二次函數的極值 | Optimum Values of Quadratic Functions

[@IDENTITYHK]

IDENTITY 網站： project-identity.hk 影片內容： 00:35 - 配方法 | completing the square 07:37 - 二次函數的極值 | optimum values of quadratic functions 13:02 - 總結 (附找出頂點的捷徑) | summary (with a shortcut of finding the vertex) Playlist： HKDSE Maths | 函數與其圖像 | Functions and Graphs kzbin.info/aero/PL1Pv0-rhcQx6tW3L1NpXLKSQepnohVDVh

[@IDENTITYHK]

喺 notification 見到有人問咗條問題，但唔知點解個 comment 冇咗，我照喺度答返啦。個問題就咁：「點解 2:21 左右嗰題嘅第二個 step 係兩邊除 2，但係 5:13 嗰度就係抽個 2 出嚟？」 其實 2:21 嗰度我係播返另一條片 (二次公式 | Quadratic Formula) 嘅其中一段去解釋咩係 completing the square 配方法 (因為我懶…)。而嗰條片嘅主題其實係 quadratic equation，咁所以 2:21 嗰個位係講緊 “equation”. Equation 係有左右兩邊，2:21 做緊嘅係首先兩邊除 2 令 2x 變成 2 好睇啲。 5:13 就唔同喇，我地唔係睇緊 equation，而係純綷睇返 2x^2 - 8x + 9 呢個 function 函數本身。你無啦啦將佢除 2 嘅話係會令佢變成唔同嘅 function。 再畀多個例子喇。例如 2x = 0 呢條 equation。因為佢係 equation，所以你可以兩邊除 2 去得出 x = 0 呢個結果；但 f(x) = 2x 同 f(x) = x 就係兩個完全唔同嘅 function。

[@千羽樱]

最後 -b/2a 個証明 好似 Q.Formula 個証明, 其實依家讀緊中二學校就教 Quadratic Equation (冇校Discriminant之後哥D) 就逼我地條條都要用配方，唔俾用 Q.Formula. 所以睇起上去易明D. 多謝曬你地D片, 令我學到好多好有趣既數又唔會覺得悶.

[@IDENTITYHK]

見到「令我學到好多好有趣既數又唔會覺得悶」呢句真係有啲感動。話說我從你一直以嚟問嘅問題都 feel 到你 keep 住有進步，繼續加油喇。

[@chauyuenleung1503]

感謝有圖，容易理解咗，好用心的老師✨❤

[@hongningsuen1348]

To prove h is x-coordinate of vertex: f(x) = y = ax^2 + bx + c, dy/dx = 2ax + b, dy/dx = 0 at vertex of parabola as the turing point of graph, hence x-coordinate of vertex from 0 = 2ax + b => x = -b/2a. By comparing coefficients of identity ax^2 + bx + c = a(x - h)^2 + k, h = -b/2a. Hence h = x-coordinate of vertex. k is y-coordinate of vertex by substituting x = h into a(x - h)^2 + k = k. By comparing coefficients of above identity c = ah^2 + k. Hence k = c - ah^2 = c - a(-b/2a)^2 = (4ac - b^2)/4a as h = -b/2a from above. 4ac - b^2 = - D where D is the discriminant. Hence k = -D/4a which can be set into a program for solution instead of substituting h into ax^2 + bx + c = 0 to find k. To find h and k by completing the squares is a waste of time. Given general form with a, b, c, find h, k of vertex form by h = -b/2a and k = -D/4a, preferrably using a program. Given a vertex form with a, h, k, find b, c of general form by b = -2ah and c = ah^2 + k (by substituting h for x and -2ah for b into ax^2 + bx + c = k). When a is not given with h and k but a point on graph with given coordinates (x1, y1) is given, it is still possible to get a, b, c by setting up 2 simultaneous equations with (h, k) and (x1, y1) to solve a, b, c in ax^2 + bx + c = 0 by reducing 3 unknowns a, b, c to 2 unknowns a, c in ax^2 - 2ahx + ah^2 + k = 0. When a is positive, vertex of upward opening parabola has minimal y i.e. minima of function. When a is negative, vertex of downward opening parabola has maximal y i.e. maxima of function.

[@chauyuenleung1503]

明咗超多,感謝您🙏

[@doridorimidori7791]

感謝上傳

[@elephantjuice7192]

:)

[@qqqhinata8349]

your opening sucks too loud