Quadratic Question & Algebraic（快）

12 by using the formula, b^2 -4ac = 0
(9k)^2 - 4(1)(9) = 0
k^2 = 4/9
k = 2/3

X-intercept = -b/2a = -6/2 = -3

13.by using the formula, b^2 -4ac = 0
(-2k)^2 - 4(4)(k+3) = 0
(k-6)(k+2) = 0 therefore, k= 6 or -2

X-intercept = -b/2a = -12/8 = - 3/2 , when k=6
or -4/4 = 1 wen k =2

16. by using the formula, b^2 -4ac = 0
(10)^2 - 4(1)(k) = 0

k = 25

X-intercept = -b/2a = -10/2 = - 5

11.
f(x)=x^2-9k+9
=(x^2-9k+81k^2/4)-81k^2/4+9
=(x-9k/2)^2+(36-81k^2)/4

If the vertex of f(x) touches the x-axis, (36-81k)/4=0.
(36-81k^2)/4 = 0
36-81k^2 = 0
k = 2/3 or -2/3
The value of k is -2/3 or 2/3.

For k = 2/3:
(9/2)(2/3)
= 3
For k = -2/3:
(9/2)(-2/3)
= -3
The x-intercept is -3 or 3.

有錯請指正!
另外, 你check一check其他幾題的題目有沒有打錯.

2007-10-19 12:57:02 補充：
11.有地方打錯了, 應該是: f(x)=x^2-9kx+9=(x^2-9kx+81k^2/4)-81k^2/4+9另外, b^2-4ac 是discriminant. 如果個graph touches x-axis at 1 point, discriminant = 0.其實, 可以用completing square方法計然後put perfect square出面的數為0 也可以計到的!

2007-10-19 13:06:09 補充：
13.f(x)=4x^2-2kx+k+3 =4(x^2-kx/2+k^2/16)-(k^2/4)+k+3 =4(x-k/4)^2-(k^2/4)+k+3If the vertex of f(x) touches the x-axis, -(k^2/4)+k+3=0.-k^2+4k+12 = 0k^2-4k-12 = 0(k-6)(k+2) = 0k = 6 or k = -2For k = 6,x = 6/4 = 3/2For k = -2,x = -2/4 = -1/2

2007-10-19 13:08:59 補充：
16.f(x)=x^2 10x k =(x^2 10x 25)-25 k =(x-5)^2-25 kIf the vertex of f(x) touches the x-axis, -25 k=0.Therefore, k = 25x-intercept is x = 5.