Definition D1
f(x) = ax² + bx + c (a ≠ 0) is called a quadratic function.
Proof.
(1)x² means x·x, u − v means u + (−v), and u/v means u·v⁻¹ (v ≠ 0). 2 means 1 + 1, and 4 means 2 × 2.convention
(2)a ≠ 0 is assumed in every block below. Hence 2a ≠ 0 and 4a ≠ 0.convention, A11, A10
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Definition D2
If f(x₀) ≤ f(x) for every x, then f(x₀) is the minimum; if f(x) ≤ f(x₀) for every x, then f(x₀) is the maximum; and the point (x₀, f(x₀)) is called the vertex.
Proof.
The vertex is not the name of a shape but the single point where the value is smallest or largest.convention
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Lemma L1
For all u, v, (u + v)² = u² + 2uv + v²
Proof.
(1)(u + v)² = (u + v)(u + v)D1
(2) = u(u + v) + v(u + v)(1), A1, A3
(3) = (u² + uv) + (vu + v²)(2), A1
(4)vu = uvA3
(5)uv + uv = (1 + 1)uv = 2uvD1, A1, A3, A8
(6)(u + v)² = u² + 2uv + v²(3), (4), (5), A4, A5
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Lemma L2
For a ≠ 0, ax² + bx + c = a(x + b/2a)² + (c − b²/4a)
Proof.
(1)Set p = b/2a, q = c − b²/4a. Here 2a ≠ 0, 4a ≠ 0.D1
(2)a(x + p)² = a(x² + 2px + p²)L1, E
(3) = ax² + a(2px) + ap²(2), A1
(4)a(2px) = (2ap)xA2, A3
(5)2ap = 2a(b/2a) = b(2a·(2a)⁻¹) = b(1), A2, A3, A9, A8
(6)4a(ap²) = (2a)(2a)(2a)⁻¹(2a)⁻¹b² = b² and 4a(b²/4a) = b²(1), D1, A2, A3, A9, A8
(7)4a ≠ 0, so ap² = b²/4a(6), (1), A13
(8)ap² + q = b²/4a + (c − b²/4a) = c(7), (1), A4, A5, A7, A6
(9)a(x + p)² + q = ax² + bx + c(3), (4), (5), (8), A4
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Theorem T1
If a > 0, then f attains its minimum c − b²/4a at x = −b/2a, and at every other x the value is larger.
Proof.
(1)Set p = b/2a, q = c − b²/4a, x₀ = −p.L2
(2)f(x₀) = a(x₀ + p)² + q = a × 0 + q = q(1), L2, A7, A12, A5, A6
(3)Suppose x ≠ x₀ and set t = x + p; then t ≠ 0.assumption, A4, A5, A6, A7
(4)f(x) = at² + q(3), L2
(5)0 < t²(3), A14
(6)0 < a, so 0 < at²assumption, (5), A15
(7)q < at² + q(6), A17, A6, A5
(8)Therefore f(x₀) < f(x)(2), (4), (7)
(9)This holds for every x ≠ x₀, so f(x₀) = q is the minimum.(8), (2), D2
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Theorem T2
If a < 0, then f attains its maximum c − b²/4a at x = −b/2a, and at every other x the value is smaller.
Proof.
(1)Set p = b/2a, q = c − b²/4a, x₀ = −p, and suppose x ≠ x₀.assumption, L2
(2)f(x₀) = q, and setting t = x + p gives t ≠ 0 and f(x) = at² + q(1), L2, A7, A12, A4, A5, A6
(3)0 < t²(2), A14
(4)a < 0, so at² < 0assumption, (3), A16
(5)at² + q < q(4), A17, A6, A5
(6)Therefore f(x) < f(x₀)(2), (5)
(7)This holds for every x ≠ x₀, so f(x₀) = q is the maximum.(6), (2), D2
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Corollary C
The vertex of the quadratic function f(x) = ax² + bx + c is (−b/2a, c − b²/4a).
Proof.
(1)If a > 0, f attains its minimum c − b²/4a at x = −b/2a.T1
(2)If a < 0, f attains its maximum c − b²/4a at x = −b/2a.T2
(3)Since a ≠ 0, these are the only two cases, and in either case the vertex is (−b/2a, c − b²/4a).(1), (2), D1, D2
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