中四數 急需

1. Because y=2x^2-3x-(k+1) doesn't touch the x-axis
so delta<0 (where delta=b^2-4ac)
(-3)^2-4(2)[-(k+1)]<0
9+8k+8<0
k<-17/8
2. (a)Because y=-4+kx-x^2 touches the x-axis
so delta=0
(k)^2-4(-1)(-4)=0
k^2-16=0
k=+4or-4(or +-4)
(b)When k>0,k=4
y=-4+(4)x-x^2
y=-(x^2-4x+4)
y=-(x-2)^2
The x-intercept is 2 as when x=2,y=0
3. k is a integer
Because the graph of y=kx^2+4x+1 opens upwards,
so k >0
Because y=kx^2+4x+1 has two distinct real roots
so delta>0
(4)^2-4(k)(1)>0------------>16-4k>0
4k<16<-------------16>4k
k<4
0<k<4,k can be 1,2,3

2008-09-26 18:47:13 補充：
2(b) The x-intercept is 2 as when x=2,y=0
The coordinates of the touching point should be(2,0)

c)k>0 開口向上
b^2-4AC>0, (2 points<>)
16-4k(1)>0
4k<16
k<4
–
- -k>0 and k<4. ie 0<k<4.
1 2 唔教你=-=

2008-09-25 21:06:53 補充：
2(b) (2,0)