What is a Quadratic Function in the context of Cambridge IGCSE Additional Mathematics?

In the Cambridge IGCSE Additional Mathematics curriculum, a quadratic function is a polynomial function of degree two, typically expressed in the form y = ax^2 + bx + c, where a, b, and c are constants, and a ≠ 0. Quadratic functions are fundamental in understanding various mathematical concepts, including graphing, solving equations, and analyzing parabolic shapes.

How do you find the vertex of a quadratic function?

The vertex of a quadratic function given by y = ax^2 + bx + c can be found using the formula for the x-coordinate of the vertex, x = -b/(2a). Once you have the x-coordinate, substitute it back into the original equation to find the y-coordinate. The vertex is a crucial point that represents the maximum or minimum value of the quadratic function, depending on the direction of the parabola.

What are the methods to solve quadratic equations in Additional Mathematics?

In Cambridge IGCSE Additional Mathematics, quadratic equations can be solved using several methods: factoring, completing the square, and using the quadratic formula. Factoring involves expressing the quadratic as a product of two binomials. Completing the square transforms the equation into a perfect square trinomial, and the quadratic formula, x = (-b ± √(b^2 - 4ac)) / (2a), provides solutions directly.

How can you determine the nature of the roots of a quadratic function?

The nature of the roots of a quadratic function can be determined using the discriminant, which is the part of the quadratic formula under the square root: b^2 - 4ac. If the discriminant is positive, the quadratic has two distinct real roots. If it is zero, there is one real root (a repeated root). If the discriminant is negative, the quadratic has two complex roots. Understanding the discriminant helps in predicting the behavior of the quadratic function's graph.

What is the significance of the axis of symmetry in a quadratic function?

The axis of symmetry of a quadratic function is a vertical line that passes through the vertex of the parabola, given by the equation x = -b/(2a). It divides the parabola into two mirror-image halves, providing a line of reflection. This concept is significant in graphing quadratic functions as it helps in accurately plotting the parabola and understanding its symmetrical properties.

Why is "Reading the discriminant inequality the wrong way" a common mistake in Quadratics Functions?

Why it happens: Confusion over > 0 vs 0 vs = 0. How to avoid it: > 0: TWO distinct real roots. = 0: ONE repeated real root. < 0: NO real roots. Memorise. Source: 0606 Examiner Reports 2022-2024

Why is "Forgetting to flip the inequality when multiplying by negative" a common mistake in Quadratics Functions?

Why it happens: Habit of treating equations. How to avoid it: When you multiply or divide both sides of an inequality by a NEGATIVE number, FLIP the inequality sign. For quadratic inequalities, sketch the parabola to read off the answer.

Why is "Halving the linear coefficient incorrectly" a common mistake in Quadratics Functions?

Why it happens: Sign or arithmetic slip. How to avoid it: For x^2 + 6x: half of 6 is 3, so (x + 3)^2 - 9. For x^2 - 6x: half of -6 is -3, so (x - 3)^2 - 9. Sign matters.

Why is "Missing the boundary cases in inequalities" a common mistake in Quadratics Functions?

Why it happens: Forgetting the difference between strict (<) and inclusive (). How to avoid it: Use OPEN brackets/circles for strict, CLOSED for inclusive. Match the inequality in the question.

Quadratics FunctionsCambridge IGCSE Additional Mathematics 0606

Quadratics Functions

Q: Why is "Forgetting to flip the inequality when multiplying by negative" a common mistake in Quadratics Functions?

Why it happens: Habit of treating equations. How to avoid it: When you multiply or divide both sides of an inequality by a NEGATIVE number, FLIP the inequality sign. For quadratic inequalities, sketch the parabola to read off the answer.

Q: Why is "Halving the linear coefficient incorrectly" a common mistake in Quadratics Functions?

Why it happens: Sign or arithmetic slip. How to avoid it: For x^2 + 6x: half of 6 is 3, so (x + 3)^2 - 9. For x^2 - 6x: half of -6 is -3, so (x - 3)^2 - 9. Sign matters.

Q: Why is "Missing the boundary cases in inequalities" a common mistake in Quadratics Functions?

Why it happens: Forgetting the difference between strict (<) and inclusive (). How to avoid it: Use OPEN brackets/circles for strict, CLOSED for inclusive. Match the inequality in the question.

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Quadratic Functions Study Notes — Cambridge IGCSE Additional Mathematics 0606 (2025-2027 syllabus)

Solving and analysing quadratics. Three solving methods (factorise, complete the square, formula). Discriminant determines root nature. Completed-square form reveals max/min and vertex. Quadratic inequalities via parabola sketches.

What you’ll learn

Mapped to the Cambridge IGCSE 0606 syllabus (2025-2027).

1.2.1 — Solve quadratic equations by factorising, completing the square, formula.
1.2.2 — Use the discriminant to determine root nature.
1.2.3 — Complete the square; find max/min.
1.2.4 — Solve quadratic inequalities.

Three ways to solve $ax^2 + bx + c = 0$

Factorise, complete the square, or use the quadratic formula.

Method 1: Factorise. Find two numbers whose PRODUCT is $ac$ and SUM is $b$ . Split the middle term and factor by grouping.

Example: $x^2 + 5x + 6 = (x + 2)(x + 3) = 0$ , giving $x = -2$ or $x = -3$ .

Method 2: Quadratic formula. $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$

Works for ANY quadratic. Especially useful when factorising is hard.

Method 3: Complete the square.

Rewrite $ax^2 + bx + c$ as $a(x + p)^2 + q$ . Then $a(x + p)^2 = -q$ , $(x + p)^2 = -q/a$ , $x + p = \pm \sqrt{-q/a}$ , $x = -p \pm \sqrt{-q/a}$ .

Cambridge tip. Try factorising first. If the discriminant isn't a perfect square, use the formula. Use completing the square when asked for max/min or vertex.

Factorise — quickest if it works.
Quadratic formula — universal.
Completing the square — also gives vertex.

Discriminant — what kind of roots?

$\Delta = b^2 - 4ac$ . Sign tells you the root count.

Discriminant $\Delta = b^2 - 4ac$ .

Three cases.

$\Delta > 0$ : TWO distinct real roots. Parabola crosses x-axis at TWO points.
$\Delta = 0$ : ONE REPEATED real root. Parabola TOUCHES x-axis at ONE point (vertex).
$\Delta < 0$ : NO real roots. Parabola does NOT cross x-axis.

Common question types.

'Find $k$ for which the equation has equal roots' → set $\Delta = 0$ .
'Find $k$ for two distinct real roots' → solve $\Delta > 0$ .
'Find $k$ for no real roots' → solve $\Delta < 0$ .

Cambridge tip. ALWAYS check whether the question wants real / equal / distinct roots — then set up the right inequality / equation involving $\Delta$ .

$\Delta = b^2 - 4ac$ .
$> 0$ — two distinct. $= 0$ — equal. $< 0$ — no real.
Match the inequality to the question.

See the full worked example for quadratics functions →

Completing the square

Rewrite $ax^2 + bx + c$ as $a(x + p)^2 + q$ . Vertex revealed.

Procedure.

Factor out $a$ from the $x^2$ and $x$ terms: $a\left(x^2 + \frac{b}{a}x\right) + c$ .
Halve the $x$ coefficient and square it: $\left(\frac{b}{2a}\right)^2$ .
Add and subtract that value inside the bracket: $a\left[x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2\right] + c$ .
Recognise the perfect square: $a\left[\left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2\right] + c$ .
Distribute and simplify: $a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}$ .

Vertex. At $x = -\frac{b}{2a}$ . Value there is $c - \frac{b^2}{4a}$ .

If $a > 0$ , this is a MINIMUM. If $a < 0$ , it's a MAXIMUM.

Worked walkthrough. $f(x) = 2x^2 - 12x + 5$ .

Factor: $2(x^2 - 6x) + 5$ .
Half of $-6$ is $-3$ , squared is $9$ . So $x^2 - 6x = (x - 3)^2 - 9$ .
$f(x) = 2[(x - 3)^2 - 9] + 5 = 2(x - 3)^2 - 18 + 5 = 2(x - 3)^2 - 13$ .
Minimum: $-13$ at $x = 3$ .

Completed-square form a(x + p)² + q gives the vertex (−p, q) directly: here (3, −13). Since a = 2 > 0, the parabola opens up so the vertex is a minimum.

Cambridge tip. When the question asks for max/min and vertex, completing the square is the most direct method.

Factor out leading coefficient first.
Half the $x$ coefficient, square, add/subtract.
Vertex at $-\frac{b}{2a}$ ; value $c - \frac{b^2}{4a}$ .

See the full worked example for quadratics functions →

Quadratic inequalities

Sketch the parabola; read off where it's above or below the x-axis.

Procedure.

Find the roots of $ax^2 + bx + c = 0$ (factorise / formula).
Sketch the parabola. Note whether it opens UP ( $a > 0$ ) or DOWN ( $a < 0$ ).
Read off the regions where the parabola is below/above the x-axis according to the inequality.

Worked walkthrough. Solve $2x^2 - 5x - 3 < 0$ .

Factorise: $(2x + 1)(x - 3) = 0$ . Roots: $x = -\frac{1}{2}$ , $x = 3$ .
$a = 2 > 0$ , parabola opens UP.
BELOW x-axis BETWEEN the roots.
Solution: $-\frac{1}{2} < x < 3$ .

Sign-table alternative. Some prefer sign analysis: pick test values left of, between, and right of the roots, evaluate the inequality, mark each region with $+$ or $-$ .

Cambridge tip. Always SKETCH or use a sign diagram. Mark scheme awards method marks for the sketch even if the inequality is read off wrong.

Factorise → roots → sketch → read off.
$a > 0$ opens up; $a < 0$ opens down.
Below x-axis BETWEEN roots (when $a > 0$ ).

See the full worked example for quadratics functions →

How it’s examined

Quadratics appear on every Paper 1. Most-tested questions: completing the square + max/min (5-7 marks), discriminant for $k$ (4-5 marks), quadratic inequality (5 marks).

Sources: Cambridge IGCSE Additional Mathematics 0606 syllabus (2025-2027); 0606 Examiner Reports 2022-2024; 0606/12 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-09.

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Step-by-step worked examples — Quadratics Functions

Step-by-step solutions to past-paper-style questions on quadratics functions, written exactly the way a tutor would explain them at the board.

1Complete the square (5 marks)
Extended• Adapted from 0606/12 May/Jun 2024 Q5• completing-square
Question
Express $f(x) = 2x^2 - 12x + 5$ in the form $a(x + b)^2 + c$ and hence find the minimum value of $f(x)$ and the value of $x$ at which it occurs. (5 marks)
Step-by-step solution
1. Step 1
  Factor out the leading coefficient (1 mark). $f(x) = 2(x^2 - 6x) + 5$ .
2. Step 2
  Complete the square inside (2 marks). Take half of $-6$ , square it: $9$ . So $x^2 - 6x = (x - 3)^2 - 9$ . Hence $f(x) = 2[(x - 3)^2 - 9] + 5 = 2(x - 3)^2 - 18 + 5 = 2(x - 3)^2 - 13$ .
3. Step 3
  Identify the minimum (2 marks). $(x - 3)^2 \ge 0$ always, equal to 0 when $x = 3$ . So the minimum of $f$ is $-13$ , occurring at $x = 3$ .
Answer
$f(x) = 2(x - 3)^2 - 13$ . Minimum value $-13$ at $x = 3$ .
2Use the discriminant (4 marks)
Extended• discriminant
Question
Find the values of $k$ for which the equation $x^2 + (k - 2)x + 9 = 0$ has equal roots. (4 marks)
Step-by-step solution
1. Step 1
  Equal roots condition (1 mark). Discriminant $= 0$ , i.e. $b^2 - 4ac = 0$ .
2. Step 2
  Substitute (1 mark). Here $a = 1$ , $b = k - 2$ , $c = 9$ . So $(k - 2)^2 - 4(1)(9) = 0$ .
3. Step 3
  Solve (2 marks). $(k - 2)^2 = 36$ , so $k - 2 = \pm 6$ , giving $k = 8$ or $k = -4$ .
Answer
$k = 8$ or $k = -4$ .
3Solve a quadratic inequality (5 marks)
Extended• inequality
Question
Solve $2x^2 - 5x - 3 < 0$ . (5 marks)
Step-by-step solution
1. Step 1
  Factorise (2 marks). $2x^2 - 5x - 3 = (2x + 1)(x - 3)$ .
2. Step 2
  Find roots (1 mark). $(2x + 1)(x - 3) = 0$ gives $x = -\frac{1}{2}$ or $x = 3$ .
3. Step 3
  Sketch a parabola (1 mark). Coefficient of $x^2$ is positive, so the parabola opens UPWARDS. It is BELOW the x-axis between the roots.
4. Step 4
  State the solution (1 mark). $-\frac{1}{2} < x < 3$ .
Answer
$-\dfrac{1}{2} < x < 3$ .
Examiner tip
Mark scheme: ALWAYS sketch the parabola or use a sign diagram for inequalities. Listing roots without checking the inequality direction loses marks.

Key Formulae — Quadratics Functions

The formulae you need to memorise for quadratics functions on the Cambridge IGCSE 0606 paper, with every variable defined in plain English and a note on when to use it.

Quadratic formula
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$a, b, c$
Coefficients of $ax^2 + bx + c = 0$ , with $a \ne 0$
When to use
Solving any quadratic that doesn't factorise easily.
Example
For $2x^2 + 3x - 5 = 0$ : $x = \dfrac{-3 \pm \sqrt{9 + 40}}{4} = \dfrac{-3 \pm 7}{4}$ , so $x = 1$ or $x = -2.5$ .
Discriminant
$\Delta = b^2 - 4ac$
$\Delta$
Discriminant
$a, b, c$
Coefficients of $ax^2 + bx + c = 0$
When to use
Determining the nature of roots: $\Delta > 0$ → two distinct real roots; $\Delta = 0$ → equal roots; $\Delta < 0$ → no real roots.
Completed-square form
$ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}$
$a, b, c$
Coefficients
When to use
Finding the maximum/minimum value, sketching, or solving via square roots. The vertex of the parabola is at $\left(-\frac{b}{2a}, c - \frac{b^2}{4a}\right)$ .

Key Definitions and Keywords — Quadratics Functions

Definitions to memorise and the exact keywords mark schemes credit for quadratics functions answers — sharpened from recent examiner reports for the 2026 0606 sitting.

Quadratic function
Examiner keyword
A function of the form $f(x) = ax^2 + bx + c$ with $a \ne 0$ . Its graph is a PARABOLA.
Discriminant
Examiner keyword
$\Delta = b^2 - 4ac$ . Tells us how many real roots the quadratic has.
Roots
Examiner keyword
The values of $x$ that satisfy $ax^2 + bx + c = 0$ . Same as the x-intercepts of the graph.
Vertex (turning point)
Examiner keyword
The maximum or minimum point of the parabola. At $x = -\frac{b}{2a}$ .
Completed square form
Examiner keyword
Writing $ax^2 + bx + c$ as $a(x + p)^2 + q$ . Reveals the vertex directly: at $(-p, q)$ .

Common Mistakes and Misconceptions — Quadratics Functions

The traps other students keep falling into on quadratics functions questions — taken from recent Cambridge IGCSE 0606 examiner reports and mark schemes — and how to avoid them.

✕Reading the discriminant inequality the wrong way
0606 Examiner Reports 2022-2024
Why it happens
Confusion over $> 0$ vs $\ge 0$ vs $= 0$ .
How to avoid it
$\Delta > 0$ : TWO distinct real roots. $\Delta = 0$ : ONE repeated real root. $\Delta < 0$ : NO real roots. Memorise.
✕Forgetting to flip the inequality when multiplying by negative
Why it happens
Habit of treating equations.
How to avoid it
When you multiply or divide both sides of an inequality by a NEGATIVE number, FLIP the inequality sign. For quadratic inequalities, sketch the parabola to read off the answer.
✕Halving the linear coefficient incorrectly
Why it happens
Sign or arithmetic slip.
How to avoid it
For $x^2 + 6x$ : half of 6 is 3, so $(x + 3)^2 - 9$ . For $x^2 - 6x$ : half of $-6$ is $-3$ , so $(x - 3)^2 - 9$ . Sign matters.
✕Missing the boundary cases in inequalities
Why it happens
Forgetting the difference between strict ( $<$ ) and inclusive ( $\le$ ).
How to avoid it
Use OPEN brackets/circles for strict, CLOSED for inclusive. Match the inequality in the question.

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