What is a quadratic function in the context of Cambridge IGCSE Mathematics?

In Cambridge IGCSE Mathematics, a quadratic function is a type of polynomial function that can be expressed in the standard form y = ax^2 + bx + c, where a, b, and c are constants, and a ≠ 0. The graph of a quadratic function is a parabola, which can open upwards or downwards depending on the sign of the coefficient a.

How do you find the vertex of a quadratic function?

To find the vertex of a quadratic function given in the form y = ax^2 + bx + c, you can use 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 the point (x, y) and represents the maximum or minimum point of the parabola.

What are the methods to solve a quadratic equation?

In Cambridge IGCSE Mathematics, quadratic equations can be solved using several methods: factoring, completing the square, and using the quadratic formula. Factoring involves expressing the quadratic equation as a product of two binomials. Completing the square involves rewriting the equation in the form (x - p)^2 = q. The quadratic formula, x = (-b ± √(b^2 - 4ac))/(2a), can be used when other methods are not feasible.

How can you determine the direction in which a parabola opens?

The direction in which a parabola opens is determined by the sign of the coefficient a in the quadratic function y = ax^2 + bx + c. If a > 0, the parabola opens upwards, resembling a U-shape. If a < 0, the parabola opens downwards, resembling an upside-down U-shape. This is an important concept in Cambridge IGCSE Mathematics when analyzing the graph of quadratic functions.

What is the significance of the discriminant in a quadratic function?

The discriminant of a quadratic function, given by the expression b^2 - 4ac, provides important information about the nature of the roots of the quadratic equation. In Cambridge IGCSE Mathematics, if the discriminant is positive, the equation has two distinct real roots. If it is zero, there is exactly one real root (a repeated root). If the discriminant is negative, the equation has no real roots, indicating that the parabola does not intersect the x-axis.

Why is "Forgetting to compute the y-coordinate of the vertex" a common mistake in Quadratic Functions?

Why it happens: Students find x = -b/2a and stop. How to avoid it: Substitute that x back into the function to get the y-coordinate. Source: 0580/42 — recurring

Why is "Wrong sign on $h$ in $(x - h)^{2}$" a common mistake in Quadratic Functions?

Why it happens: y = (x + 3)^2 has vertex (-3, 0), not (3, 0). How to avoid it: Match to the form (x - h)^2. If you see (x + 3), that's h = -3.

Why is "Concluding "no roots" when the discriminant is positive" a common mistake in Quadratic Functions?

Why it happens: Sign error or misreading = 0 as 0 → two real roots. = 0 → one repeated. < 0 → none.

Why is "Calling a parabola "max" when $a > 0$ (it's a min)" a common mistake in Quadratic Functions?

Why it happens: Confusing positive-coefficient parabolas (smiley) with negative ones (frowny). How to avoid it: Positive a = upward parabola = vertex is a MINIMUM. Negative a = downward = MAXIMUM.

FunctionsCambridge IGCSE Maths 0580 Extended

Quadratic Functions

Q: What is the significance of the discriminant in a quadratic function?

The discriminant of a quadratic function, given by the expression b^2 - 4ac, provides important information about the nature of the roots of the quadratic equation. In Cambridge IGCSE Mathematics, if the discriminant is positive, the equation has two distinct real roots. If it is zero, there is exactly one real root (a repeated root). If the discriminant is negative, the equation has no real roots, indicating that the parabola does not intersect the x-axis.

Q: Why is "Forgetting to compute the y-coordinate of the vertex" a common mistake in Quadratic Functions?

Why it happens: Students find x = -b/2a and stop. How to avoid it: Substitute that x back into the function to get the y-coordinate. Source: 0580/42 — recurring

Q: Why is "Wrong sign on $h$ in $(x - h)^{2}$" a common mistake in Quadratic Functions?

Why it happens: y = (x + 3)^2 has vertex (-3, 0), not (3, 0). How to avoid it: Match to the form (x - h)^2. If you see (x + 3), that's h = -3.

Q: Why is "Concluding "no roots" when the discriminant is positive" a common mistake in Quadratic Functions?

Why it happens: Sign error or misreading = 0 as 0 → two real roots. = 0 → one repeated. < 0 → none.

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Quadratic Functions — Cambridge IGCSE 0580 Maths Extended (2026)

Parabolas, vertices, axes of symmetry, roots and intercepts. Once you can sketch any quadratic from its three forms (standard, factorised, completed-square), every Paper 4 question on the topic is a few minutes' work.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E2.10 — Sketch the curve $y = ax^{2} + bx + c$ and identify roots, vertex, and axis of symmetry.

Three forms of a quadratic

Same parabola, three ways to write it. Each form makes one feature easy to read.

Standard form. $y = ax^{2} + bx + c$ .

Easy to read: $y$ -intercept $= c$ .
Vertex (calculate): $x = -\dfrac{b}{2a}$ , then $y =$ that value substituted.
Roots (calculate): use factorising or the quadratic formula.

Factorised form. $y = a(x - p)(x - q)$ .

Easy to read: $x$ -intercepts $= p, q$ .
Axis of symmetry: $x = \dfrac{p + q}{2}$ (midpoint of roots).
$y$ -intercept (calculate): substitute $x = 0$ .

Completed-square form. $y = a(x - h)^{2} + k$ .

Easy to read: vertex $= (h, k)$ .
Axis of symmetry: $x = h$ .
$y$ -intercept (calculate): substitute $x = 0$ .

Conversions.

Standard ↔ factorised: factorise the trinomial.
Standard ↔ completed-square: complete the square.
Factorised ↔ completed-square: usually go via standard.

Worked. Express $y = x^{2} - 6x + 5$ in completed-square form.

Half $b$ : $-3$ . Square: $9$ .
$y = (x - 3)^{2} - 9 + 5 = (x - 3)^{2} - 4$ .
Vertex: $(3, -4)$ . Axis: $x = 3$ .

Every quadratic question asks for some of these five features.

Standard: $y$ -intercept easy.
Factorised: roots easy.
Completed-square: vertex easy.
Convert by factorising or completing the square.

Sketching a quadratic

Mark the vertex, the axis of symmetry, and any intercepts. Connect with a smooth parabola.

Five-point method.

Direction: opens up if $a > 0$ , down if $a < 0$ .
$y$ -intercept: substitute $x = 0$ .
$x$ -intercepts (roots): solve $y = 0$ — may be 2, 1, or 0 real roots (use discriminant to check).
Vertex (turning point): $x = -\dfrac{b}{2a}$ ; substitute back for $y$ .
Axis of symmetry: vertical line through the vertex.

Worked. Sketch $y = x^{2} - 4x - 5$ .

$a = 1 > 0$ : opens up.
$y$ -intercept: $-5$ .
Roots: factorise → $(x - 5)(x + 1)$ . Roots: $x = 5, -1$ .
Vertex: $x = -\dfrac{-4}{2} = 2$ ; $y = 4 - 8 - 5 = -9$ . So $(2, -9)$ .
Axis: $x = 2$ .

Plot the four anchor points (intercepts plus vertex) and draw a smooth U through them.

No real roots case. $y = x^{2} + 4x + 5$ . Discriminant $= 16 - 20 = -4 < 0$ . The parabola does not cross the $x$ -axis. Vertex: $x = -2, y = 4 - 8 + 5 = 1$ . Sketch shows a U above the $x$ -axis with minimum at $(-2, 1)$ .

Positive a gives a valley (minimum); negative a gives a hill (maximum).

Direction first: $a$ 's sign.
$y$ -intercept: $x = 0$ .
Roots: solve $y = 0$ (or note 'no real roots' if $\Delta < 0$ ).
Vertex: axis of symmetry, then substitute.
Smooth parabola through the anchor points.

Finding the formula from given features

If the question hands you the roots, the vertex, or other points, build the formula from the form that matches.

Given roots and a third point. Use factorised form.

Worked. A quadratic has roots $x = -2$ and $x = 5$ , and passes through $(1, -12)$ .

$y = a(x + 2)(x - 5)$ .
Use the third point: $-12 = a(3)(-4) = -12a \Rightarrow a = 1$ .
$y = (x + 2)(x - 5) = x^{2} - 3x - 10$ .

Given vertex and a point. Use completed-square form.

Worked. Vertex at $(3, -4)$ , passes through $(0, 5)$ .

$y = a(x - 3)^{2} - 4$ .
$5 = a(9) - 4 \Rightarrow a = 1$ .
$y = (x - 3)^{2} - 4$ (same as our earlier example).

Given any three points. Use standard form, set up three equations, solve.

Roots given → factorised form.
Vertex given → completed-square form.
Three arbitrary points → standard form, three equations.
Always use one extra point to fix $a$ .

How it’s examined

Quadratic functions appear most years on Paper 4 as a 5-7 mark sketch-and-solve item. Cambridge gives you the formula, asks for vertex / roots / sketch / range. Examiner reports flag dropped negative signs in the axis formula ( $-b/2a$ ) and forgetting to plot the vertex when sketching.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E2.10); 0580/42 Oct/Nov 2024 — Q14 (sketch quadratic); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

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Step-by-step worked examples — Quadratic Functions

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

1Find the vertex by completing the square
Extended• Adapted from 0580/42 May/Jun 2024 Q14• vertex, completing the square
Question
Find the coordinates of the vertex of $y = x^{2} - 6x + 11$ .
Step-by-step solution
1. Step 1
  Complete the square.
  $y = (x - 3)^{2} - 9 + 11 = (x - 3)^{2} + 2$
2. Step 2
  Vertex is at $(h, k)$ in $(x - h)^{2} + k$ .
Answer
$(3, 2)$ — minimum (parabola opens upward)
2Axis of symmetry from coefficients
Extended• axis of symmetry
Question
Find the axis of symmetry of $y = 2x^{2} + 8x - 1$ .
Step-by-step solution
1. Step 1
  $x = -\dfrac{b}{2a}$ .
  $x = -\dfrac{8}{2 \times 2} = -2$
Answer
$x = -2$
3Find x-intercepts and y-intercept
Extended• intercepts
Question
Find where the graph of $y = x^{2} - 5x + 6$ crosses the axes.
Step-by-step solution
1. Step 1
  y-intercept: substitute $x = 0$ .
  $y = 6 \to (0, 6)$
2. Step 2
  x-intercepts: solve $y = 0$ .
  $(x - 2)(x - 3) = 0 \to x = 2, 3$
Answer
y-intercept: $(0, 6)$ . x-intercepts: $(2, 0)$ and $(3, 0)$ .
4Use the discriminant to count roots
Extended• Adapted from 0580/42 Oct/Nov 2024 Q13• discriminant
Question
How many real roots does $y = 3x^{2} - 4x + 5$ have?
Step-by-step solution
1. Step 1
  Compute $b^{2} - 4ac$ .
  $(-4)^{2} - 4(3)(5) = 16 - 60 = -44$
2. Step 2
  Negative → no real roots.
Answer
No real roots (discriminant $= -44 < 0$ )
5Write a quadratic given its roots
Core• roots, factor form
Question
A quadratic has roots $x = -2$ and $x = 5$ , and passes through $(0, -10)$ . Find the equation in the form $y = ax^{2} + bx + c$ .
Step-by-step solution
1. Step 1
  Write the factor form using the roots.
  $y = a(x + 2)(x - 5)$
2. Step 2
  Substitute $(0, -10)$ to find $a$ .
  $-10 = a(2)(-5) = -10a \implies a = 1$
3. Step 3
  Expand.
  $y = (x + 2)(x - 5) = x^{2} - 3x - 10$
Answer
$y = x^{2} - 3x - 10$
Examiner tip
The examiner report flags candidates often forget the leading coefficient $a$ . Always use a third point to fix $a$ — without it the quadratic isn't unique.
6y-intercept and turning point from a quadratic
Core• intercepts, turning point
Question
For $y = -x^{2} + 4x - 1$ , write down the y-intercept and the coordinates of the turning point.
Step-by-step solution
1. Step 1
  y-intercept: substitute $x = 0$ .
  $y = -1 \to (0, -1)$
2. Step 2
  Axis of symmetry: $x = -\dfrac{b}{2a} = -\dfrac{4}{2(-1)} = 2$ .
3. Step 3
  Substitute $x = 2$ for the $y$ -coordinate of the vertex.
  $y = -(2)^{2} + 4(2) - 1 = -4 + 8 - 1 = 3$
Answer
y-intercept $(0, -1)$ ; turning point (maximum) at $(2, 3)$
7Read the roots and vertex from a sketched parabola
Extended• Adapted from 0580/22 May/Jun 2023 Q12• roots, vertex, graph
Question
A parabola $y = ax^{2} + bx + c$ crosses the $x$ -axis at $(-1, 0)$ and $(7, 0)$ , with minimum value $-16$ . Find $a$ , $b$ and $c$ .
Step-by-step solution
1. Step 1
  Roots $-1$ and $7$ give the factor form.
  $y = a(x + 1)(x - 7)$
2. Step 2
  Axis of symmetry midway between the roots: $x = 3$ . Substitute to find the minimum value.
  $-16 = a(3 + 1)(3 - 7) = a(4)(-4) = -16a \implies a = 1$
3. Step 3
  Expand.
  $y = (x + 1)(x - 7) = x^{2} - 6x - 7$
Answer
$a = 1,\ b = -6,\ c = -7$
Examiner tip
The mark scheme awards method marks for setting up $y = a(x - r_{1})(x - r_{2})$ even before $a$ is found. Showing this factor form is the safest route on graph-interpretation questions.
8Identify a horizontal translation
Extended• transformations
Question
The graph of $y = x^{2}$ is transformed to $y = (x + 3)^{2} - 2$ . Describe the transformation fully.
Step-by-step solution
1. Step 1
  $f(x + 3)$ shifts the graph LEFT by $3$ units (replacing $x$ with $x + 3$ requires $x$ to be $3$ smaller to give the same value).
2. Step 2
  Subtracting $2$ outside the bracket shifts DOWN by $2$ units.
3. Step 3
  Combine.
  $\text{Translation by } \begin{pmatrix} -3 \\ -2 \end{pmatrix}$
Answer
Translation by $\begin{pmatrix} -3 \\ -2 \end{pmatrix}$
Examiner tip
The examiner report flags candidates often shift in the wrong direction for $(x + a)$ . Memorise: $(x + 3)$ inside the bracket means $3$ units LEFT.
9Maximise area with a fixed perimeter
Challenge• Adapted from 0580/42 Oct/Nov 2023 Q9• optimisation, application, vertex
Question
A rectangular pen is built against a long wall, so fencing is needed on only three sides. There are $60$ m of fencing available. Letting the side perpendicular to the wall be $x$ m, find the maximum area of the pen.
Step-by-step solution
1. Step 1
  Two sides of length $x$ and one side parallel to the wall use the fencing.
  $2x + y = 60 \implies y = 60 - 2x$
2. Step 2
  Area as a function of $x$ .
  $A(x) = x(60 - 2x) = 60x - 2x^{2}$
3. Step 3
  Maximum is at the vertex of this downward parabola: $x = -\dfrac{60}{2(-2)} = 15$ .
4. Step 4
  Evaluate.
  $A(15) = 15(60 - 30) = 15 \times 30 = 450$
Answer
Maximum area $= 450\ \text{m}^{2}$ , achieved when $x = 15$ m.
Examiner tip
The mark scheme awards method marks for forming a quadratic in one variable, then for identifying the vertex. Bare answers without the model lose multiple marks.
10Find $k$ so that a line is tangent to a parabola
Challenge• discriminant, tangent, simultaneous
Question
Find the value(s) of $k$ for which the line $y = 2x + k$ is tangent to the parabola $y = x^{2} - 4x + 7$ .
Step-by-step solution
1. Step 1
  Set the expressions equal and rearrange.
  $x^{2} - 4x + 7 = 2x + k \implies x^{2} - 6x + (7 - k) = 0$
2. Step 2
  A tangent intersects the curve in exactly one point, so $\Delta = 0$ .
  $(-6)^{2} - 4(1)(7 - k) = 0$
3. Step 3
  Solve.
  $36 - 28 + 4k = 0 \implies 4k = -8 \implies k = -2$
Answer
$k = -2$
Examiner tip
Tangent-to-curve conditions reduce to a discriminant equation. The examiner report flags candidates who set $\Delta < 0$ (no intersection) instead of $\Delta = 0$ (single touch).

Key Formulae — Quadratic Functions

The formulae you need to memorise for quadratic functions on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Axis of symmetry
$x = -\dfrac{b}{2a}$
$a, b$
coefficients in $y = ax^{2} + bx + c$
When to use
Quickest way to find the vertex's $x$ -coordinate.
Vertex form
$y = a(x - h)^{2} + k,\quad \text{vertex } (h, k)$
When to use
Found by completing the square — gives vertex coordinates directly.
Discriminant
$\Delta = b^{2} - 4ac$
When to use
$\Delta > 0$ : two real roots. $\Delta = 0$ : one repeated root. $\Delta < 0$ : no real roots.

Key Definitions and Keywords — Quadratic Functions

Definitions to memorise and the exact keywords mark schemes credit for quadratic functions answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Quadratic function
Examiner keyword
A function of the form $f(x) = ax^{2} + bx + c$ with $a \neq 0$ . Its graph is a parabola.
Vertex / turning point
Examiner keyword
The lowest (minimum) or highest (maximum) point of a parabola. Lies on the axis of symmetry.
Axis of symmetry
Examiner keyword
The vertical line through the vertex that divides the parabola into two mirror-image halves.
Roots / x-intercepts
Values of $x$ for which $y = 0$ — where the parabola crosses the x-axis.
Direction of opening
$a > 0$ → opens upward (smile). $a < 0$ → opens downward (frown).

Common Mistakes and Misconceptions — Quadratic Functions

The traps other students keep falling into on quadratic functions questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Forgetting to compute the y-coordinate of the vertex
0580/42 — recurring
Why it happens
Students find $x = -\frac{b}{2a}$ and stop.
How to avoid it
Substitute that $x$ back into the function to get the $y$ -coordinate.
✕Wrong sign on $h$ in $(x - h)^{2}$
Why it happens
$y = (x + 3)^{2}$ has vertex $(-3, 0)$ , not $(3, 0)$ .
How to avoid it
Match to the form $(x - h)^{2}$ . If you see $(x + 3)$ , that's $h = -3$ .
✕Concluding "no roots" when the discriminant is positive
Why it happens
Sign error or misreading $\Delta = 0$ as $\Delta < 0$ .
How to avoid it
$\Delta > 0$ → two real roots. $\Delta = 0$ → one repeated. $\Delta < 0$ → none.
✕Calling a parabola "max" when $a > 0$ (it's a min)
Why it happens
Confusing positive-coefficient parabolas (smiley) with negative ones (frowny).
How to avoid it
Positive $a$ = upward parabola = vertex is a MINIMUM. Negative $a$ = downward = MAXIMUM.

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FunctionsCambridge IGCSE Maths 0580 Extended

Quadratic Functions

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Quadratic Functions — Cambridge IGCSE 0580 Maths Extended (2026)

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E2.10 — Sketch the curve $y = ax^{2} + bx + c$ and identify roots, vertex, and axis of symmetry.

Three forms of a quadratic

Same parabola, three ways to write it. Each form makes one feature easy to read.

Standard form. $y = ax^{2} + bx + c$ .

Easy to read: $y$ -intercept $= c$ .
Vertex (calculate): $x = -\dfrac{b}{2a}$ , then $y =$ that value substituted.
Roots (calculate): use factorising or the quadratic formula.

Factorised form. $y = a(x - p)(x - q)$ .

Easy to read: $x$ -intercepts $= p, q$ .
Axis of symmetry: $x = \dfrac{p + q}{2}$ (midpoint of roots).
$y$ -intercept (calculate): substitute $x = 0$ .

Completed-square form. $y = a(x - h)^{2} + k$ .

Easy to read: vertex $= (h, k)$ .
Axis of symmetry: $x = h$ .
$y$ -intercept (calculate): substitute $x = 0$ .

Conversions.

Standard ↔ factorised: factorise the trinomial.
Standard ↔ completed-square: complete the square.
Factorised ↔ completed-square: usually go via standard.

Worked. Express $y = x^{2} - 6x + 5$ in completed-square form.

Half $b$ : $-3$ . Square: $9$ .
$y = (x - 3)^{2} - 9 + 5 = (x - 3)^{2} - 4$ .
Vertex: $(3, -4)$ . Axis: $x = 3$ .

Every quadratic question asks for some of these five features.

Standard: $y$ -intercept easy.
Factorised: roots easy.
Completed-square: vertex easy.
Convert by factorising or completing the square.

Sketching a quadratic

Mark the vertex, the axis of symmetry, and any intercepts. Connect with a smooth parabola.

Five-point method.

Direction: opens up if $a > 0$ , down if $a < 0$ .
$y$ -intercept: substitute $x = 0$ .
$x$ -intercepts (roots): solve $y = 0$ — may be 2, 1, or 0 real roots (use discriminant to check).
Vertex (turning point): $x = -\dfrac{b}{2a}$ ; substitute back for $y$ .
Axis of symmetry: vertical line through the vertex.

Worked. Sketch $y = x^{2} - 4x - 5$ .

$a = 1 > 0$ : opens up.
$y$ -intercept: $-5$ .
Roots: factorise → $(x - 5)(x + 1)$ . Roots: $x = 5, -1$ .
Vertex: $x = -\dfrac{-4}{2} = 2$ ; $y = 4 - 8 - 5 = -9$ . So $(2, -9)$ .
Axis: $x = 2$ .

Plot the four anchor points (intercepts plus vertex) and draw a smooth U through them.

Positive a gives a valley (minimum); negative a gives a hill (maximum).

Direction first: $a$ 's sign.
$y$ -intercept: $x = 0$ .
Roots: solve $y = 0$ (or note 'no real roots' if $\Delta < 0$ ).
Vertex: axis of symmetry, then substitute.
Smooth parabola through the anchor points.

Finding the formula from given features

If the question hands you the roots, the vertex, or other points, build the formula from the form that matches.

Given roots and a third point. Use factorised form.

Worked. A quadratic has roots $x = -2$ and $x = 5$ , and passes through $(1, -12)$ .

$y = a(x + 2)(x - 5)$ .
Use the third point: $-12 = a(3)(-4) = -12a \Rightarrow a = 1$ .
$y = (x + 2)(x - 5) = x^{2} - 3x - 10$ .

Given vertex and a point. Use completed-square form.

Worked. Vertex at $(3, -4)$ , passes through $(0, 5)$ .

$y = a(x - 3)^{2} - 4$ .
$5 = a(9) - 4 \Rightarrow a = 1$ .
$y = (x - 3)^{2} - 4$ (same as our earlier example).

Given any three points. Use standard form, set up three equations, solve.

Roots given → factorised form.
Vertex given → completed-square form.
Three arbitrary points → standard form, three equations.
Always use one extra point to fix $a$ .

How it’s examined

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E2.10); 0580/42 Oct/Nov 2024 — Q14 (sketch quadratic); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

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Step-by-step worked examples — Quadratic Functions

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

1Find the vertex by completing the square
Extended• Adapted from 0580/42 May/Jun 2024 Q14• vertex, completing the square
Question
Find the coordinates of the vertex of $y = x^{2} - 6x + 11$ .
Step-by-step solution
1. Step 1
  Complete the square.
  $y = (x - 3)^{2} - 9 + 11 = (x - 3)^{2} + 2$
2. Step 2
  Vertex is at $(h, k)$ in $(x - h)^{2} + k$ .
Answer
$(3, 2)$ — minimum (parabola opens upward)
2Axis of symmetry from coefficients
Extended• axis of symmetry
Question
Find the axis of symmetry of $y = 2x^{2} + 8x - 1$ .
Step-by-step solution
1. Step 1
  $x = -\dfrac{b}{2a}$ .
  $x = -\dfrac{8}{2 \times 2} = -2$
Answer
$x = -2$
3Find x-intercepts and y-intercept
Extended• intercepts
Question
Find where the graph of $y = x^{2} - 5x + 6$ crosses the axes.
Step-by-step solution
1. Step 1
  y-intercept: substitute $x = 0$ .
  $y = 6 \to (0, 6)$
2. Step 2
  x-intercepts: solve $y = 0$ .
  $(x - 2)(x - 3) = 0 \to x = 2, 3$
Answer
y-intercept: $(0, 6)$ . x-intercepts: $(2, 0)$ and $(3, 0)$ .
4Use the discriminant to count roots
Extended• Adapted from 0580/42 Oct/Nov 2024 Q13• discriminant
Question
How many real roots does $y = 3x^{2} - 4x + 5$ have?
Step-by-step solution
1. Step 1
  Compute $b^{2} - 4ac$ .
  $(-4)^{2} - 4(3)(5) = 16 - 60 = -44$
2. Step 2
  Negative → no real roots.
Answer
No real roots (discriminant $= -44 < 0$ )
5Write a quadratic given its roots
Core• roots, factor form
Question
A quadratic has roots $x = -2$ and $x = 5$ , and passes through $(0, -10)$ . Find the equation in the form $y = ax^{2} + bx + c$ .
Step-by-step solution
1. Step 1
  Write the factor form using the roots.
  $y = a(x + 2)(x - 5)$
2. Step 2
  Substitute $(0, -10)$ to find $a$ .
  $-10 = a(2)(-5) = -10a \implies a = 1$
3. Step 3
  Expand.
  $y = (x + 2)(x - 5) = x^{2} - 3x - 10$
Answer
$y = x^{2} - 3x - 10$
Examiner tip
The examiner report flags candidates often forget the leading coefficient $a$ . Always use a third point to fix $a$ — without it the quadratic isn't unique.
6y-intercept and turning point from a quadratic
Core• intercepts, turning point
Question
For $y = -x^{2} + 4x - 1$ , write down the y-intercept and the coordinates of the turning point.
Step-by-step solution
1. Step 1
  y-intercept: substitute $x = 0$ .
  $y = -1 \to (0, -1)$
2. Step 2
  Axis of symmetry: $x = -\dfrac{b}{2a} = -\dfrac{4}{2(-1)} = 2$ .
3. Step 3
  Substitute $x = 2$ for the $y$ -coordinate of the vertex.
  $y = -(2)^{2} + 4(2) - 1 = -4 + 8 - 1 = 3$
Answer
y-intercept $(0, -1)$ ; turning point (maximum) at $(2, 3)$
7Read the roots and vertex from a sketched parabola
Extended• Adapted from 0580/22 May/Jun 2023 Q12• roots, vertex, graph
Question
A parabola $y = ax^{2} + bx + c$ crosses the $x$ -axis at $(-1, 0)$ and $(7, 0)$ , with minimum value $-16$ . Find $a$ , $b$ and $c$ .
Step-by-step solution
1. Step 1
  Roots $-1$ and $7$ give the factor form.
  $y = a(x + 1)(x - 7)$
2. Step 2
  Axis of symmetry midway between the roots: $x = 3$ . Substitute to find the minimum value.
  $-16 = a(3 + 1)(3 - 7) = a(4)(-4) = -16a \implies a = 1$
3. Step 3
  Expand.
  $y = (x + 1)(x - 7) = x^{2} - 6x - 7$
Answer
$a = 1,\ b = -6,\ c = -7$
Examiner tip
The mark scheme awards method marks for setting up $y = a(x - r_{1})(x - r_{2})$ even before $a$ is found. Showing this factor form is the safest route on graph-interpretation questions.
8Identify a horizontal translation
Extended• transformations
Question
The graph of $y = x^{2}$ is transformed to $y = (x + 3)^{2} - 2$ . Describe the transformation fully.
Step-by-step solution
1. Step 1
  $f(x + 3)$ shifts the graph LEFT by $3$ units (replacing $x$ with $x + 3$ requires $x$ to be $3$ smaller to give the same value).
2. Step 2
  Subtracting $2$ outside the bracket shifts DOWN by $2$ units.
3. Step 3
  Combine.
  $\text{Translation by } \begin{pmatrix} -3 \\ -2 \end{pmatrix}$
Answer
Translation by $\begin{pmatrix} -3 \\ -2 \end{pmatrix}$
Examiner tip
The examiner report flags candidates often shift in the wrong direction for $(x + a)$ . Memorise: $(x + 3)$ inside the bracket means $3$ units LEFT.
9Maximise area with a fixed perimeter
Challenge• Adapted from 0580/42 Oct/Nov 2023 Q9• optimisation, application, vertex
Question
A rectangular pen is built against a long wall, so fencing is needed on only three sides. There are $60$ m of fencing available. Letting the side perpendicular to the wall be $x$ m, find the maximum area of the pen.
Step-by-step solution
1. Step 1
  Two sides of length $x$ and one side parallel to the wall use the fencing.
  $2x + y = 60 \implies y = 60 - 2x$
2. Step 2
  Area as a function of $x$ .
  $A(x) = x(60 - 2x) = 60x - 2x^{2}$
3. Step 3
  Maximum is at the vertex of this downward parabola: $x = -\dfrac{60}{2(-2)} = 15$ .
4. Step 4
  Evaluate.
  $A(15) = 15(60 - 30) = 15 \times 30 = 450$
Answer
Maximum area $= 450\ \text{m}^{2}$ , achieved when $x = 15$ m.
Examiner tip
The mark scheme awards method marks for forming a quadratic in one variable, then for identifying the vertex. Bare answers without the model lose multiple marks.
10Find $k$ so that a line is tangent to a parabola
Challenge• discriminant, tangent, simultaneous
Question
Find the value(s) of $k$ for which the line $y = 2x + k$ is tangent to the parabola $y = x^{2} - 4x + 7$ .
Step-by-step solution
1. Step 1
  Set the expressions equal and rearrange.
  $x^{2} - 4x + 7 = 2x + k \implies x^{2} - 6x + (7 - k) = 0$
2. Step 2
  A tangent intersects the curve in exactly one point, so $\Delta = 0$ .
  $(-6)^{2} - 4(1)(7 - k) = 0$
3. Step 3
  Solve.
  $36 - 28 + 4k = 0 \implies 4k = -8 \implies k = -2$
Answer
$k = -2$
Examiner tip
Tangent-to-curve conditions reduce to a discriminant equation. The examiner report flags candidates who set $\Delta < 0$ (no intersection) instead of $\Delta = 0$ (single touch).

Key Formulae — Quadratic Functions

The formulae you need to memorise for quadratic functions on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Axis of symmetry
$x = -\dfrac{b}{2a}$
$a, b$
coefficients in $y = ax^{2} + bx + c$
When to use
Quickest way to find the vertex's $x$ -coordinate.
Vertex form
$y = a(x - h)^{2} + k,\quad \text{vertex } (h, k)$
When to use
Found by completing the square — gives vertex coordinates directly.
Discriminant
$\Delta = b^{2} - 4ac$
When to use
$\Delta > 0$ : two real roots. $\Delta = 0$ : one repeated root. $\Delta < 0$ : no real roots.

Key Definitions and Keywords — Quadratic Functions

Definitions to memorise and the exact keywords mark schemes credit for quadratic functions answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Quadratic function
Examiner keyword
A function of the form $f(x) = ax^{2} + bx + c$ with $a \neq 0$ . Its graph is a parabola.
Vertex / turning point
Examiner keyword
The lowest (minimum) or highest (maximum) point of a parabola. Lies on the axis of symmetry.
Axis of symmetry
Examiner keyword
The vertical line through the vertex that divides the parabola into two mirror-image halves.
Roots / x-intercepts
Values of $x$ for which $y = 0$ — where the parabola crosses the x-axis.
Direction of opening
$a > 0$ → opens upward (smile). $a < 0$ → opens downward (frown).

Common Mistakes and Misconceptions — Quadratic Functions

The traps other students keep falling into on quadratic functions questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Forgetting to compute the y-coordinate of the vertex
0580/42 — recurring
Why it happens
Students find $x = -\frac{b}{2a}$ and stop.
How to avoid it
Substitute that $x$ back into the function to get the $y$ -coordinate.
✕Wrong sign on $h$ in $(x - h)^{2}$
Why it happens
$y = (x + 3)^{2}$ has vertex $(-3, 0)$ , not $(3, 0)$ .
How to avoid it
Match to the form $(x - h)^{2}$ . If you see $(x + 3)$ , that's $h = -3$ .
✕Concluding "no roots" when the discriminant is positive
Why it happens
Sign error or misreading $\Delta = 0$ as $\Delta < 0$ .
How to avoid it
$\Delta > 0$ → two real roots. $\Delta = 0$ → one repeated. $\Delta < 0$ → none.
✕Calling a parabola "max" when $a > 0$ (it's a min)
Why it happens
Confusing positive-coefficient parabolas (smiley) with negative ones (frowny).
How to avoid it
Positive $a$ = upward parabola = vertex is a MINIMUM. Negative $a$ = downward = MAXIMUM.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Quadratic Functions — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Quadratic Functions

Short Study Notes in the form of Slides

Short Study Notes — Quadratic Functions

Detailed Notes

What you’ll learn

How it’s examined

1Find the vertex by completing the square

2Axis of symmetry from coefficients

3Find x-intercepts and y-intercept

4Use the discriminant to count roots

5Write a quadratic given its roots

6y-intercept and turning point from a quadratic

7Read the roots and vertex from a sketched parabola

8Identify a horizontal translation

9Maximise area with a fixed perimeter

10Find kkk so that a line is tangent to a parabola

Axis of symmetry

Vertex form

Discriminant

Quadratic function

Vertex / turning point

Axis of symmetry

Roots / x-intercepts

Direction of opening

✕Forgetting to compute the y-coordinate of the vertex

✕Wrong sign on hhh in (x−h)2(x - h)^{2}(x−h)2

✕Concluding "no roots" when the discriminant is positive

✕Calling a parabola "max" when a>0a > 0a>0 (it's a min)

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Quadratic Functions — frequently asked questions

Quadratic Functions

Short Study Notes in the form of Slides

Short Study Notes — Quadratic Functions

Detailed Notes

What you’ll learn

How it’s examined

1Find the vertex by completing the square

2Axis of symmetry from coefficients

3Find x-intercepts and y-intercept

4Use the discriminant to count roots

5Write a quadratic given its roots

6y-intercept and turning point from a quadratic

7Read the roots and vertex from a sketched parabola

8Identify a horizontal translation

9Maximise area with a fixed perimeter

10Find kkk so that a line is tangent to a parabola

Axis of symmetry

Vertex form

Discriminant

Quadratic function

Vertex / turning point

Axis of symmetry

Roots / x-intercepts

Direction of opening

✕Forgetting to compute the y-coordinate of the vertex

✕Wrong sign on hhh in (x−h)2(x - h)^{2}(x−h)2

✕Concluding "no roots" when the discriminant is positive

✕Calling a parabola "max" when a>0a > 0a>0 (it's a min)

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Quadratic Functions — frequently asked questions

10Find $k$ so that a line is tangent to a parabola

✕Wrong sign on $h$ in $(x - h)^{2}$

✕Calling a parabola "max" when $a > 0$ (it's a min)

10Find $k$ so that a line is tangent to a parabola

✕Wrong sign on $h$ in $(x - h)^{2}$

✕Calling a parabola "max" when $a > 0$ (it's a min)