f.4數學1題

(a)
y=ax^2+bx-3 which passes through P(1,-4)
put P(1,-4) into equation
-4=a+b-3
a+b=-1...(1)
one of the x-intercepts of the graph is 3.
so, put (3,0) is on the line
9a+3b-3=0
3a+b=1...(2)
from (1)
a=-1-b
put it into (2)
3(-1-b)+b=1
-2b=4
b=-2
So a=1
(b)
Now y=x^2-2x-3
put x=2, the corresponding value of y is
y=4-4-3=-3


2007-10-17 17:42:27 補充：
so, put (3,0) is on the line應寫成so, (3,0) is on the graph, put (3,0) into the equation

P(1,-4) when x=1 y=-4
-4=a(1)^2+(b)(1)-3
-4=a+b-3
a+b=-1
a=-b-1 ............(1)
x-intercepts of the graph is 3
when x=3 y=0
9a+3b-3=0
3a+b-1=0..........(2)
sub (1) into (2) , 3(-b-1)+b-1=0
-3b-3+b-1=0
-2b-4=0
b=-2.......(3)
sub (3) into (1), a=-(-2)-1
a=1
so the graph of the equation y=x^2-2x-3
(b). when x=2
y= 2^2-2(2)-3
y=4-4-3
y=-3

2007-10-17 17:42:43 補充：
俾10分我啦

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