maths holiday的hw的Q22的ANS
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因為我們覺得這條數很難
希望可以給你們作一個參考
Q22
as 2 and 4 are x-intercepts of the graph
so,
0 = a*2^2 + b*2 + c ---- (1), i.e. c = -4a - 2b ---- (3)
0 = a*4^2 + b*4 + c ---- (2)
(2) - (1):
12a + 2b = 0
6a + b = 0, b = -6a ---- (4)
The coordinate of vertex = ( -b/2a, (4ac - b^2)/4a )
and as the x-coordinate is equal to the value of its y-coordinate, so
-b/2a = (4ac-b^2) / 4a
-2b = 4ac - b^2
b^2 - 2b = 4a(-4a-2b) , by (3)
b^2 - 2b = -16a^2 - 8ab
(-6a)^2 - 2(-6a) = -16a2 - 8a(-6a), by (4)
36a^2 + 12a = -16a^2 + 48a^2
4a^2 + 12a = 0
a(a+3) = 0
a = 0(rejected) or -3 as the graph have two x-intercepts, it must be quadratic
(4): b = -6a = -6(-3) = 18
(3): c = -4a -2b = -4(-3) - 2(18) = -24
hence y = -3x^2 + 18x - 24
BY S!N 蒨ISIS
Wish y0u All have a nice ChristmaS"
