Quadratic Graphs — AQA GCSE Mathematics

SPEC A11 • ALGEBRA — QUADRATIC GRAPHS

Quadratic Graphs: Roots, Intercepts, Turning Points & Symmetry

Every parabola has four key features — and every exam question asks about at least one of them.

The Parabola: y = ax² + bx + c

A quadratic equation produces a parabola — a symmetric U-shaped curve. The sign of a (the coefficient of x²) decides which way up it sits.

Check the sign of a first

a > 0 → parabola opens upwards, turning point is a minimum.

a < 0 → parabola opens downwards, turning point is a maximum.

The Four Key Features

For any parabola, AQA asks you to identify (at most) four things. Here they all are on one graph — y = x² − 4x − 5:

The four features

● Roots (where curve crosses x-axis) · ● y-intercept (where it crosses y-axis) · ● Turning point (max/min vertex) · ┆ Line of symmetry (vertical line through the vertex).

Feature 1: The y-Intercept

The y-intercept is where the graph crosses the y-axis — that's where x = 0.

y-Intercept Rule

In y = ax² + bx + c, substitute x = 0:

y = a(0)² + b(0) + c = c.

So c is the y-intercept — read it straight off the equation.

Examples

y = x² − 4x − 5 → y-intercept is −5. Point: (0, −5).

y = 2x² + 3x + 7 → y-intercept is 7. Point: (0, 7).

y = x² − 9 → y-intercept is −9. Point: (0, −9).

y = x² → y-intercept is 0. The parabola passes through the origin.

Feature 2: The Roots

The roots are where the graph crosses the x-axis — that's where y = 0. You find them by solving the quadratic equation.

Finding Roots Algebraically

Set y = 0 and solve ax² + bx + c = 0.

Method: factorise

Example: y = x² − 4x − 5.

Set y = 0: x² − 4x − 5 = 0.

Factorise: (x − 5)(x + 1) = 0.

Each factor = 0 → x = 5 or x = −1.

How many roots?

Two roots → parabola crosses x-axis twice.

One (repeated) root → parabola just touches x-axis.

No real roots → parabola doesn't reach the x-axis.

⚠️ Roots are x-values, not y-values. They're points of the form (x, 0) — on the x-axis. Do not write them as (0, x).

Feature 3: Turning Point & Line of Symmetry

A parabola is symmetric. The vertical line through its vertex is the line of symmetry, and the vertex sitting on it is the turning point (a minimum or maximum).

x = −b / (2a)

This gives both the x-coordinate of the turning point and the equation of the line of symmetry.

Procedure

Compute x = −b/(2a) using the coefficients from y = ax² + bx + c.

Substitute that x back into the equation to find the corresponding y.

The turning point is (x, y). The line of symmetry is x = that value.

Symmetry shortcut

If you already know the roots p and q, the turning point's x-coordinate is simply (p + q)/2 — the midpoint of the roots. No −b/(2a) needed.

⚠️ Don't drop the minus. The formula is x = −b/(2a), not b/(2a). The sign of b matters — track it through carefully.

From Factorised Form: y = (x − p)(x − q)

When the quadratic is already factorised, every feature comes out almost for free.

From y = (x − p)(x − q)

Roots: x = p and x = q (read directly from the brackets).

y-intercept: multiply out the constants: y = (0 − p)(0 − q) = pq. Alternatively, expand to y = x² − (p+q)x + pq — y-intercept is pq.

Turning point x-coord: (p + q)/2 — midpoint of the roots.

Turning point y-coord: substitute (p+q)/2 back in.

Example: y = (x − 1)(x − 5)

Roots: x = 1 and x = 5 (each bracket set to zero).

y-intercept: substitute x = 0 → y = (0 − 1)(0 − 5) = (−1)(−5) = 5. Point (0, 5).

TP x-coord: (1 + 5)/2 = 3. TP y: (3 − 1)(3 − 5) = 2 × (−2) = −4. Turning point (3, −4), minimum.

From Completed-Square Form HIGHER

The completed-square form reveals the turning point at a glance. This is the AQA Higher-only method for deducing turning points.

Completed-Square Form

y = a(x − h)² + kTurning point is (h, k) directly.

Minimum if a > 0; maximum if a < 0.

Worked conversion: y = x² + 6x + 2

Take half of 6 → 3. Square it → 9. Rewrite:

y = (x + 3)² − 9 + 2 = (x + 3)² − 7.

Read off: h = −3 (note the sign!), k = −7. Turning point (−3, −7), minimum.

⚠️ Sign of h. The form is (x − h)², so y = (x + 3)² corresponds to h = −3, not +3. The TP x-coord is negated from what's inside the bracket.

Worked Example — All Four Features01

For y = x² − 4x − 5, find the roots, y-intercept, turning point, and state whether the turning point is a maximum or minimum. [4 marks]

M

Method: (1) factorise for roots; (2) substitute x = 0 for y-intercept; (3) use x = −b/(2a) for TP x-coord, substitute for y; (4) check sign of a for max/min.

C

Roots: x² − 4x − 5 = (x − 5)(x + 1) = 0 → x = 5 or x = −1.

y-intercept: y = 0 − 0 − 5 = −5 → point (0, −5).

Turning point: x = −(−4)/(2·1) = 2. y = 4 − 8 − 5 = −9. Point (2, −9).

Max/min: a = 1 > 0, so turning point is a minimum.

E

Evaluate: Roots x = −1 and x = 5; y-intercept (0, −5); minimum at (2, −9). Sanity check — the TP x-coord 2 is the midpoint of roots (−1 + 5)/2 = 2 ✓.

Worked Example — From Factorised Form02

For y = (x − 2)(x − 6), find the roots, y-intercept, and turning point. [3 marks]

M

Method: read roots directly from brackets; substitute x = 0 for y-intercept; take midpoint of roots for TP x-coord; substitute for TP y.

C

Roots: x = 2 and x = 6 (each bracket set to zero).

y-intercept: y = (0 − 2)(0 − 6) = (−2)(−6) = 12 → point (0, 12).

TP x-coord: (2 + 6)/2 = 4. TP y-coord: (4 − 2)(4 − 6) = 2 × (−2) = −4. Turning point (4, −4).

E

Evaluate: Roots 2 and 6; y-intercept (0, 12); minimum at (4, −4). Check via expansion — y = (x − 2)(x − 6) = x² − 8x + 12, so a = 1 (minimum ✓), c = 12 (y-intercept ✓), x = −(−8)/2 = 4 (TP ✓).

Reading Features From a Graph

AQA often gives you the graph and asks you to read off features. Here's y = x² − 2x − 3:

Read-off procedure

Roots: find where the curve meets the x-axis. Here: x = −1 and x = 3.

y-intercept: find where the curve meets the y-axis. Here: (0, −3).

Turning point: find the lowest (or highest) point of the curve. Here: (1, −4) — a minimum.

Examiner Alert: Min vs Max

⚠️ EXAMINER ALERT: Students confuse minimum and maximum under exam stress — especially when a is negative. The direction of the parabola depends entirely on the sign of a, the coefficient of x².

EquationSign of aShapeTurning Point Type

y = x² − 4x − 5a = +1 > 0U (opens up)Minimum

y = −x² + 4x + 5a = −1 < 0∩ (opens down)Maximum

y = 3x² + 2x − 1a = +3 > 0U (opens up)Minimum

y = −2x² − x + 6a = −2 < 0∩ (opens down)Maximum

Build the habit

Before anything else, look at the sign of a. Circle it on your working. Then you can't forget whether it's a min or max when you write your final answer.

Examiner Alert: Roots Notation

⚠️ EXAMINER ALERT: A real AQA Jun24 Paper 3 question asked for the roots of f(x) = 0 from a graph. The mark scheme (Jun24 P3 Q13) explicitly rejects "(0, −2.2) and (0, 1.5)" with a B0. Why? Because that's the form for y-intercepts — on the y-axis — not roots.

How students write itMark awardedWhy

x = −2.2, x = 1.5B2 ✓Correct — x-values stated directly.

(−2.2, 0) and (1.5, 0)B2 ✓Correct — (x, 0) coordinates on the x-axis.

(−2.2, 1.5)B2 ✓Accepted by AQA — pair of x-values is given the benefit of the doubt.

(0, −2.2) and (0, 1.5)B0 ✗Wrong axis — these would be y-intercepts, not roots.

Safe answer

Write roots as "x = p, x = q" or as coordinate pairs "(p, 0) and (q, 0)". Never put the x-values second inside coordinate brackets.

The MCE Framework for Quadratic Graph Questions

Any "find the features" or "sketch the parabola" question follows the same three-step structure.

MMethod: Check the sign of a (min or max?). Pick the right approach for each feature — factorise, substitute, −b/(2a), or read off.

CCalculate: Work through each feature in order — y-intercept, roots, turning point. Track signs carefully, especially the minus in −b/(2a).

EEvaluate: Sanity-check: midpoint of roots should equal TP x-coord. TP y should be below the x-axis for a U with roots (and vice versa).

Typical mark breakdown

A 4-mark "find all features" question usually marks: B1 roots · B1 y-intercept · M1 x = −b/(2a) or symmetry · A1 turning point. The MCE steps map 1-to-1 onto the mark scheme.

Practice Question 1

A quadratic function is defined as y = x² − 6x + 5.

(a) Write down the y-intercept. [1 mark]

(b) Find the roots of y = 0. [2 marks]

(c) Find the coordinates of the turning point and state whether it is a maximum or a minimum. [3 marks]

M

Method: (a) read c. (b) factorise and set each bracket = 0. (c) x = −b/(2a), then substitute for y; check sign of a.

C

(a) y-intercept = c = 5, so point (0, 5).

(b) x² − 6x + 5 = (x − 5)(x − 1) = 0 → x = 5 or x = 1.

(c) x = −(−6)/(2·1) = 3. y = 9 − 18 + 5 = −4. Turning point (3, −4). Since a = 1 > 0, it's a minimum.

E

Evaluate: (a) B1. (b) B2 both roots. (c) M1 x-coord, A1 y-coord, A1 minimum. Midpoint check: (1 + 5)/2 = 3 ✓.

Practice Question 2

The diagram shows the graph of y = f(x), where f(x) is a quadratic function.

(a) Write down the roots of f(x) = 0. [2 marks]

(b) Write down the coordinates of the point where the graph crosses the y-axis. [1 mark]

M

Method: (a) roots = x-values where the curve crosses the x-axis. (b) y-intercept = where the curve crosses the y-axis.

C

(a) The curve crosses the x-axis at x = −2 and x = 3.

(b) The curve crosses the y-axis at (0, −6).

E

Evaluate: (a) B2 both correct x-values. Write them as "x = −2, x = 3" — not as (0, −2) and (0, 3), which AQA would mark B0. (b) B1 coordinate pair (0, −6).

Practice Question 3 HIGHER

A quadratic function is given in completed-square form:

y = 2(x − 4)² − 11

(a) Write down the coordinates of the turning point. [1 mark]

(b) State whether the turning point is a maximum or a minimum, giving a reason. [1 mark]

(c) Work out the y-intercept of the curve. [2 marks]

M

Method: (a) read h and k directly from y = a(x − h)² + k. (b) examine the sign of a. (c) substitute x = 0 into the given expression and evaluate.

C

(a) Comparing to y = a(x − h)² + k: h = 4, k = −11. Turning point (4, −11).

(b) a = 2 > 0, so the parabola opens upward → minimum.

(c) At x = 0: y = 2(0 − 4)² − 11 = 2(16) − 11 = 32 − 11 = 21. y-intercept is (0, 21).

E

Evaluate: (a) B1. (b) B1 with "a > 0 so minimum" reason. (c) M1 substitute x = 0, A1 21. Sanity check — minimum at y = −11, y-intercept at y = 21, so the curve rises steeply from the vertex ✓.

Summary

The Four Features of y = ax² + bx + c

Shape: a > 0 → U, minimum. a < 0 → ∩, maximum.

y-intercept: the constant c (set x = 0).

Roots: solve ax² + bx + c = 0 (factorise, or read from graph).

Turning point: x = −b/(2a), substitute for y. Equivalently, midpoint of the two roots.

Line of symmetry: x = −b/(2a).

Special forms

Factorised y = (x − p)(x − q): roots p, q.

Completed-square y = a(x − h)² + k H: TP (h, k).

Avoid

Writing roots as (0, x) — wrong axis.

Dropping the minus in x = −b/(2a).

Calling a minimum a maximum (or vice versa).

Exam Tip

For any quadratic question, write the four features in a list before drawing anything: shape, y-intercept, roots, turning point. That's your entire answer structure. Each feature is a separate mark.