Lesson 2: Quadratics and Their Forms

The quadratic function f(x) = ax² + bx + c (a ≠ 0) is the hardest-working function in the AA course (Topic 2). Its graph is a parabola, opening upwards when a > 0 and downwards when a < 0, with a vertical axis of symmetry. IB questions rarely ask you merely to "solve a quadratic"; they ask you to choose the *form* of the quadratic that reveals the information you need, and to interpret that information in context. Three forms dominate:

• Standard form ax² + bx + c — reveals the y-intercept c at a glance.
• Vertex (completed square) form a(x − h)² + k — reveals the vertex (h, k), hence the maximum or minimum value.
• Factored form a(x − p)(x − q) — reveals the x-intercepts (roots) p and q, and the axis x = (p + q)/2 runs midway between them.

The discriminant and the nature of roots

The discriminant Δ = b² − 4ac, taken from the quadratic formula x = (−b ± √(b² − 4ac)) / (2a), decides the number of real roots before you solve anything:

Examiners particularly like the reverse direction: "the equation has two distinct real roots / a repeated root / no real roots, find the range of k." Translate the words into Δ > 0, Δ = 0 or Δ < 0 and solve the resulting inequality — this is a Reasoning-mark favourite in Paper 1 Section B.

Worked Example 1 — choosing the form. A quadratic has vertex (3, −4) and passes through (0, 5). Find its equation in standard form and its x-intercepts.

Solution. Start from the vertex form, which is built for exactly this data: f(x) = a(x − 3)² − 4. The point (0, 5) gives 9a − 4 = 5, so a = 1 and f(x) = (x − 3)² − 4 = x² − 6x + 5. Now the factored form appears for free: f(x) = (x − 1)(x − 5), so the x-intercepts are 1 and 5 — consistent with the axis x = 3 lying midway between them. One piece of data (vertex and a point) unlocked all three forms.

Worked Example 2 — discriminant in reverse (past-paper style). The equation x² + kx + 9 = 0 has two distinct real roots. Find the set of values of k.

Solution. Two distinct real roots ⟺ Δ > 0. Here Δ = k² − 36, so k² − 36 > 0, giving k < −6 or k > 6. On Paper 1 this must be done by hand: solve k² = 36, then reason with the sign of the upward parabola in k. A very common examiner complaint is candidates answering −6 < k < 6, the inequality reversed — always sanity-check with one test value (k = 0 gives x² + 9 = 0, no real roots, so 0 must be excluded — it is, in the correct answer).

Completing the square and optimisation

Completing the square converts ax² + bx + c into vertex form and is the engine behind optimisation, the vertex formula x = −b/(2a), and (at HL) the derivation of the quadratic formula itself. The algorithm: halve the coefficient of x, square it, add and subtract. For 2x² − 12x + 7: factor the 2 from the x-terms, 2(x² − 6x) + 7 = 2[(x − 3)² − 9] + 7 = 2(x − 3)² − 11, vertex (3, −11), minimum value −11 since a > 0.

Worked Example 3 — a modelling optimisation. A farmer has 60 m of fencing for three sides of a rectangular pen against a wall. Find the maximum area.

Solution. Let each side perpendicular to the wall be x m; the side parallel to the wall is 60 − 2x. Area A(x) = x(60 − 2x) = −2x² + 60x, a downward parabola, with contextual domain 0 < x < 30. Complete the square: A(x) = −2(x − 15)² + 450. The maximum area is 450 m² at x = 15 m. Technology check (Paper 2): graph A(x) on the GDC and use the maximum-finder — but on Paper 1 the completed square *is* the method, and it also proves the value is a maximum, not just a candidate.

Intersections and systems

Where a line meets a parabola, substitute to get a single quadratic; the discriminant of that quadratic counts the intersections (two points, tangent, or none). A line tangent to a parabola is the Δ = 0 case — a bridge to the calculus of Lesson 11, where tangency is handled by derivatives instead. With the GDC (Paper 2), use the intersection solver, but state the equations you solved: method marks follow your setup, not the button presses.

Common mistakes

• Sign errors completing the square, especially with a negative b or when factoring out a coefficient — write the intermediate line, don't compress.
• Discriminant used, then forgotten: finding Δ > 0 but reporting the roots instead of the condition on k.
• "Repeated root" answered with the root's value when the question asked for k.
• Vertex read as (h, k) from a(x + h)² + k — the sign of h flips: (x + 3)² has vertex at x = −3.
• Context dropped: optimisation answers without units, or with x outside its sensible domain (negative lengths).
• Premature rounding: keep exact values (fractions, surds) through the working; round only the final answer, to three significant figures.

IB Exam Experience

Question 1 [2 marks]. Write 2x² + 8x + 5 in the form a(x + h)² + k. *Marking notes:* M1 for a valid completing-the-square step (2(x + 2)² …); A1 for 2(x + 2)² − 3.

Question 2 [3 marks]. The equation 3x² − 5x + k = 0 has two distinct real roots. Find the range of k. *Marking notes:* M1 for Δ = 25 − 12k; M1 for setting Δ > 0; A1 for k < 25/12. Exact fraction required — a decimal here loses the accuracy mark on Paper 1.

Question 3 [4 marks]. Let f(x) = x² − 4x − 5. (a) Write down the equation of the axis of symmetry. [1] (b) Find the coordinates of the vertex. [3] *Marking notes:* (a) A1: x = 2. (b) M1 for completing the square or substituting x = 2; A1 for f(x) = (x − 2)² − 9 (or f(2) = −9); A1 for vertex (2, −9). Note the axis can also be read off the roots 5 and −1 — any correct route earns full credit.

Extended problem (Paper 2 style) [10 marks]. A ball is thrown upward; its height is modelled by h(t) = −4.9t² + 19.6t + 2 metres, t seconds after launch. (a) Find the height from which the ball was thrown. [1] (b) Find the maximum height reached and the time at which it occurs. [4] (c) Find when the ball hits the ground. [3] (d) State the domain of the model and one limitation of it. [2] *Marking notes:* (a) A1: h(0) = 2 m. (b) M1 for the vertex method (axis t = −19.6/(2 × −4.9) = 2, or GDC maximum); A1 for t = 2 s; M1 for substitution; A1 for h = 21.6 m. (c) M1 for setting h(t) = 0 and solving (quadratic formula or GDC); A1 for the positive root t ≈ 4.10 s; A1 for rejecting the negative root with a reason — in-context rejection is what examiners look for. (d) A1: 0 ≤ t ≤ 4.10; A1: any sensible limitation (air resistance ignored; the model allows the ball below ground after impact). Technology note: on Paper 2 the GDC earns the marks if the equations solved are written down; on Paper 1, parts (b) and (c) need completion of the square / the formula by hand.

Exploration (IA) idea: model a real projectile — a basketball free throw filmed on your phone — fit a quadratic to measured (t, h) data, and evaluate the fit against the physics of constant acceleration. Clean data, honest evaluation, real personal engagement.

Going deeper: quadratics meet lines and parameters

Section B loves the hybrid: a line depending on a parameter, a parabola, and a tangency condition. Tangency ⟺ the substituted quadratic has Δ = 0 — the discriminant doing geometry.

Extra worked example. Find the values of k for which the line y = 2x + k is tangent to the parabola y = x² − 4x + 7.

Solution. Intersections satisfy x² − 4x + 7 = 2x + k, i.e. x² − 6x + (7 − k) = 0. Tangency needs Δ = 0: 36 − 4(7 − k) = 0 → 8 + 4k = 0 → k = −2. Check: with k = −2 the equation becomes x² − 6x + 9 = (x − 3)² = 0, touching at x = 3, point (3, 4) ✓ — and the point lies on both curves (9 − 12 + 7 = 4 ✓; 6 − 2 = 4 ✓).

Extra exam practice [4 marks]. The equation x² − 6x + k = 0 has roots α and β with α + β = 6 and αβ = k (no working needed — these are the sum and product of roots). Given also that α² + β² = 20, find k. *Marking notes:* M1 for α² + β² = (α + β)² − 2αβ; M1: 36 − 2k = 20; A1: k = 8; A1 for a solvability sanity check: roots are 2 and 4 (Δ = 36 − 32 = 4 > 0 ✓). Sum-and-product of roots (α + β = −b/a, αβ = c/a) is the fastest route through a whole family of Paper 1 questions — drill it.