Functions

a)
y=b-(x+a)^2

when x=0, y=-3
y=b-(x+a)^2
-3=b-(0+a)^2
-3=b-(a^2)
b=(a^2)-3 ---(1)

when x=1, y=0
y=b-(x+a)^2
0=b-(1+a)^2
b=(1+a)^2

b=(a^2)-3=(1+a)^2
(a^2)-3=(1+a)^2
(a^2)-[(1+a)^2]-3=0
(a^2)-[(a^2)+2a+1]-3=0
(a^2)-(a^2)-2a-1-3=0
-2a-4=0
-2a=4
a=-2

sub a=-2 into (1)
b=(a^2)-3
b=[(-2)^2]-3
b=1

when x=c ,y=0
y=b-(x+a)^2
y=1-[x+(-2)]^2
0=1-[c+(-2)]^2
0=1-(c-2)^2
[(c-2)^2]=1
c-2=±√(1)
c-2=±1
c=2±1
c=2+1 or 2-1
c=3 or 1

∴a=-2, b=1, c=3,1

b)
y=b-(x+a)^2
y=1-[x+(-2)]^2
y=-[(x-2)^2]+1

∴the greatest value of y is 1 when x=2

2006-10-23 17:05:54 補充：
a) ∴a=-2, b=1, c=3 or 1

2006-10-23 18:52:15 補充：
c=1 (rejected) because a point (1,0) is above or c=3so c=3

a) y=b-(x+a)^2 passes through the points (0,-3)(y-intercept),(1,0) and (c,0)
when x=0, y=-3
-3=b-a^2
when x=1, y=0
0=b-(1+a)^2
b=a^2+2a+1
Put b=a^2+2a+1 into -3=b-a^2
-3=a^2+2a+1-a^2
2a=-4
a=-2
put a=-2 into -3=b-a^2
-3=b-(-2)^2
b=1

so, y=1-(x-2)^2

when x=c, y=0
0=1-(c-2)^2
c^2+4c-5=0
(c-1)(c+5)=0
c=1 or c=-5(rej beacuse c is greater than 0)

so c=1

b) y=1-(x-2)^2
because (x-2)^2 must equal or greater than 0
So, to obtain the greatest value of y
(x-2)^2 must be equals to 0
then the greatest value of y=1-0=1

2006-10-23 17:17:09 補充：
the greatest value of y=1 when x=2

2006-10-23 17:23:40 補充：
have some wrong calculationsa) when x=c, y=00=1-(c-2)^2c^2-4c 3=0(c-1)(c-3)=0c=1 (rej as there is a point (1,0) mentioned above) or c=3so c=3