math mix sensations

1. h=30t-t^2
(a) h=30t-t^2
h=-(t^2-30t+225)+225
h=-(t-15)^2+225
Since -(t-15)^2≤0, the maximum value of h=0+225=225 which occurs when t=15.
The ball will attain its maximum height 15 seconds after being kicked.
(b) from (a) the max. height = 225, half max. height=225/2.
225/2=-(t-15)^2+225
(t-15)^2-225/2=0
(t-15)^2-(15/sqrt 2)^2=0
(t-15+15/sqrt 2)(t-15-15/sqrt 2)=0
t=15-15/sqrt 2 or t=15+15/sqrt 2
t=15-(15 sqrt 2)/2 or t=15+(15 sqrt 2)/2
The ball will reach its half maximum height 15-(15 sqrt 2)/2 or 15+(15 sqrt 2)/2 seconds after being kicked.

2. y=-2x^2+bx+c
Given: graph opens downwards and passes through (0,0) and (p,0). The MAX, value of y is 8.
(a) Substituting (0,0) into the graph equation:
0=-2(0)^2+b(0)+c
c=0
Therefore y=-2x^2+bx
y=-2[x^2+(b/2)x+(b^2)/4]+(b^2)/4
y=-2(x+b/2)^2+(b^2)/4
Since -2(x+b/2)^2≤0, the maximum value of y=(b^2)/4=8
b=sqrt 32=4 sqrt 2
The values of b and c are 4 sqrt 2 and 0 respectively.
(b) Substituting (p,0) into the graph equation:
0=-2(p)^2+b(p)
b=2p
p=b/2=(4 sqrt 2)/2=2 sqrt 2
The coordinates of P are (2 sqrt 2,0)

3. f(x) is a quadratic function. f(1)=f(3) and f(2)=-1
(a) y has same values at f(1) and f(3), therefore y has an axis of symmetry on x=(1+3)/2=2
The axis of symmetry of the graph is x=2.
(b) Let f(x)=ax^2+bx+c where a,b,c are constants.
f(1)=a(1)^2+b(1)+c=a+b+c ……..[1]
f(3)=a(3)^2+b(3)+c=9a+3b+c ……..[2]
f(2)=a(2)^2+b(2)+c=4a+2b+c=-1 ……..[3]
f(1)=f(3), therefore [1]=[2]
a+b+c=9a+3b+c
8a+2b=0
b=-4a
from [3]:
4a+2b+c=-1
4a+2(-4a)+c=-1
c=4a-1
This gives f(x)=ax^2-4ax+4a-1 = a(x^2-4x+4)-1 = a(x-2)^2-1
A possible algebraic expression for f(x) is f(x)=a(x-2)^2-1

Q4 and Q5 in comment due to word limit exceeded.

2014-10-30 18:17:57 補充：
Please open another request for Q4 and Q5, the word limit for reply has been exceeded and the word limit for comment (300 words) is insufficient to hold the answers.

okay!
thanks Eddie for answering my question first����