Quadratic Functions

Quadratic functions are polynomials of degree 2 with the general form f(x) = ax² + bx + c, where a ≠ 0. This lesson covers solving quadratics, completing the square, the discriminant, and sketching parabolas — all essential for the Edexcel 9MA0 specification.

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Forms of a Quadratic

Form	Expression	Key Information
General form	ax² + bx + c	Coefficients visible; y-intercept is c
Factorised form	a(x − p)(x − q)	Roots are x = p and x = q
Completed square	a(x − h)² + k	Vertex at (h, k); line of symmetry x = h

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Solving Quadratic Equations

Method 1: Factorisation

If ax² + bx + c can be written as a product of two linear factors, set each factor to zero.

Example: Solve x² − 5x + 6 = 0.

• (x − 2)(x − 3) = 0
• x = 2 or x = 3

Example: Solve 6x² + x − 2 = 0.

• (2x − 1)(3x + 2) = 0
• x = 1/2 or x = −2/3

Method 2: The Quadratic Formula

For ax² + bx + c = 0, the solutions are:

x = (−b ± √(b² − 4ac)) / 2a

This formula always works, whether or not the quadratic factorises.

Example: Solve 2x² − 3x − 4 = 0.

• a = 2, b = −3, c = −4
• b² − 4ac = 9 + 32 = 41
• x = (3 ± √41) / 4
• x = (3 + √41)/4 or x = (3 − √41)/4

Method 3: Completing the Square

Write ax² + bx + c in the form a(x + p)² + q.

For x² + bx + c:

• Half the coefficient of x: p = b/2
• Square it and adjust: x² + bx + c = (x + b/2)² − (b/2)² + c

Example: Write x² + 6x + 2 in completed square form.

• (x + 3)² − 9 + 2 = (x + 3)² − 7

For ax² + bx + c where a ≠ 1, first factor out a:

• 2x² + 12x + 5 = 2(x² + 6x) + 5 = 2((x + 3)² − 9) + 5 = 2(x + 3)² − 13

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The Discriminant

The expression b² − 4ac is called the discriminant (often written as Δ or D). It determines the nature of the roots.

Discriminant	Nature of Roots	Graphical Meaning
b² − 4ac > 0	Two distinct real roots	Parabola crosses x-axis twice
b² − 4ac = 0	One repeated real root	Parabola touches x-axis (tangent)
b² − 4ac < 0	No real roots	Parabola does not cross x-axis

Example: For what values of k does kx² + 4x + 1 = 0 have two distinct real roots?

• b² − 4ac > 0
• 16 − 4k > 0
• k < 4
• But also k ≠ 0 (otherwise it is not quadratic), so k < 4 and k ≠ 0.

Exam Tip: When the question says "equal roots", set b² − 4ac = 0. When it says "no real roots", set b² − 4ac < 0. Always state the condition clearly before substituting.

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Sketching Quadratic Graphs

A quadratic y = ax² + bx + c produces a parabola.

• If a > 0, the parabola opens upwards (U-shape) — the vertex is a minimum.
• If a < 0, the parabola opens downwards (∩-shape) — the vertex is a maximum.

Steps for Sketching

1. y-intercept: Set x = 0, giving y = c.
2. x-intercepts (roots): Solve ax² + bx + c = 0.
3. Vertex: Use completed square form or the formula x = −b/(2a), then find y.
4. Shape: Check the sign of a.

Example: Sketch y = x² − 4x + 3.

• y-intercept: (0, 3)
• Roots: x² − 4x + 3 = (x − 1)(x − 3) = 0, so x = 1, x = 3
• Vertex: x = −(−4)/(2 × 1) = 2, y = 4 − 8 + 3 = −1, so vertex (2, −1)
• a > 0 so U-shape

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Quadratic Inequalities

To solve a quadratic inequality, first solve the corresponding equation, then use a sketch to determine the solution set.

Example: Solve x² − 5x + 6 > 0.

• Roots: x = 2 and x = 3
• The parabola (U-shape, a > 0) is above the x-axis when x < 2 or x > 3.
• Solution: x < 2 or x > 3

Example: Solve x² − 5x + 6 ≤ 0.

• The parabola is on or below the x-axis when 2 ≤ x ≤ 3.

Exam Tip: Always sketch the graph when solving quadratic inequalities. A common error is writing x > 3 and x > 2 instead of x < 2 or x > 3 for a > 0 case. The sketch prevents this.

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Key Terms

Term	Definition
Quadratic	A polynomial of degree 2 in the form ax² + bx + c
Discriminant	The value b² − 4ac, which determines the nature of the roots
Completing the square	Rewriting a quadratic in the form a(x + p)² + q
Vertex	The turning point of a parabola
Parabola	The U-shaped (or ∩-shaped) curve produced by a quadratic function

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Summary

• Quadratic equations can be solved by factorisation, the quadratic formula or completing the square.
• The discriminant b² − 4ac tells you how many real roots the equation has.
• Completing the square reveals the vertex and line of symmetry.
• Always sketch a quadratic when solving inequalities — identify the roots first, then use the shape of the parabola to determine the solution set.

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A-Level Deep Dive: Quadratic Functions

Spec mapping

Edexcel 9MA0-01 specification section 2 — Algebra and functions, sub-strands 2.3 to 2.6 covers work with quadratic functions and their graphs; the discriminant of a quadratic function, including the conditions for real and repeated roots; completing the square; solution of quadratic equations including those that reduce to a quadratic equation in another variable (refer to the official specification document for exact wording). This is one of the most heavily examined topics on Paper 1, and it is foundational for section 7 (Differentiation — finding turning points), section 8 (Integration) and sections 14–16 (Mechanics — projectile motion) in 9MA0-03. The Edexcel formula booklet provides the quadratic-formula solution under "Pure Mathematics — quadratic equations".

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Worked example with full mark scheme

Question (9 marks): The quadratic function $f(x) = 2x^2 + (k-3)x + (k+5)$f(x)=2x2+(k−3)x+(k+5) has two distinct real roots, where $k$k is a real constant.

(a) Show that $k$k satisfies the inequality $k^2 - 14k - 31 > 0$k2−14k−31>0. (3)

(b) Hence find the set of possible values of $k$k, giving your answer in exact form. (4)

(c) For the case $k = 1$k=1, write $f(x)$f(x) in completed-square form and state the coordinates of the vertex of the curve $y = f(x)$y=f(x). (2)

Solution with mark scheme:

(a) For two distinct real roots, the discriminant must be strictly positive:

$b^2 - 4ac > 0$b2−4ac>0

B1 — stating the correct condition. Note the strict inequality; $\geq 0$≥0 would lose this mark because it permits the repeated-root boundary.

Here $a = 2$a=2, $b = (k - 3)$b=(k−3), $c = (k + 5)$c=(k+5):

$(k - 3)^2 - 4(2)(k + 5) > 0$(k−3)2−4(2)(k+5)>0

M1 — substituting correctly into the discriminant. The most common error is writing $b = k - 3$b=k−3 but then computing $b^2 = k^2 - 3$b2=k2−3 instead of expanding the bracket fully.

Expanding: $(k - 3)^2 = k^2 - 6k + 9$(k−3)2=k2−6k+9 and $4 \cdot 2 \cdot (k + 5) = 8k + 40$4⋅2⋅(k+5)=8k+40.

$k^2 - 6k + 9 - 8k - 40 > 0$k2−6k+9−8k−40>0 $k^2 - 14k - 31 > 0$k2−14k−31>0

A1 — printed answer obtained. Examiners are strict on "show that" outputs: the inequality sign and the exact form of the constants must match.

(b) Solve $k^2 - 14k - 31 = 0$k2−14k−31=0 to find the critical values.

Using the quadratic formula: $k = \dfrac{14 \pm \sqrt{196 + 124}}{2} = \dfrac{14 \pm \sqrt{320}}{2}$k=214±196+124​​=214±320​​.

M1 — applying the quadratic formula to the inequality boundary. Equivalently, completing the square on $k^2 - 14k - 31$k2−14k−31 gives $(k - 7)^2 - 80 = 0$(k−7)2−80=0, so $k = 7 \pm \sqrt{80}$k=7±80​.

Simplify: $\sqrt{320} = \sqrt{64 \cdot 5} = 8\sqrt{5}$320​=64⋅5​=85​, so $k = \dfrac{14 \pm 8\sqrt{5}}{2} = 7 \pm 4\sqrt{5}$k=214±85​​=7±45​.

A1 — both critical values in simplified surd form. Leaving the answer as $k = (14 \pm \sqrt{320})/2$k=(14±320​)/2 is not sufficient; the spec demands fully-simplified surds.

M1 (AO2) — recognising that since $k^2 - 14k - 31$k2−14k−31 is a U-shaped parabola in $k$k, the inequality $> 0$>0 is satisfied outside the critical values.

$k < 7 - 4\sqrt{5} \quad \text{or} \quad k > 7 + 4\sqrt{5}$k<7−45​ork>7+45​

A1 — final answer with both branches explicitly stated. Writing only one branch loses this mark.

(c) When $k = 1$k=1: $f(x) = 2x^2 + (1 - 3)x + (1 + 5) = 2x^2 - 2x + 6$f(x)=2x2+(1−3)x+(1+5)=2x2−2x+6.

Factor out the leading coefficient on the $x$x-terms: $2(x^2 - x) + 6$2(x2−x)+6.

Complete the square inside: $x^2 - x = (x - \tfrac{1}{2})^2 - \tfrac{1}{4}$x2−x=(x−21​)2−41​.

So $f(x) = 2\left[(x - \tfrac{1}{2})^2 - \tfrac{1}{4}\right] + 6 = 2(x - \tfrac{1}{2})^2 - \tfrac{1}{2} + 6 = 2\left(x - \tfrac{1}{2}\right)^2 + \tfrac{11}{2}$f(x)=2[(x−21​)2−41​]+6=2(x−21​)2−21​+6=2(x−21​)2+211​.

M1 — completed-square method shown clearly with bracket expansion.

A1 — vertex at $\left(\tfrac{1}{2},\, \tfrac{11}{2}\right)$(21​,211​), read directly from the form $a(x - h)^2 + k$a(x−h)2+k.

Total: 9 marks (B1 M3 A4 M1 = 3 + 4 + 2 split as shown).

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Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (8 marks): $g(x) = 2x^2 + 3x - 5$g(x)=2x2+3x−5.

(a) Express $g(x)$g(x) in the form $a(x + b)^2 + c$a(x+b)2+c, where $a$a, $b$b and $c$c are constants to be determined. (3)

(b) Hence sketch the graph of $y = g(x)$y=g(x), showing the coordinates of the vertex and the points where the curve meets the coordinate axes. (3)

(c) Use your completed-square form to write down the equation of the line of symmetry of $y = g(x)$y=g(x). (1)

(d) Find the range of $g$g, given the domain is $x \in \mathbb{R}$x∈R. (1)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — factoring out 2 from the $x$x-terms: $2(x^2 + \tfrac{3}{2}x) - 5$2(x2+23​x)−5.
• M1 (AO1.1b) — completing the square inside: $x^2 + \tfrac{3}{2}x = (x + \tfrac{3}{4})^2 - \tfrac{9}{16}$x2+23​x=(x+43​)2−169​.
• A1 (AO1.1b) — final form $2(x + \tfrac{3}{4})^2 - \tfrac{49}{8}$2(x+43​)2−849​.

(b)

• B1 (AO1.1b) — vertex at $\left(-\tfrac{3}{4},\, -\tfrac{49}{8}\right)$(−43​,−849​).
• B1 (AO1.1b) — y-intercept $(0, -5)$(0,−5) and x-intercepts at $x = 1$x=1 and $x = -\tfrac{5}{2}$x=−25​ (from factorising $(2x + 5)(x - 1)$(2x+5)(x−1)).
• B1 (AO2.4) — correct U-shape with all key features labelled.

(c)

• B1 (AO2.2a) — line of symmetry $x = -\tfrac{3}{4}$x=−43​.

(d)

• B1 (AO2.5) — range $g(x) \geq -\tfrac{49}{8}$g(x)≥−849​, written using inequality or interval notation.

Total: 8 marks split AO1 = 5, AO2 = 3. This is a textbook AO1-heavy question; the AO2 marks come from "translation" — converting algebraic facts (vertex coordinates) into geometric language (range, line of symmetry).

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Synoptic links

Connects to:

1. Section 7 — Differentiation: the vertex of a quadratic is its turning point. Setting $f'(x) = 0$f′(x)=0 on $f(x) = ax^2 + bx + c$f(x)=ax2+bx+c gives $2ax + b = 0$2ax+b=0, so $x = -b/(2a)$x=−b/(2a) — exactly matching the completed-square $x$x-coordinate. Calculus is the systematic generalisation of "find the turning point".

2. Section 8 — Integration: the area between a parabola and the $x$x-axis between its roots is computed by integrating, e.g. $\int_p^q (x - p)(x - q)\,dx = -\tfrac{1}{6}(q - p)^3$∫pq​(x−p)(x−q)dx=−61​(q−p)3 — a deep result first encountered as an integration exercise.

3. Section 12 — Vectors and Section 16 — Projectile motion (Mechanics 9MA0-03): projectile trajectories under gravity are quadratic in time. The maximum height (vertex of the parabolic path) is found by completing the square on the height-time equation $h(t) = u\sin\theta\, t - \tfrac{1}{2}gt^2$h(t)=usinθt−21​gt2.

4. Section 5 — Trigonometric identities: equations like $2\cos^2 x - 3\cos x + 1 = 0$2cos2x−3cosx+1=0 are quadratic in $\cos x$cosx. The substitution $u = \cos x$u=cosx converts them to standard quadratics, then back-substitute and solve in the required interval.

5. Section 11 — Differential equations (Year 2): the auxiliary equation of a second-order linear ODE is a quadratic; its discriminant determines whether the solution is exponential, oscillatory or critically damped — directly importing the trichotomy you learned at A-Level.

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Mark-scheme literacy

Quadratic-function questions on 9MA0-01 typically split AO marks as follows: