equation are two views of the same thing.

55 min Quadratics

The big picture

The roots of $ax^{2} + bx + c = 0$ are exactly the $x$ -intercepts of $y = ax^{2} + bx + c$ . That is why a quadratic equation can have two, one or no solutions: the parabola can cross the $x$ -axis twice, touch it once, or miss it entirely.

OCR tests quadratic graphs in three ways: completing a table and plotting, sketching from algebra with key points labelled, and using a graph to solve equations such as $x^{2} - 2x - 8 = 3$ by drawing a horizontal line.

GCSE Maths - What are Quadratic Graphs? (2026/27 exams)Curated by Kepler, not produced by us. Watch it, then read the notes below.

Open on YouTube

Before you start

Try a short calculation from the preceding topic without the notes.
Check the tier and grade band shown in this lesson.

Watch, pause, try

Video by Cognito. The video’s own board label is retained; follow the specification reference and scope notes on this lesson for assessed content.

Viewing focus: Plot quadratic graphs from a table of values

After watching, attempt these questions before opening the answers. You can also use the written lesson below.

After watching 1: What is the $y$ -intercept of $y = x^{2} - 3x + 7$ ?

Show answer and reasoning

$(0, 7)$ . Substitute $x = 0$ .

After watching 2: What are the roots of $y = (x - 3)(x + 5)$ ?

Show answer and reasoning

$x = 3$ and $x = -5$ . Each bracket equals zero.

What you'll learn

Plot quadratic graphs from a table of values
Identify roots, the $y$ -intercept, the line of symmetry and the turning point
Sketch quadratic graphs from factorised and completed-square forms
Solve quadratic equations graphically and interpret the number of solutions

Plotting from a table

Substitute each $x$ value into the equation carefully, especially negative values: for $y = x^{2} - 2x - 8$ at $x = -3$ , $y = 9 + 6 - 8 = 7$ .

Plot the points and join them with a $\textbf{smooth curve}$ — never straight line segments, and never a flat bottom between two points.

The $y$ -values are symmetric about the turning point, which is a quick check on your table.

Example 1Complete a table for

y = x^{2} - 2x - 8

from

x = -3

x = 5

x = -3, -2, -1, 0, 1, 2, 3, 4, 5

y = 7, 0, -5, -8, -9, -8, -5, 0, 7

3The values mirror around

x = 1

AnswerRoots at

x = -2

and

x = 4

; minimum at

(1, -9)

Your turnSame method, new numbers — try it before moving on.

For $y = x^2 - 2x - 8$ , find $y$ when $x = -3$ .

Your answer

Tip — If the $y$ -values in your table are not symmetric, recheck the substitutions before plotting — one has almost certainly gone wrong.

Key features

$\textbf{Shape}$ : if $a > 0$ the parabola is U-shaped with a minimum; if $a < 0$ it is ∩-shaped with a maximum.

$\textbf{y-intercept}$ : substitute $x = 0$ , giving $(0, c)$ .

$\textbf{Roots}$ : solve $ax^{2} + bx + c = 0$ . Factorised form $y = (x - r)(x - s)$ shows them directly as $x = r$ and $x = s$ .

$\textbf{Line of symmetry}$ and $\textbf{turning point}$ : halfway between the roots, or from completed-square form $y = (x + p)^{2} + q$ , which has turning point $(-p, q)$ .

x = \dfrac{r + s}{2}

Line of symmetry, halfway between the roots.

Your turnSame method, new numbers — try it before moving on.

Find the turning point of $y = x^2 - 4x + 1$ .

Your answer

Type coordinates like (2, -3).

Why this works

For $y = x^{2} - 2x - 8 = (x - 4)(x + 2)$ , the roots $-2$ and $4$ average to $1$ , so the line of symmetry is $x = 1$ and the turning point is at $y = 1 - 2 - 8 = -9$ . No table needed.

Sketching

A sketch shows the correct shape and labels the key points; it does not need to be to scale.

Label every point where the curve meets the axes, and the turning point, with coordinates.

Example 1Sketch

y = -x^{2} + 6x - 5

a < 0

, so the curve is ∩-shaped.

y

-intercept

(0, -5)

3Roots:

-(x^{2} - 6x + 5) = -(x - 1)(x - 5)

, so

x = 1

and

x = 5

4Symmetry at

x = 3

; maximum

y = -9 + 18 - 5 = 4

Answer∩-shaped, crossing

(1, 0)

and

(5, 0)

, through

(0, -5)

, maximum at

(3, 4)

Your turnSame method, new numbers — try it before moving on.

Where does $y = x^2 - 7x + 10$ cross the $x$ -axis? Give the $x$ -values.

Your answer

Separate with a comma.

Solving equations graphically

The solutions of $ax^{2} + bx + c = 0$ are the $x$ -coordinates where the curve crosses the $x$ -axis.

To solve $ax^{2} + bx + c = k$ , draw the horizontal line $y = k$ and read the $x$ -coordinates of the intersections.

To solve a rearranged equation, first rewrite it so one side matches the plotted curve.

Example 1Using the graph of

y = x^{2} - 2x - 8

, estimate the solutions of

x^{2} - 2x - 8 = 4

1Draw

y = 4

2It meets the curve at about

x = -2.6

and

x = 4.6

Answer

x \approx -2.6

x \approx 4.6

(exactly

1 \pm \sqrt{13}

, i.e.

-2.61

and

4.61

)

Tip — Graphical solutions are estimates. Give them to the accuracy the graph allows, usually 1 decimal place.

Think like an examiner

Join plotted points with a smooth curve through every point.
Label axis intercepts and the turning point with coordinates on sketches.
Show the line $y = k$ used to solve an equation graphically.

Parabolas

y = (x - r)(x - s)

Roots at

x = r

and

x = s

y = (x + p)^{2} + q

Turning point

(-p, q)

(0,\ c)

y-intercept of

ax^{2} + bx + c

Watch out for these

Evaluating

(-3)^{2} - 2(-3)

-9 + 6

(-3)^{2} = 9

and

-2(-3) = +6

Joining points with straight lines or a flat bottom.

Draw one smooth curve.

Reading the

y

-intercept as a root.

Roots are where

y = 0

, on the

x

-axis.

Sketching

y = -x^{2} + \dots

as U-shaped.

A negative

x^{2}

coefficient gives a ∩ shape.

Stretch yourself

The graph of $y = x^{2} + 4x + k$ touches the $x$ -axis at exactly one point. Find $k$ , and the coordinates of that point.

Hint — Touching the axis once means the equation $x^{2} + 4x + k = 0$ has one repeated root. Completing the square helps.

Questions students ask

Why is a quadratic graph symmetric?

Values of

x

equally far either side of the turning point give the same

y

, because of the squared term.

How do I find the turning point without completing the square?

Find the line of symmetry halfway between the roots, then substitute that

x

value into the equation.

What if the graph doesn’t cross the

x

-axis?

The equation

ax^{2} + bx + c = 0

has no real solutions.

Key takeaways

$a > 0$ gives a U-shape; $a < 0$ gives a ∩-shape.
The $y$ -intercept is $(0, c)$ ; roots are the solutions of $y = 0$ .
The line of symmetry is halfway between the roots and passes through the turning point.
Completed-square form gives the turning point directly.
Solve $f(x) = k$ graphically by drawing $y = k$ .

Exam-style questions

Practise it the way it is examined

3 original questions · 6 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 43 marks

This question is about $y = x^2 + 2x - 15$ .

(a)Find the roots.[2]
(b)Write down the coordinates of the $y$ -intercept.[1]

Question 2Grade 51 mark

A quadratic graph crosses the $x$ -axis at $-3$ and $7$ . Write down the equation of its line of symmetry.[1]

Question 3Grade 72 marks

The graph of $y = x^2 - 4x$ has been drawn. What straight line should be drawn to solve $x^2 - 5x + 2 = 0$ ?[2]

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Return to this tomorrow

Close the notes and answer from memory. Revisit any explanation you cannot yet reconstruct.

What is the line of symmetry of $y = (x - 1)(x - 7)$ ?
At $x = -2$ , what is $y$ for $y = 2x^{2} - x - 3$ ?
What is the turning point of $y = x^{2} + 8x + 10$ ?
The curve $y = x^{2} - 4$ and the line $y = k$ meet at exactly one point. What is $k$ ?

Test yourself

Answer a few questions, then review anything you missed.

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