{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-quadratic-function-01",
  "mko_id": "mko-quadratic-function",
  "title": {
    "zh-Hant": "頂點式二次函數：求值、對稱、全域最大值與限制區間",
    "en": "Quadratic functions in vertex form: evaluation, symmetry, global maximum and a restricted domain"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-quadratic-function-01",
    "source": {
      "mko_id": "mko-quadratic-function",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "439528a9247668c73092d919fb031abee74709f0633c254061b2256ed99f6441",
      "english_sha256": "302ee0311ea4ba78b81f3c48951bf1dc495a2b158c4c0e46b3f8b3b0ff72bf67"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "square_nonneg",
      "order_basics",
      "distributive",
      "neg_mul",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "q:ℝ→ℝ，q(x)=−2(x−3)²+5（a=−2、h=3、k=5）。依序求 q(1) 及其對稱點；證明對所有實數 x 有 q(6−x)=q(x)；證明 q 的全域最大值為 5、只在 x=3 取得，頂點為 (3,5)，且 q 沒有全域最小值；再說明限制在 [4,6] 時最大值是 3（在 x=4），以及 a=0 時結論不能沿用。",
      "en": "q:ℝ→ℝ, q(x)=−2(x−3)²+5 (a=−2, h=3, k=5). Evaluate q(1) step by step and find its mirror point; prove q(6−x)=q(x) for every real x; prove that the global maximum of q is 5, attained only at x=3, that the vertex is (3,5), and that q has no global minimum; then show that on the restricted domain [4,6] the maximum is 3 (at x=4), and that the conclusions do not carry over when a=0."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "q:ℝ→ℝ，q(x)=−2(x−3)²+5；a=−2、h=3、k=5 為整數，a≠0，定義域為全部實數（MKO a1、a2）。",
          "en": "q:ℝ→ℝ, q(x)=−2(x−3)²+5; a=−2, h=3, k=5 are integers, a≠0, and the domain is all reals (MKO a1, a2)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "限制區間題：D=[4,6]，即 4≤x≤6。",
          "en": "Restricted-domain question: D=[4,6], that is 4≤x≤6."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "對照：r(x)=0·(x−3)²+5（a=0，不屬於本條目的二次函數）。",
          "en": "Comparison: r(x)=0·(x−3)²+5 (a=0, not a quadratic in this entry)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "mko_quadratic"
        ],
        "before_tex": "q(x)=-2(x-3)^2+5",
        "after_tex": "a=-2\\neq0,\\ h=3,\\ k=5,\\ \\mathrm{dom}\\,q=\\mathbb{R}",
        "relation": "definition",
        "rule": "mko_quadratic",
        "conditions": [
          "a\\neq0"
        ],
        "explanation": {
          "zh-Hant": "對照頂點式 a(x−h)²+k 讀出 a=−2、h=3、k=5。a≠0，所以 q 是本條目所說的二次函數；定義域是全部實數。MKO 預告的結論是：頂點 (3,5)、對稱軸 x=3，因 a<0 全域最大值為 5，在 x=3 取得；本稿逐一證明。",
          "en": "Matching the vertex form a(x−h)²+k gives a=−2, h=3, k=5. a≠0, so q is a quadratic in the sense of this entry; its domain is all reals. The MKO's announced conclusions are: vertex (3,5), axis x=3, and since a<0 the global maximum 5, attained at x=3; this guide proves them one by one."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_quadratic"
          ],
          "explanation": {
            "zh-Hant": "只讀出參數並確認 a≠0；尚未推出任何極值性質。",
            "en": "This only reads off the parameters and confirms a≠0; no extremum property is derived yet."
          }
        },
        "leaf_basis": "mko_quadratic + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "S01"
        ],
        "before_tex": "q(1)",
        "after_tex": "-3",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 的代入次序：先算 x−h，再平方，再乘 a，最後加 k；三個子步 S02.1 至 S02.3 給出 q(1)=−3。",
          "en": "Follow the MKO's order of substitution: first x−h, then square, then multiply by a, finally add k; the three sub-steps S02.1 to S02.3 give q(1)=−3."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "每個子步都是精確的等值計算，串起來兩邊同值。",
            "en": "Each sub-step is an exact evaluation of equals, so the chain gives the same value on both sides."
          }
        },
        "leaf_basis": "numeric_evaluation + eq_replace"
      },
      {
        "id": "S02.1",
        "uses": [
          "S01"
        ],
        "before_tex": "x-h\\ \\text{at}\\ x=1",
        "after_tex": "1-3=-2",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "先算括號內：1−3=−2。",
          "en": "First the bracket: 1−3=−2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "整數減法，可逆向核對 −2+3=1。",
            "en": "Integer subtraction, checkable backwards: −2+3=1."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S02.2",
        "uses": [
          "S02.1"
        ],
        "before_tex": "(-2)^2",
        "after_tex": "4",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "平方：(−2)²=(−2)(−2)=4，負數的平方為正。",
          "en": "Square: (−2)²=(−2)(−2)=4; the square of a negative number is positive."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation",
            "neg_mul"
          ],
          "explanation": {
            "zh-Hant": "精確計算，符號規則 (−u)(−v)=uv。",
            "en": "Exact arithmetic with the sign rule (−u)(−v)=uv."
          }
        },
        "leaf_basis": "numeric_evaluation + neg_mul"
      },
      {
        "id": "S02.3",
        "uses": [
          "S02.2"
        ],
        "before_tex": "a\\cdot4+k",
        "after_tex": "-2\\cdot4+5=-3",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "乘以 a=−2 得 −8，再加 k=5 得 −3。負的 a 保持原樣，不取絕對值。",
          "en": "Multiply by a=−2 to get −8, then add k=5 to get −3. The negative a is kept as is, not replaced by its absolute value."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確整數計算，可逆向核對。",
            "en": "Exact integer arithmetic, checkable backwards."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "S02"
        ],
        "before_tex": "x=1,\\ 2h-x",
        "after_tex": "2h-x=5,\\ q(5)=-2(5-3)^2+5=-3=q(1)",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x=1 關於對稱軸 x=3 的對稱點是 2·3−1=5；兩點與軸的距離都是 2。代入得 q(5)=−2·4+5=−3，與 q(1) 相同，所以 (1,−3) 與 (5,−3) 互為對稱點。",
          "en": "The mirror of x=1 across the axis x=3 is 2·3−1=5; both points are at distance 2 from the axis. Substituting gives q(5)=−2·4+5=−3, the same as q(1), so (1,−3) and (5,−3) are mirror points."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確計算；這只是一個實例，一般對稱性在 S04 證明。",
            "en": "Exact arithmetic; this is only one instance, and the general symmetry is proved in S04."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "S01"
        ],
        "before_tex": "q(2h-x)=a((2h-x)-h)^2+k",
        "after_tex": "a(x-h)^2+k=q(x)",
        "relation": "equivalent",
        "rule": "L_mirror",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_mirror，((2h−x)−h)²=(x−h)²，代入後 q(2h−x)=q(x) 對所有實數 x 成立（本例為 q(6−x)=q(x)）。x=h 時 2h−h=h，頂點的對稱點就是它自己。引理在 S04.1、S04.2 證明。",
          "en": "By Lemma L_mirror, ((2h−x)−h)²=(x−h)², so substituting gives q(2h−x)=q(x) for every real x (here q(6−x)=q(x)). For x=h, 2h−h=h, so the vertex is its own mirror. The lemma is proved in S04.1 and S04.2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_mirror",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "等值替換括號內的平方，兩邊是同一個數，雙向成立。",
            "en": "Replacing the squared bracket by an equal expression leaves the same number on both sides, so it holds both ways."
          }
        },
        "leaf_basis": "add_assoc + add_inverse + neg_mul + eq_replace"
      },
      {
        "id": "S04.1",
        "uses": [
          "add_assoc",
          "add_inverse"
        ],
        "before_tex": "(2h-x)-h",
        "after_tex": "h-x",
        "relation": "equivalent",
        "rule": "add_assoc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(2h−x)−h=(h+h−x)−h=h−x（加法結合、交換與反元素）。",
          "en": "(2h−x)−h=(h+h−x)−h=h−x (associativity, commutativity and inverses for addition)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "add_assoc",
            "add_comm",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "對所有實數成立的恆等式，雙向成立。",
            "en": "An identity for all reals; it holds both ways."
          }
        },
        "leaf_basis": "add_assoc + add_comm + add_inverse + add_zero"
      },
      {
        "id": "S04.2",
        "uses": [
          "S04.1",
          "neg_mul"
        ],
        "before_tex": "(h-x)^2",
        "after_tex": "(-(x-h))^2=(x-h)^2",
        "relation": "equivalent",
        "rule": "neg_mul",
        "conditions": [],
        "explanation": {
          "zh-Hant": "h−x=−(x−h)，而 (−u)(−u)=u·u，所以 (h−x)²=(x−h)²。S04.1 與 S04.2 合起來證得引理 L_mirror。",
          "en": "h−x=−(x−h), and (−u)(−u)=u·u, so (h−x)²=(x−h)². S04.1 and S04.2 together prove Lemma L_mirror."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "neg_mul"
          ],
          "explanation": {
            "zh-Hant": "符號規則是恆等式，雙向成立。",
            "en": "The sign rule is an identity and holds both ways."
          }
        },
        "leaf_basis": "neg_mul + distributive"
      },
      {
        "id": "S05",
        "uses": [
          "square_nonneg"
        ],
        "before_tex": "x\\in\\mathbb{R}",
        "after_tex": "(x-3)^2\\ge0",
        "relation": "implies",
        "rule": "square_nonneg",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對任意實數 x，(x−3)² 是平方，因此非負；等號恰在 x−3=0 時成立。",
          "en": "For every real x, (x−3)² is a square and therefore nonnegative; equality holds exactly when x−3=0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "square_nonneg"
          ],
          "explanation": {
            "zh-Hant": "直接套用平方非負；這需要 x 是實數（MKO a1）。",
            "en": "A direct use of nonnegativity of squares; it needs x to be real (MKO a1)."
          }
        },
        "leaf_basis": "square_nonneg"
      },
      {
        "id": "S06",
        "uses": [
          "S05",
          "order_basics"
        ],
        "before_tex": "(x-3)^2\\ge0,\\ a=-2<0",
        "after_tex": "-2(x-3)^2\\le0",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "-2<0"
        ],
        "explanation": {
          "zh-Hant": "分兩種情形：若 (x−3)²=0，乘積為 0；若 (x−3)²>0，把 0<(x−3)² 兩邊乘以負數 −2，不等號反向，得 0>−2(x−3)²。兩種情形都有 −2(x−3)²≤0。正因 a<0，開口向下。",
          "en": "Two cases: if (x−3)²=0 the product is 0; if (x−3)²>0, multiplying 0<(x−3)² by the negative −2 reverses the inequality, giving 0>−2(x−3)². In both cases −2(x−3)²≤0. It is because a<0 that the parabola opens downwards."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "case_analysis",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "乘負數反向是關鍵；若 a>0，結論會變成 ≥0，給出最小值而非最大值。",
            "en": "Reversal under multiplication by a negative is the key; with a>0 the conclusion would become ≥0, giving a minimum instead of a maximum."
          }
        },
        "leaf_basis": "order_basics + case_analysis + mul_zero"
      },
      {
        "id": "S07",
        "uses": [
          "S06",
          "order_basics"
        ],
        "before_tex": "-2(x-3)^2\\le0",
        "after_tex": "q(x)=-2(x-3)^2+5\\le5\\ \\ \\forall x\\in\\mathbb{R}",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩邊同加 5，順序保持：q(x)≤5。x 是任意實數，所以 5 是 q 在全部實數上的一個上界。",
          "en": "Add 5 to both sides, which keeps the order: q(x)≤5. Since x is an arbitrary real, 5 is an upper bound of q on all reals."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "加法保序；推廣到所有 x 只因 x 任意。上界還不是最大值，需 S08 證明被取得。",
            "en": "Addition preserves order; the generalization to all x relies only on x being arbitrary. An upper bound is not yet a maximum; S08 shows it is attained."
          }
        },
        "leaf_basis": "order_basics + forall_intro"
      },
      {
        "id": "S08",
        "uses": [
          "S07",
          "S01"
        ],
        "before_tex": "q(3)=-2\\cdot0^2+5",
        "after_tex": "q(3)=5=\\max_{\\mathbb{R}}q,\\ \\text{vertex}\\ (3,5)",
        "relation": "implies",
        "rule": "mko_quadratic",
        "conditions": [],
        "explanation": {
          "zh-Hant": "q(3)=−2·0+5=5，上界 5 被取得，所以 5 是全域最大值，在 x=3 取得。頂點是坐標對 (3,5)，最大值是數 5，取得處是 x=3，三者不同。",
          "en": "q(3)=−2·0+5=5, so the upper bound 5 is attained and is the global maximum, attained at x=3. The vertex is the coordinate pair (3,5), the maximum value is the number 5, and the place where it is attained is x=3; these are three different things."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "numeric_evaluation",
            "mko_quadratic"
          ],
          "explanation": {
            "zh-Hant": "最大值需要「上界」與「被取得」兩者；S07 給上界，本步給取得點。",
            "en": "A maximum needs both an upper bound and attainment; S07 gives the bound and this step the point of attainment."
          }
        },
        "leaf_basis": "mko_quadratic + numeric_evaluation + and_intro_elim"
      },
      {
        "id": "S09",
        "uses": [
          "S01",
          "zero_product",
          "square_nonneg"
        ],
        "before_tex": "q(x)=5",
        "after_tex": "x=3",
        "relation": "equivalent",
        "rule": "zero_product",
        "conditions": [
          "-2\\neq0"
        ],
        "explanation": {
          "zh-Hant": "q(x)=5 ⇔ −2(x−3)²=0（兩邊減 5）⇔ (x−3)²=0（零積性質，−2≠0）⇔ x=3（平方為零當且僅當底數為零）。所以最大值只在 x=3 取得。",
          "en": "q(x)=5 ⇔ −2(x−3)²=0 (subtract 5) ⇔ (x−3)²=0 (zero product, −2≠0) ⇔ x=3 (a square is zero exactly when its base is). So the maximum is attained only at x=3."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_congruence",
            "zero_product",
            "square_nonneg"
          ],
          "explanation": {
            "zh-Hant": "三個環節都是等價：同減 5 可逆、零積性質雙向、平方為零雙向。",
            "en": "All three links are equivalences: subtracting 5 is reversible, the zero-product property and the zero-square property are two-way."
          }
        },
        "leaf_basis": "eq_congruence + zero_product + square_nonneg"
      },
      {
        "id": "S10",
        "uses": [
          "S01"
        ],
        "before_tex": "x_0\\in\\mathbb{R}",
        "after_tex": "\\exists x:\\ q(x)<q(x_0)",
        "relation": "implies",
        "rule": "L_unbounded",
        "conditions": [],
        "explanation": {
          "zh-Hant": "q 沒有全域最小值：任取候選點 x₀，依引理 L_unbounded 取 M=q(x₀)，存在 x 使 q(x)<q(x₀)，所以 q(x₀) 不是最小值。引理在 S10.1、S10.2 證明；這對應 MKO 所說「另一方向沒有全域極值」。",
          "en": "q has no global minimum: for any candidate x₀, Lemma L_unbounded with M=q(x₀) gives an x with q(x)<q(x₀), so q(x₀) is not the minimum. The lemma is proved in S10.1 and S10.2; this is the MKO's 'there is no global extremum in the other direction'."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_unbounded",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "x₀ 任意，所以沒有任何點是最小值點；這依賴定義域是全部實數。",
            "en": "x₀ is arbitrary, so no point is a minimizer; this relies on the domain being all reals."
          }
        },
        "leaf_basis": "order_basics + forall_intro"
      },
      {
        "id": "S10.1",
        "uses": [
          "S01",
          "eq_replace"
        ],
        "before_tex": "q(3+t)",
        "after_tex": "-2t^2+5",
        "relation": "equivalent",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "令 x=3+t：x−3=t，所以 q(3+t)=−2t²+5。",
          "en": "Put x=3+t: x−3=t, so q(3+t)=−2t²+5."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_replace",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "代換 x=3+t 與 t=x−3 互逆，兩式對每個 t 同值。",
            "en": "The substitutions x=3+t and t=x−3 are inverse to each other, and the two expressions agree for every t."
          }
        },
        "leaf_basis": "eq_replace + add_inverse"
      },
      {
        "id": "S10.2",
        "uses": [
          "S10.1",
          "order_basics",
          "abs_rules"
        ],
        "before_tex": "M\\in\\mathbb{R},\\ t=|5-M|+1",
        "after_tex": "q(3+t)=5-2t^2<M",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "t\\ge1"
        ],
        "explanation": {
          "zh-Hant": "給定 M，取 t=|5−M|+1≥1。把 t≥1 乘以 t>0 得 t²≥t；又 t>|5−M|≥5−M，所以 t²>5−M；而 t²≥0 使 2t²≥t²>5−M。移項得 5−2t²<M。S10.1 與 S10.2 合起來證得引理 L_unbounded。",
          "en": "Given M, take t=|5−M|+1≥1. Multiplying t≥1 by t>0 gives t²≥t; also t>|5−M|≥5−M, so t²>5−M; and t²≥0 makes 2t²≥t²>5−M. Rearranging gives 5−2t²<M. S10.1 and S10.2 together prove Lemma L_unbounded."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "abs_rules",
            "square_nonneg"
          ],
          "explanation": {
            "zh-Hant": "用到 |u|≥u、正數乘法保序與傳遞性；t 的選法明確，不依賴任何有限核對。",
            "en": "It uses |u|≥u, order-preservation under multiplication by positives, and transitivity; t is chosen explicitly, with no reliance on a finite check."
          }
        },
        "leaf_basis": "order_basics + abs_rules + square_nonneg"
      },
      {
        "id": "S11",
        "uses": [
          "G2"
        ],
        "before_tex": "4\\le x\\le6",
        "after_tex": "x-3\\ge1\\ \\Rightarrow\\ (x-3)^2\\ge1",
        "relation": "implies",
        "rule": "L_sq_ge_one",
        "conditions": [
          "x\\in[4,6]"
        ],
        "explanation": {
          "zh-Hant": "轉到限制區間 D=[4,6]（MKO a1 要求另行檢查）。x≥4 兩邊減 3 得 x−3≥1；依引理 L_sq_ge_one，(x−3)²≥1。證明在 S11.1。",
          "en": "Turn to the restricted domain D=[4,6] (MKO a1 asks for a separate check). Subtracting 3 from x≥4 gives x−3≥1; by Lemma L_sq_ge_one, (x−3)²≥1. The proof is in S11.1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_sq_ge_one",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "只用到 D 的下端 x≥4；這是單向推論。",
            "en": "Only the lower end x≥4 of D is used; this is a one-way inference."
          }
        },
        "leaf_basis": "order_basics"
      },
      {
        "id": "S11.1",
        "uses": [
          "order_basics"
        ],
        "before_tex": "u\\ge1",
        "after_tex": "u^2\\ge u\\ge1",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "u\\ge1"
        ],
        "explanation": {
          "zh-Hant": "u≥1>0。把 u≥1 兩邊乘以正數 u 得 u²≥u，再由傳遞性 u²≥u≥1。此步證得引理 L_sq_ge_one。",
          "en": "u≥1>0. Multiplying u≥1 by the positive number u gives u²≥u, and by transitivity u²≥u≥1. This step proves Lemma L_sq_ge_one."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "mul_one"
          ],
          "explanation": {
            "zh-Hant": "需要 u>0 才能保序相乘；對 u≤−1 結論 u²≥1 仍成立但需另證，本稿不需要。",
            "en": "It needs u>0 to multiply while keeping the order; for u≤−1 the conclusion u²≥1 still holds but needs a separate argument, which this guide does not need."
          }
        },
        "leaf_basis": "order_basics + mul_one"
      },
      {
        "id": "S12",
        "uses": [
          "S11",
          "S09",
          "G2"
        ],
        "before_tex": "(x-3)^2\\ge1\\ \\ (x\\in[4,6])",
        "after_tex": "q(x)\\le-2+5=3,\\ q(4)=3=\\max_{[4,6]}q",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "-2<0"
        ],
        "explanation": {
          "zh-Hant": "把 (x−3)²≥1 乘以 −2<0，不等號反向：−2(x−3)²≤−2；加 5 得 q(x)≤3。q(4)=−2·1+5=3，所以 D 上最大值是 3，在 x=4 取得。全域最大值 5 只在 x=3 取得（S09），而 3∉D，所以 5 在 D 上不會出現。",
          "en": "Multiplying (x−3)²≥1 by −2<0 reverses the inequality: −2(x−3)²≤−2; adding 5 gives q(x)≤3. q(4)=−2·1+5=3, so the maximum on D is 3, attained at x=4. The global maximum 5 is attained only at x=3 (S09), and 3∉D, so 5 never occurs on D."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "S09",
            "order_basics",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "需要上界（本步）與取得點 x=4∈D 兩者；不宣稱 D 上的最小值。",
            "en": "It needs both the upper bound (this step) and the point of attainment x=4∈D; the minimum on D is not claimed."
          }
        },
        "leaf_basis": "order_basics + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "G3"
        ],
        "before_tex": "r(x)=0\\cdot(x-3)^2+5",
        "after_tex": "r(x)=5\\ \\ \\forall x",
        "relation": "implies",
        "rule": "mul_zero",
        "conditions": [],
        "explanation": {
          "zh-Hant": "a=0 時 0·(x−3)²=0，r 是常數函數 5：每一點都是最大值點，也都是最小值點，沒有唯一頂點或對稱軸。這不是本條目的二次函數，S08、S09 的「只在 x=3 取得」不能沿用（MKO a2）。",
          "en": "With a=0, 0·(x−3)²=0, so r is the constant function 5: every point is a maximizer and also a minimizer, and there is no unique vertex or axis. This is not a quadratic of this entry, and the 'attained only at x=3' of S08 and S09 does not carry over (MKO a2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mul_zero",
            "G3"
          ],
          "explanation": {
            "zh-Hant": "只用到乘以 0 得 0；結論是對照，說明 a≠0 這個前提不可省略。",
            "en": "Only multiplication by 0 is used; the conclusion is a contrast showing that the hypothesis a≠0 cannot be dropped."
          }
        },
        "leaf_basis": "mul_zero + forall_intro"
      },
      {
        "id": "S14",
        "uses": [
          "S02",
          "S03",
          "S04",
          "S08",
          "S09",
          "S10",
          "S12",
          "S13"
        ],
        "before_tex": "\\text{S02 to S13}",
        "after_tex": "q(1)=q(5)=-3;\\ q(6-x)=q(x);\\ \\max_{\\mathbb{R}}q=5\\ \\text{only at}\\ 3;\\ \\text{no min};\\ \\max_{[4,6]}q=3",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "整理：q(1)=−3，對稱點 5 也有 q(5)=−3；q(6−x)=q(x) 對所有 x 成立；全域最大值 5 只在 x=3 取得，頂點 (3,5)，沒有全域最小值；限制在 [4,6] 時最大值是 3（x=4）；a=0 時結論失效。",
          "en": "Summary: q(1)=−3 and its mirror 5 has q(5)=−3; q(6−x)=q(x) for every x; the global maximum 5 is attained only at x=3, the vertex is (3,5), and there is no global minimum; on [4,6] the maximum is 3 (at x=4); with a=0 the conclusions fail."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S02",
            "S03",
            "S04",
            "S08",
            "S09",
            "S10",
            "S12",
            "S13",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G 為 goal 的全部要求；足夠條件 H 為各步結論的合取，皆已正向證成，H⇒G 只用合取引入。",
            "en": "Goal G is everything asked in the goal; the sufficient condition H is the conjunction of the step results, all proved forwards, and H⇒G uses only conjunction introduction."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "q(1)=−3，其關於 x=3 的對稱點 5 也有 q(5)=−3，且對所有實數 q(6−x)=q(x)。因 (x−3)²≥0 且 a=−2<0，q(x)≤5 對所有 x 成立，q(3)=5，所以全域最大值為 5，只在 x=3 取得，頂點為 (3,5)；q 沒有全域最小值。限制在 [4,6] 時最大值為 3（在 x=4），全域最大值 5 不出現。a=0 時函數為常數 5，結論不適用。",
      "en": "q(1)=−3, its mirror point 5 across x=3 also has q(5)=−3, and q(6−x)=q(x) for every real x. Since (x−3)²≥0 and a=−2<0, q(x)≤5 for all x and q(3)=5, so the global maximum is 5, attained only at x=3, and the vertex is (3,5); q has no global minimum. On [4,6] the maximum is 3 (at x=4), and the global maximum 5 does not occur. With a=0 the function is the constant 5 and the conclusions do not apply."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "q 的最大值是 (3,5)",
          "en": "The maximum of q is (3,5)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "(3,5) 是頂點（坐標對）；最大值是數 5，在 x=3 取得。",
          "en": "(3,5) is the vertex (a coordinate pair); the maximum is the number 5, attained at x=3."
        }
      },
      {
        "claim": {
          "zh-Hant": "在 [4,6] 上 q 的最大值是 5",
          "en": "On [4,6] the maximum of q is 5"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "q(x)=5 只在 x=3，而 3∉[4,6]；[4,6] 上最大值為 q(4)=3。",
          "en": "q(x)=5 only at x=3, and 3∉[4,6]; the maximum on [4,6] is q(4)=3."
        }
      },
      {
        "claim": {
          "zh-Hant": "q 也有全域最小值，因為拋物線在頂點取極值",
          "en": "q also has a global minimum, since a parabola takes an extremum at its vertex"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "a<0 時頂點給最大值；對任意 M 都有 x 使 q(x)<M（S10），所以沒有最小值。",
          "en": "With a<0 the vertex gives the maximum; for every M some x has q(x)<M (S10), so there is no minimum."
        }
      },
      {
        "claim": {
          "zh-Hant": "r(x)=0·(x−3)²+5 的最大值只在 x=3 取得",
          "en": "The maximum of r(x)=0·(x−3)²+5 is attained only at x=3"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "r 是常數 5，每個 x 都取得最大值；a=0 不屬於本條目。",
          "en": "r is the constant 5, so every x attains the maximum; a=0 is outside this entry."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "square_nonneg",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數平方非負：對所有實數 u，u²≥0，且 u²=0 ⇔ u=0。",
          "en": "Squares are nonnegative: for every real u, u²≥0, and u²=0 ⇔ u=0."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "distributive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。",
          "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu."
        }
      },
      {
        "id": "neg_mul",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "負號與乘法：(−u)v=−(uv)，(−u)(−v)=uv。",
          "en": "Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "eq_replace",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。",
          "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true."
        }
      },
      {
        "id": "add_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法結合律：(u+v)+w=u+(v+w)。",
          "en": "Associativity of addition: (u+v)+w=u+(v+w)."
        }
      },
      {
        "id": "add_inverse",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。",
          "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)."
        }
      },
      {
        "id": "add_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法交換律：u+v=v+u。",
          "en": "Commutativity of addition: u+v=v+u."
        }
      },
      {
        "id": "add_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "id": "case_analysis",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。",
          "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility."
        }
      },
      {
        "id": "mul_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。",
          "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0."
        }
      },
      {
        "id": "forall_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。",
          "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain."
        }
      },
      {
        "id": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "zero_product",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "零乘積性質：在 ℤ、ℚ、ℝ、ℂ（或任何整域）中，uv=0 ⇔ u=0 或 v=0。",
          "en": "Zero-product property: in ℤ, ℚ, ℝ, ℂ (or any integral domain), uv=0 ⇔ u=0 or v=0."
        }
      },
      {
        "id": "eq_congruence",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等式同用：若 u=v，對兩邊做同一個有定義的運算 f，得 f(u)=f(v)。只對一邊做運算不屬於此規則。",
          "en": "Congruence of equality: if u=v, applying the same defined operation f to both sides gives f(u)=f(v). Operating on one side only is not an instance of this rule."
        }
      },
      {
        "id": "abs_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "絕對值（實數）：|u|=u（u≥0）、|u|=−u（u<0）；|u|≥0；|uv|=|u||v|；三角不等式 |u+v|≤|u|+|v|；|u|<r ⇔ −r<u<r。",
          "en": "Absolute value (reals): |u|=u if u≥0 and |u|=−u if u<0; |u|≥0; |uv|=|u||v|; triangle inequality |u+v|≤|u|+|v|; |u|<r ⇔ −r<u<r."
        }
      },
      {
        "id": "mul_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "id": "mko_quadratic",
        "kind": "cited",
        "source": {
          "mko_id": "mko-quadratic-function",
          "sha256": "439528a9247668c73092d919fb031abee74709f0633c254061b2256ed99f6441",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "令 q:R→R，q(x)=a(x−h)²+k，其中 a、h、k 為實數且 a≠0。其頂點為 (h,k)，對稱軸為 x=h；a>0 時全域最小值為 k，a<0 時全域最大值為 k，均在 x=h 取得。圖上 (x,q(x)) 關於該軸的對稱點為 (2h−x,q(x))。",
          "en": "Let q:R→R be q(x)=a(x−h)²+k with real a,h,k and a≠0. Its vertex is (h,k) and its axis is x=h. When a>0, the global minimum is k; when a<0, the global maximum is k, both attained at x=h. The point (2h−x,q(x)) mirrors (x,q(x)) across the axis."
        }
      },
      {
        "id": "L_mirror",
        "kind": "lemma",
        "proved_in": [
          "S04.1",
          "S04.2"
        ],
        "statement": {
          "zh-Hant": "對所有實數 x、h：((2h−x)−h)²=(x−h)²。",
          "en": "For all reals x, h: ((2h−x)−h)²=(x−h)²."
        }
      },
      {
        "id": "L_unbounded",
        "kind": "lemma",
        "proved_in": [
          "S10.1",
          "S10.2"
        ],
        "statement": {
          "zh-Hant": "對每個實數 M，存在實數 x 使 q(x)<M（q(x)=−2(x−3)²+5）。",
          "en": "For every real M there is a real x with q(x)<M (q(x)=−2(x−3)²+5)."
        }
      },
      {
        "id": "L_sq_ge_one",
        "kind": "lemma",
        "proved_in": [
          "S11.1"
        ],
        "statement": {
          "zh-Hant": "對實數 u≥1：u²≥u≥1。",
          "en": "For a real u≥1: u²≥u≥1."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02.1",
        "lhs": "1-3",
        "rhs": "-2"
      },
      {
        "id": "N2",
        "step": "S02.2",
        "lhs": "(-2)**2",
        "rhs": "4"
      },
      {
        "id": "N3",
        "step": "S02.3",
        "lhs": "-2*4+5",
        "rhs": "-3"
      },
      {
        "id": "N4",
        "step": "S03",
        "lhs": "2*3-1",
        "rhs": "5"
      },
      {
        "id": "N5",
        "step": "S03",
        "lhs": "-2*(5-3)**2+5",
        "rhs": "-3"
      },
      {
        "id": "N6",
        "step": "S04",
        "vars": [
          "x",
          "h"
        ],
        "degree": 2,
        "lhs": "((2*h-x)-h)**2",
        "rhs": "(x-h)**2"
      },
      {
        "id": "N7",
        "step": "S04",
        "vars": [
          "x"
        ],
        "degree": 2,
        "lhs": "-2*((2*3-x)-3)**2+5",
        "rhs": "-2*(x-3)**2+5"
      },
      {
        "id": "N8",
        "step": "S08",
        "lhs": "-2*(3-3)**2+5",
        "rhs": "5"
      },
      {
        "id": "N9",
        "step": "S10.1",
        "vars": [
          "t"
        ],
        "degree": 2,
        "lhs": "-2*((3+t)-3)**2+5",
        "rhs": "5-2*t**2"
      },
      {
        "id": "N10",
        "step": "S10.2",
        "lhs": "5-2*106**2",
        "relation": "<",
        "rhs": "-100"
      },
      {
        "id": "N11",
        "step": "S12",
        "lhs": "-2*(4-3)**2+5",
        "rhs": "3"
      },
      {
        "id": "N12",
        "step": "S12",
        "lhs": "-2*(6-3)**2+5",
        "relation": "<=",
        "rhs": "3"
      },
      {
        "id": "N13",
        "step": "S07",
        "vars": [
          "x"
        ],
        "range": [
          -20,
          20
        ],
        "lhs": "-2*(x-3)**2+5",
        "relation": "<=",
        "rhs": "5"
      },
      {
        "id": "N14",
        "step": "S13",
        "vars": [
          "x"
        ],
        "degree": 2,
        "lhs": "0*(x-3)**2+5",
        "rhs": "5"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 範圍內：自變數遍及全部實數、a≠0、整數參數、精確代入（a1、a2）。對稱性對一般 h 證明；極值結論只對 a=−2<0 的這個函數證明，a>0 的情形同理但未寫出。限制區間只處理 [4,6] 上的最大值。不處理一般式 ax²+bx+c 的配方、二次方程式的解或微分；不改動 canonical MKO 或英文 edition。",
      "en": "Inside the MKO: the input ranges over all reals, a≠0, integer parameters, exact substitution (a1, a2). Symmetry is proved for general h; the extremum results are proved for this function with a=−2<0, and the case a>0 is analogous but not written out. On the restricted domain only the maximum on [4,6] is treated. Completing the square from ax²+bx+c, solving quadratic equations and differentiation are not treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "q(x)=a(x-h)^2+k",
        "頂點式二次函數",
        "a quadratic in vertex form"
      ],
      [
        "a=-2,\\ h=3,\\ k=5",
        "本稿的參數",
        "this guide's parameters"
      ],
      [
        "(h,k)=(3,5)",
        "頂點（坐標對）",
        "the vertex (a coordinate pair)"
      ],
      [
        "x=h",
        "對稱軸",
        "the axis of symmetry"
      ],
      [
        "2h-x",
        "橫坐標 x 關於對稱軸反射後的值",
        "the x-coordinate after reflection across the symmetry axis"
      ]
    ],
    "md_error_controls": [
      {
        "title": "頂點不是極值 / The vertex is not the extreme value",
        "zh": "MKO 的誤解之一是混用頂點與極值。頂點是坐標對 (3,5)；最大值是數 5；取得最大值的位置是 x=3（S08）。說「最大值是 (3,5)」或「最大值是 3」都錯。",
        "en": "One misconception identified in the MKO is conflating the vertex with the extreme value. The vertex is the coordinate pair (3,5); the maximum value is the number 5; the place where it is attained is x=3 (S08). Saying 'the maximum is (3,5)' or 'the maximum is 3' is wrong."
      },
      {
        "title": "限制區間要另行檢查 / A restricted domain needs its own check",
        "zh": "另一個誤解是在限制區間上直接沿用全實數結論。在 [4,6] 上 x=3 不在區間內，最大值是 3（x=4），不是 5（S12）。零係數 a=0 也不能沿用：r(x)=5 是常數（S13）。",
        "en": "The other misconception is to carry the all-reals conclusion over to a restricted domain. On [4,6], x=3 is not in the interval and the maximum is 3 (at x=4), not 5 (S12). Nor does it carry over to a=0: r(x)=5 is constant (S13)."
      },
      {
        "title": "負係數不能取絕對值 / A negative coefficient may not be replaced by its absolute value",
        "zh": "若把 a=−2 換成 2，函數變成 2(x−3)²+5，開口向上，最小值 5、沒有最大值，結論恰好相反。S02.3 與 S06 都保留 a 的負號。",
        "en": "Replacing a=−2 by 2 gives 2(x−3)²+5, which opens upwards with minimum 5 and no maximum, the opposite conclusion. S02.3 and S06 both keep the negative sign of a."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "N13 只是在 −20 到 20 的 41 個整數點上的有限核對，不證明 S07；S07 由 S05、S06 的論證證明。",
        "en": "N13 is only a finite check at the 41 integers from −20 to 20 and does not prove S07; S07 is proved by the argument of S05 and S06."
      },
      {
        "zh-Hant": "a>0 的最小值情形與本稿對稱，未逐步寫出；由一般式配方成頂點式不在本稿範圍。",
        "en": "The minimum case a>0 is symmetric to this guide and not written out; converting a general quadratic to vertex form is outside this guide."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "66a236eb7dc1563609f389b65f6d95f1730346ffad6bd1bf7d19007ea7af878b",
    "original_source_sha256": "66a236eb7dc1563609f389b65f6d95f1730346ffad6bd1bf7d19007ea7af878b",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}