{"schema_version":"ocme-site-results-v1","locale":"en","page_type":"derivations","facts":{"questions":36000,"reviewed":36000,"display_candidates":0,"knowledge":302,"guides":299,"paths":59,"batches":360,"families":1441,"atlas":302,"core":80,"evidence":9,"formal_evidence":5,"formal_subjects":4,"lean_files":6,"lean_toolchain":"leanprover/lean4:v4.30.0","question_state":"candidate","english_state":"ai_candidate","question_Lean":"NOT_RUN","canonical_truth_promotion":false},"filters":{"q":"mko-quadratic-function"},"kind":"guides","total":1,"pages":1,"page":1,"entries":[{"id":"derivation-mko-quadratic-function-01","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"},"summary":{"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."},"mko_ids":["mko-quadratic-function"]}],"title":"Concept derivation guides"}