標題:
5分4條 5分4條 5分4條
發問:
5分4條 http://i293.photobucket.com/albums/mm67/zaza520/0004-34.jpg
最佳解答:
(7a) Sub x = 0, y = 5 5 = k - 4 k = 9 (b) A and B are the roots of -x2 + 4x + 5 = 0 and then x2 - 4x - 5 = 0 (x - 5)(x + 1) = 0 x = 5 or -1 (c) Using completing square method: y = -x2 + 4x + 5 = -(x2 - 4x) + 5 = -(x - 2)2 + 9 So when x = 2, y is max. which is 9. (8) (a) x2 - 5x - 7 = 0 (x2 + 2) - 5x - 9 = 0 y = 5x + 9 (b) 5 - 4x + x2 = 0 (x2 + 2) - 4x + 3 = 0 y = 4x - 3 (c) 2x2 - 12x + 13 = 0 2(x2 + 2) - 12x + 9 = 0 2y = 12x - 9 (9) (a) Sub x = 0, y = 5 So C = 5 Sub x = 1, y = 8 8 = A + B + 5 A + B = 3 ... (1) Sub x = 6, y = -7 -7 = 36A + 6B + 5 6A + B = -2 ... (2) Solving we have A = -1 and B = 4 (b) The equation is -x2 + 4x + 5 = 0 x2 - 4x - 5 = 0 (x - 5)(x + 1) = 0 x = 5 or -1 (c) y = -x2 + 4x + 5 = -(x2 - 4x) + 5 = -(x - 2)2 + 9 So when x = 2, y is max. at 9 and thus the max. point is (2, 9). (d) -x2 + 4x + 5 = x + 2 x2 - 3x - 3 = 0 x = (3 + √21)/2 or (3 - √21)/2 (10) (a) Sub x = 2, y = 1 - p, so 1 - p = 4 + 2p + q 3p + q = -3 ... (1) Sub x = 6, y = - 5p/2, so - 5p/2 = 36 + 6p + q 17p/2 + q = -36 ... (2) Solving we have p = -6 and q = 15 (b) (i) From the given we have a + b = m a + (a2 - 6a + 15) = m m = a2 - 5a + 15 (ii) m = a2 - 5a + 15 = (a - 5/2)2 + 15 - 25/4 = (a - 5/2)2 + 35/4 So when a = 5/2, m is min. at 35/4 and also b = m - a = 25/4
5分4條 5分4條 5分4條
發問:
5分4條 http://i293.photobucket.com/albums/mm67/zaza520/0004-34.jpg
最佳解答:
(7a) Sub x = 0, y = 5 5 = k - 4 k = 9 (b) A and B are the roots of -x2 + 4x + 5 = 0 and then x2 - 4x - 5 = 0 (x - 5)(x + 1) = 0 x = 5 or -1 (c) Using completing square method: y = -x2 + 4x + 5 = -(x2 - 4x) + 5 = -(x - 2)2 + 9 So when x = 2, y is max. which is 9. (8) (a) x2 - 5x - 7 = 0 (x2 + 2) - 5x - 9 = 0 y = 5x + 9 (b) 5 - 4x + x2 = 0 (x2 + 2) - 4x + 3 = 0 y = 4x - 3 (c) 2x2 - 12x + 13 = 0 2(x2 + 2) - 12x + 9 = 0 2y = 12x - 9 (9) (a) Sub x = 0, y = 5 So C = 5 Sub x = 1, y = 8 8 = A + B + 5 A + B = 3 ... (1) Sub x = 6, y = -7 -7 = 36A + 6B + 5 6A + B = -2 ... (2) Solving we have A = -1 and B = 4 (b) The equation is -x2 + 4x + 5 = 0 x2 - 4x - 5 = 0 (x - 5)(x + 1) = 0 x = 5 or -1 (c) y = -x2 + 4x + 5 = -(x2 - 4x) + 5 = -(x - 2)2 + 9 So when x = 2, y is max. at 9 and thus the max. point is (2, 9). (d) -x2 + 4x + 5 = x + 2 x2 - 3x - 3 = 0 x = (3 + √21)/2 or (3 - √21)/2 (10) (a) Sub x = 2, y = 1 - p, so 1 - p = 4 + 2p + q 3p + q = -3 ... (1) Sub x = 6, y = - 5p/2, so - 5p/2 = 36 + 6p + q 17p/2 + q = -36 ... (2) Solving we have p = -6 and q = 15 (b) (i) From the given we have a + b = m a + (a2 - 6a + 15) = m m = a2 - 5a + 15 (ii) m = a2 - 5a + 15 = (a - 5/2)2 + 15 - 25/4 = (a - 5/2)2 + 35/4 So when a = 5/2, m is min. at 35/4 and also b = m - a = 25/4
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