MATHS 1 ,2 (A) (B) 2條 中4級

1. y = -x^2 + 6x + c
y = -(x^2 - 6x) + c
y = -[(x)^2 - 2(x)(3) + (3)^2] + c + (3)^2
y = -(x - 3)^2 + (c + 9)

As the max value of the equation is 7, c + 9 = 7, i.e. c = -2

2a) y = a(x - h)^2 + k
As the vertex is (-2 , -2), therefore k = -2 and h = -2

the equation becomes: y = a(x + 2)^2 - 2

Put (0 , 1) into the equation
1 = 4a - 2
3 = 4a
a = 3/4

2b) y = a(x - h)^2 + k
As the vertex is (4, 7), therefore k = 7 and h = 4

The equation becomes: y = a(x - 4)^2 + 7
Put (0 , 3) into the equation
3 = 16a + 7
-4 = 16a
a = -1/4

答o左你2a先：
Since the minimun point is -2, hence k = -2

When x = -2, y = -2. Hence, we will have
-2 = a( -2 - h)^2 + k
-2 = a( -2 - h)^2 - 2 (as above, k = -2 )
0 = a( -2 - h)^2 (add 2 on both sides)
0 = ( -2 - h)^2 (divid both sides by a)
0 = -2 - h (square root both sides)
h = -2 (add h on both sides)

When x = 0, y = 1. Hence, we will have
1 = a ( 0 - h)^2 + k
1 = a ( 0 + 2)^2 - 2 (as above, k = -2 and h = -2)
3 = a (2)^2 (add 2 on both sides)
3 = a * 4 (2^2 = 4)
3/4 = a (divide both sides by 4)
a = 3/4

2007-11-26 15:41:02 補充：
上邊有人答o左，我唔答其他o個oD la