1.(a)
Given that f(x)=(x-a)(x-b) and a-b =k, prove that the minimum value of f(x) is k^2/a
—I would like to ask why the first step should be (a+b/2-a)(a+b/2-b)?
Would you explain that?Is it about the mid-pt thm or something else?
(b)Let g(x)=x^2-6x+5 and h(x-c)(x-3),where c＜3
-Use (a) if g(x) and h(x) have the same minimum value of C.
-Hence,if h(x)=g(x+m)=n,find the values of m and n.
Maths
2009-11-05 8:08 am
回答 (1)
2009-11-05 6:28 pm
✔ 最佳答案
1a.a and b are the x - intercepts of the function, so the axis of symmetry is
x = (a + b)/2, that means the x - coordinate of the vertex is (a + b)/2, when sub this value into the function will get the y - coordinate of the vertex which is also the minimum value of f(x).
So min. f(x) = [(a + b)/2 - a][(a + b)/2 - b] = (b - a)(a- b)/4 = - k^2/4.
1b.
g(x) = (x - 3)^2 - 9 + 5
so min. g(x) = - 9 + 5 = - 4.
min. h(x) = - (c -3)^2/4
so - (c - 3)^2/4 = - 4
(c - 3)^2 = 16
c - 3 = +/- 4
c = 7 (rej.) or - 1
so h(x) = (x + 1)(x - 3) = x^2 - 2x - 3
The axis of symmetry of h(x) is x = 1
The axis of symmetry of g(x) is x = 3
so if g(x) shift to the left by 2 units, g(x) will be equal to h(x),
that means x + m = x + 2, so m = 2.
h(x) = x^2 - 2x - 3 = n
so value of n depends on value of x.
收錄日期: 2021-04-25 22:37:46
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091105000051KK00018