S.4 Function

👨🏻‍🏫 學術台數學

[匿名]

2011-10-22 10:31 pm

It is given that f(x)= x^2 + 3k and g(x) = x^2 +k
(a) Find f(x+3) and g(x-3).
(b) If f(x+3) - g(x-3) = 12x + 1, find the value of k.

***我計左(a)
***答案如下：(a) f(x+3) = x^2 + 6x + 9 +3k, g(x-3) = x^2 - 6x + 9 + k
(b) 1/2

更新1:

原來我計錯左少少野，Thx a lot!! :)

回答 (2)

myisland8132

2011-10-22 10:37 pm

✔ 最佳答案

(a) f(x + 3)

= (x + 3)^2 + 3k

= x^2 + 6x + 9 + 3k

g(x - 3)

= (x - 3)^2 + k

= x^2 - 6x + 9 + k

(b) f(x + 3) - g(x - 3) = 12x + 1

(x^2 + 6x + 9 + 3k) - (x^2 - 6x + 9 + k) = 12x + 1

12x + 2k = 12x + 1

k = 1/2

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fighting gravity

2011-10-22 10:48 pm

f(x)= x^2 + 3k and g(x) = x^2 +k
(a)
f(x+3)=(x+3)^2+3k
f(x+3)=x^2+6x+9+3k
g(x-3)=(x-3)^2+k
q(x-3)=x^2-6x+9+k
(b)
f(x+3) - g(x-3) = 12x + 1
x^2+6x+9+3k-(x^2-6x+9+k)=12x+1
x^2+6x+9+3k-x^2+6x-9-k=12x+1
12x+2k=12x+1
2k=1
k=1/2

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