f.4 mathz...

It is given that f(x) = ax^2 + 2x + 1 and g(x) = x^2 + 3x.

a) If f(-2)=1, find the value of a.
f(-2) = 1 and f(x) = ax^2 + 2x + 1
then：
a(-2)^2 + 2(-2) + 1 = 1
4a – 4 + 1 + 3 = 1 + 3
4a ÷ 4 = 4 ÷ 4
a = 1


b) If h(x)=g(x-1), find h(x) in terms of x.
Since h(x)=g(x-1) and g(x) = x^2 + 3x
then：
h(x) = (x – 1)^2 + 3(x – 1)
= (x - 1)(x – 1 + 3)
= (x – 1)(x + 2)
= x^2 + x – 2


c) Solve the equation f(x)=h(x)+2.
Since f(x) = ax^2 + 2x + 1、h(x) = x^2 + x – 2 and a =1
then：
x^2 + 2x + 1 = x^2 + x – 2
x^2 – x^2 + 2x – x + 1 + 2 = x^2 – x^2 + x – x – 2 + 2
x + 3 – 3 = 0 – 3
x = -3

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2008-10-19 13:42:00 補充：
對不起, c)部份睇漏了[ +2 ], 現更正c)部份:

c) Solve the equation f(x)=h(x)+2.
Since f(x) = ax^2 + 2x + 1、h(x) = x^2 + x – 2 and a =1
then：
x^2 + 2x + 1 = (x^2 + x – 2) + 2
x^2 – x^2 + 2x – x + 1 + 2 = x^2 – x^2 + x – x – 2 + 2 + 2
x + 3 – 3 = 2 – 3
x = -1

(a)
f(-2) = a(-2)^2 +2(-2)+1 = 1
4a -3 = 1
4a = 4
a = 1
(b) h(x) = g(x-1)
= (x-1)^2 +3(x-1)
= x^2 +x -2
(c) f(x) = h(x) + 2
f(x) = x^2 + 2x+1 = h(x) + 2 = x^2 + x - 2 + 2
x^2 + 2x + 1 = x^2 + x
x + 1 = 0
x = -1

a)
1= a(-2)^2 +2(-2) +1 (because(x)=(-2) so -2=x)
1=4a -4 +1
1=4a -3
4a=4
a=1
b)
h(x)=(x-1)^2 +3(x-1)
h(x)=x^2 -2x +1 +3x -3
h(x)=x^2 +x -2

c)
x^2 +2x +1=x^2 +x -2 +2
2x+1=x
x=-1

a)
f(-2) = a(-2)^2 + 2(-2) + 1 = 1
4a-4 = 0
a = 1

b)
h(x)=g(x-1)
h(x) = (x-1)^2 + 3(x-1)
h(x) = x^2 + x - 2

c)
f(x)=h(x)+2
x^2 + 2x + 1 = (x^2 + x - 2) + 2
x = -1

(a)
f(-2) = a(-2)^2 +2(-2)+1 = 1
4a -3 = 1
4a = 4
a = 1
(b) h(x) = g(x-1)
= (x-1)^2 +3(x-1)
= x^2 +x -2
(c) f(x) = h(x) + 2
f(x) = x^2 + 2x+1 = h(x) + 2 = x^2 + x - 2 + 2
x^2 + 2x + 1 = x^2 + x
x + 1 = 0
x = -1