SketchMath 數學團隊 SketchMath Maths Team ｜ 更新 Updated 2026-10-20
配方法把 x² + bx + c 化為 (x + b/2)² + c − (b/2)²。對於 y = a(x − h)² + k，頂點是 (h, k)，對稱軸是 x = h，而 k 是最小值（a > 0）或最大值（a < 0）。Completing the square rewrites x² + bx + c as (x + b/2)² + c − (b/2)². For y = a(x − h)² + k the vertex is (h, k), the axis is x = h, and k is the minimum (a > 0) or maximum (a < 0).
x² + bx = (x + b/2)² − (b/2)²ax² + bx + c = a(x + b/2a)² + c − b²/4ay = a(x − h)² + k 的頂點Vertex of y = a(x − h)² + k(h, k)(x − h)² = −k/a gives x = h ± √(−k/a)把 x² − 6x + 5 寫成 (x − h)² + k 的形式，並寫出最小值。Write x² − 6x + 5 in the form (x − h)² + k, and state its minimum value.
−6 的一半是 −3：x² − 6x + 5 = (x − 3)² − 9 + 5Half of −6 is −3: x² − 6x + 5 = (x − 3)² − 9 + 5= (x − 3)² − 4答案： Answer: (x − 3)² − 4；最小值 −4（當 x = 3）(x − 3)² − 4; minimum value −4 at x = 3
把 2x² + 8x − 3 配方，並求頂點。Complete the square for 2x² + 8x − 3 and find the vertex.
2(x² + 4x) − 32[(x + 2)² − 4] − 3 = 2(x + 2)² − 8 − 3答案： Answer: 2(x + 2)² − 11；頂點 (−2, −11)2(x + 2)² − 11; vertex (−2, −11)
解 x² + 4x − 1 = 0。Solve x² + 4x − 1 = 0.
(x + 2)² − 4 − 1 = 0, so (x + 2)² = 5x + 2 = ± √5答案： Answer: x = −2 ± √5
2x² + 8x − 3，只把 2 從含 x 的項提出。在括號內加減 (b/2)²，之後把減去的部分乘以 2。For 2x² + 8x − 3 take the 2 out of the x terms only. Add and subtract (b/2)² inside the bracket, then multiply the subtracted part by 2.(x + 2)² 代表 h = −2，頂點的 x 坐標是 −2。(x + 2)² means h = −2, so the vertex is at x = −2.溫習方法：點擊以下任何題目，即時在坐標網格載入並核對計算步驟。How to study: click any question to load it on the coordinate grid and check your steps.
SketchMath 題庫內的 DSE 風格練習，選擇題及長題目均附解答。 DSE-style practice from the SketchMath question bank, with solutions for every MC and long question.