(數學中四)二次函數問題?

2010-03-09 8:23 pm
已知(2,5)和(-1,11)是二次函數y=ax二次+bx+1的圖像上的兩點.

求常數a和b的值.

回答 (2)

2010-03-09 8:54 pm
✔ 最佳答案
y = ax^2 + bx + 1
5 = a(2)^2 + b(2) + 1
5 = 4a + 2b + 1
4 = 4a + 2b
2a + b = 2......(1)
11 = a(-1)^2 + b(-1) + 1
11 = a - b + 1
a - b = 10......(2)
(1) + (2) :
3a = 2 + 10
a = 4
By (2) : 4 - b = 10
b = - 6
2010-03-09 10:25 pm
已知(2,5)和(-1,11)是二次函數y=ax二次+bx+1的圖像上的兩點.求常數a和b的值.
Sol
f(x)=a(x-2)(x+1)+p(x-2)+5
f(-1)=p(-3)+5=11
p=-2
f(x)=a(x-2)(x+1)-2(x-2)+5
=a(x^2-x-2)-2x+9
=ax^2+(-a-2)x+(9-2a)
9-2a=1
a=4
-a-2=b
b=-6


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