Quadratic functions, equations and inequalities

Maths: Analysis and Approaches · Functions. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

The graph of a quadratic is a parabola. It opens upwards when a > 0 and downwards when a < 0. Each of the three forms shows something different: ax² + bx + c shows the y-intercept c, a(x − p)(x − q) shows the x-intercepts p and q, and a(x − h)² + k shows the vertex (h, k).

The axis of symmetry passes through the vertex, halfway between the x-intercepts.

The discriminant tells you how many real roots there are without solving. For an inequality, first find the roots, then use a sketch of the parabola to decide whether the answer is between the roots or outside them.

Rules to remember

Axis of symmetry: x = −b/(2a); from factorised form, x = (p + q)/2
Vertex form a(x − h)² + k: vertex (h, k). Note the sign: (x − 3)² gives h = 3
Completing the square: x² + bx + c = (x + b/2)² − (b/2)² + c
Quadratic formula: x = (−b ± √(b² − 4ac))/(2a)
Discriminant Δ = b² − 4ac: Δ > 0 two distinct real roots; Δ = 0 two equal real roots; Δ < 0 no real roots
For a > 0 with roots p < q: ax² + bx + c > 0 when x < p or x > q; < 0 when p < x < q
A line and a curve: set them equal; Δ < 0 no meeting, Δ = 0 tangent, Δ > 0 two points

Common mistake

Solving a quadratic inequality as if it were an equation and writing x > p, x > q. Find the roots, sketch the parabola, then read off the correct region.

Worked example

Find the values of k for which x² + kx + (k + 3) = 0 has two distinct real roots.

Two distinct real roots need Δ > 0: k² − 4 × 1 × (k + 3) > 0.
k² − 4k − 12 > 0, which factorises to (k − 6)(k + 2) > 0.
The roots are k = −2 and k = 6. The parabola opens upwards, so it is positive outside the roots.

Answer: k < −2 or k > 6

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Topic 2: Functions

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