函數一條~~~急

a) f(x+1)=x^2+4k+k and f(2)=8
所以要搵到k,首先要代x+1=2
咁x就即係=1
f(2)=f(1+1)=1^2+4(1)+k=8
所以, k=8-5=3

so, f(x+1)=x^2+4x+3

b) f(-1)=f(-2+1)=(-2)^2+4(-2)+3
= -1

c) 想搵f(x)
就以f(x+1)=x^2+4k+3 泥搵
我地可以設y=x+1
咁即係話將x=y-1代入上面既formulae
即係f(x+1)=f(y-1+1)=f(y)=(y-1)^2+4(y-1)+3
= y^2-2y+1+4y-4+3
= y^2+2y

因為x, y都係代數
所以將y轉返晒做x都無問題
so, f(x)=x^2+2x

d) f(x)=x^2+2x=3
x^2+2x-3=0
(x-1)(x+3)=0
x=1 or x=-3

希望你明個step點做

f(x+1)=x^2+4k+k 係咪 f(x+1)=x^2+4x+k 呀?
a)
f(2) = 8 so x+1 = 2 -> x=1
f(2)=1+4+k
8=5+k
k=3

b)
when x+1 = -1. x = -2
f(-1)
= (-2)^2+4(-1)+3
= 4-4+3
= 3

c) let y=x+1, x=y-1
f(y)
= (y-1)^2 + 4(y-1) + 3
= y^2 -2y +1 +4y -4 + 3
= y^2 +2y
= y(y+2)
so, f(x) = x(x+2)

d)
f(x)=3
x(x+2)=3
x^2+2x-3=0
(x-1)(x+3)=0
so, x=1 or -3