Definition D1
The graph G of a function f is the set of all ordered pairs (x, f(x)). That is, (x, y) ∈ G ⟺ y = f(x)
Proof.
The line x = p is the set of all points whose x-coordinate is p, a vertical line.convention
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Definition D2
The reflection P′ of a point P across a line m is the point such that the midpoint of the segment PP′ lies on m and the segment PP′ is perpendicular to m. If P lies on m, define P′ = P.
Proof.
The name is not attached first. Whether a correspondence is a reflection is verified by this condition.convention
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Definition D3
A set of points S is symmetric about a line m when σ(S) = S, where σ is the correspondence sending each point to its reflected point.
Proof.
This means the reflection of every point of S lies in S, and every point of S is the reflection of some point of S.convention, D2
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Lemma L1
Define σₚ(x, y) = (2p − x, y). Then σₚ is the reflection across the line x = p.
Proof.
(1)For every u, 2u = u(1 + 1) = u + uA12, A9, A6, A10
(2)If x = p, then 2p − x = (p + p) − p = p, so σₚ(P) = P and P lies on x = p.(1), A7, A8, D1
(3)Now suppose x ≠ p.assumption
(4)If 2p − x = x, then 2p = x + x = 2x, and 2 ≠ 0, so p = x. Therefore the two points are distinct.(1), (3), A7, A8, A12, A9, A11, A10
(5)x + (2p − x) = 2p, so the midpoint of the two x-coordinates is 2p/2 = pA7, A8, A12, A9, A11, A10, A3
(6)The y-coordinates of both points are y, so the midpoint's y-coordinate is (y + y)/2 = 2y/2 = y(1), A3, A12, A9, A11, A10
(7)Therefore the midpoint is (p, y) and lies on the line x = p.(5), (6), D1
(8)The two points have the same y-coordinate, so the segment is horizontal, and x = p is a vertical line, so they are perpendicular.D1, A4
(9)Therefore σₚ(P) is the reflection of P across the line x = p.(2), (4), (7), (8), D2
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Lemma L2
If f(x) = a(x − p)² + q, then for every t, f(p + t) = f(p − t)
Proof.
(1)(p + t) − p = tA7, A8
(2)f(p + t) = a t² + q(1), A13
(3)(p − t) − p = −tA7, A8
(4)f(p − t) = a(−t)² + q(3), A13
(5)(−t)² = t², so f(p − t) = a t² + q(4), A1, A13
(6)Therefore f(p + t) = f(p − t)(2), (5)
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Theorem T
The graph G of f(x) = a(x − p)² + q is symmetric about the line x = p.
Proof.
(1)Let (x, y) ∈ G. That is, y = f(x).D1
(2)Let t = x − p. Then x = p + tA7, A8
(3)2p − x = (p + p) − (p + t) = p − t(2), A12, A9, A6, A10, A7, A8
(4)f(2p − x) = f(p − t) = f(p + t) = f(x) = y(1), (2), (3), L2, A13
(5)Therefore σₚ(x, y) = (2p − x, y) ∈ G(4), L1, D1
(6)Conversely, σₚ(2p − x, y) = (2p − (2p − x), y) = (x, y), so each point of G is the reflection of some point of G.(5), L1, A7, A8
(7)The two sets have the same elements, so σₚ(G) = G(5), (6), A5
(8)σₚ is the reflection across x = p, so G is symmetric about the line x = p.(7), L1, D3
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Corollary C
The graph of y = ax² + bx + c (a ≠ 0) is symmetric about the line x = −b/2a.
Proof.
(1)ax² + bx + c = a(x − p)² + q, where p = −b/2aA2
(2)Therefore its graph is symmetric about the line x = p.(1), T
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