maths problem (20)

[匿名]

2010-04-25 12:26 am

I don"t know how to do it.
And it was very important for me.
Can anyone help me ?

If the graph of y = (x-4p)(x-2)+p has only one x-intercept,
(a) find the possible value(s) of p,
(b) find the x-intercept of the gragh for each value of p in (a).

很急...如果有人識的話可以在這裡教教我嗎...
20點!!!

回答 (1)

用戶2053

2010-04-25 12:48 am

✔ 最佳答案

a) Since the graph of y = (x-4p)(x-2)+p has only one x-intercept,
So 0 = (x-4p)(x-2)+p have one solution ,
x^2 - 4px - 2x + 8p + p = 0 have one solution ,
x^2 - 2(2p+1)x + 9p = 0 have one solution ,
△ = [-2(2p+1)]^2 - 4(9p) = 0
(2p+1)^2 - 9p = 0
4p^2 + 4p + 1 - 9p = 0
4p^2 - 5p + 1 = 0
(4p - 1)(p - 1) = 0
p = 1/4 or p = 1
b)When p = 1/4 ,
x^2 - 2(2p+1)x + 9p = 0
x^2 - 2(2/4 + 1)x + 9/4 = 0
x^2 - 3x + 9/4 = 0
(x - 3/2)^2 = 0
x = 3/2
x-intercept = 3/2

When p = 1 ,
x^2 - 2(2p+1)x + 9p = 0
x^2 - 6x + 9 = 0
(x - 3)^2 = 0
x = 3
x-intercept = 3

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