請數學達人幫手一條quadratic function

[匿名]

2009-09-18 6:22 am

C is the graph of the following function.

f(x)= [4x^2+(6+k)x+3]/(1-k) ,where k≠1

Prove that for all values of k except 1, C must pass through two fix points and find these two points.

two fix points not in teams of k

急~
要有詳細步驟

更新1:

Rewriting C as: f(x) = [4x^2 + 6x + 3 - k(- x)]/(1 - k) So it must pass through the points of intersection between the curves y = 4x^2 + 6x + 3 and y = - x

更新2:

Equating them togehter: 4x^2 + 6x + 3 = - x 4x^2 + 7x + 3 = 0 (4x + 3)(x + 1) = 0 x = - 3/4 or - 1 y = 3/4 or 1 So C must pass through (-3/4, 3/4) and (-1, 1) for all values of k NOT equal to 1. If answer it is ,i don't know y = - x

更新3:

why y = - x

回答 (1)

用戶10012

2009-09-18 6:39 am

✔ 最佳答案

Let f(x) = y
(1 - k)y = 4x^2 + (6 + k)x + 3
Let k = m
(1 - m)y = 4x^2 + (6 + m)x + 3 ...............(1)
Let k = n
( 1 - n)y = 4x^2 + (6 + n)x + 3 ............(2)
(1) - (2)
y - my - y + ny = 4x^2 + 6x + mx + 3 - 4x^2 - 6x - nx - 3
ny - my = mx - nx
y(n - m) = -(n -m)x
so y = - x.............(3)
That means for any value of k ( = m or = n) , eqt(3) holds.
So put y = f(x) = -x, we get
(1 - k)(-x) = 4x^2 + (6 + k)x + 3
-x + kx = 4x^2 + 6x + kx + 3
4x^2 + 7x + 3 = 0
(4x + 3)(x + 1) = 0
x = -3/4 or - 1
so y = 3/4 or 1.
So the 2 fixed points are (-3/4, 3/4) and (-1,1).

2009-09-17 22:43:37 補充：
Another explanation of y = - x is that f(x) is the family of curves passing through the intersection of y = 4x^2 + 6x + 3 and y = -x.

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請數學達人幫手 一條quadratic function

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請數學達人幫手一條quadratic function