Functions and Graphs

1.
f(x)=ax+b , f(1-x)=1-2x
a(1-x)+b = 1 - 2x
-ax + a + b = 1 - 2x
compare the coefficient of x and x^0
a = 2 a+ b= 1
b = -1


y=k(x-2)^2+k^2+2k-3
-b/2a = 2
the maximum point is found at x = 2
(Remark: 2次方程中 y = ax^2 + bx + c
最大點/最小點的x座標可以公式 -b/2a 找出)

put x = 2, y = 5
k^2 + 2k - 3 = 5
k^2 + 2k - 8 = 0
k = 2 or -4

2.a) 用返第一題的方法
the vertex is found at x = 1
- b/2a = 1
-b = 2a
2a + b = 0

b) minimum value of y is 2
(1,2)呢點可能係極大點或極小點
因此y=2可能係極大值
若a>0,(1,2)就是極大,反之就是極小

3. f(x)=x^2+x-1
-b/2a = -1
the vertex is found at x = -1
f(-1) = (-1)^2 -1 -1
=-1
the vertex is (-1,-1)

Dear leung, you are right... thanks for the correction. Below is the modified answer.

Q1)
f(x) = ax + b
f(1-x) = a(1-x) + b
f(1-x) = a - ax + b

f(1-x) = 1 - 2x

By equating the coefficients,
x term:
-a = -2
a = 2

constant term:
1 = a + b
b = 1 - 2
b = -1

2009-01-06 20:54:23 補充：
Q2)
y = k(x-2)^2 + k^2 + 2k - 3
Given that max. value: y = 5, such that:

5 = k^2 + 2k - 3
0 = k^2 + 2k - 8
0 = ( k+4 )( k-2 )
k = -4 or k = 2 ( rejected, as there should be only max. point )
Therefore, k = -4

2009-01-06 20:54:36 補充：
Q3)
(1, 2) is the vertex of y = ax^2 + bx + c... such that
dy/dx = 2ax + b, and when dy/dx = 0,

2a + b = 0 .... Therefore, statement (a) is correct.

Also, vertex (1, 2) shows the min. value of y = 2

Therefore, statement (b) is correct.

2009-01-06 20:54:44 補充：
Q4)
f(x) = x^2 + x - 1
f'(x) = 2x + 1

when f'(x) = 0,
0 = 2x + 1
x = -1/2

when x = -1/2,
f(x) = (-1/2)^2 +(-1/2) - 1
f(x) = 1/4 - 1/2 - 1
f(x) = -3/4

Therefore, the vertex of f(x) is ( -1/2, -3/4 )

mr_gary_cheung的答案有d錯

首先,k應該= 2 or -4 ,不過相信只係打漏左個負號

另外,k=2應該要rejected

因為如果k=2,條curve會open upward,沒有maximun pt,只有minimun pt

所以k=2要rejected

p.s. 如有任何問題,請指出,thanks~