2.Let F(x)=x^2-6x+k
the curve y=F(x) cut the x-axis at A(x1,0) and B(x2,0),V is the vertex of the curve.
2.a.1.Find,in terms of k ,the coordinates of V.
2.a.2.the length of AB.
2.a.3.If the area of triangle VAB is 8 units,show that k=5.
2.b.the curve is shifted vertically downwards by m units to form the curve y=G(x).The curve y=G(x) cuts the x-axis at A' and B'
Suppose k=5,
2.b.1.Find, in terms of m , the length of A'B'
2.b.2.If A'B'=2AB,find the length of m.
難度偏高F4 a maths 1條 10分[急]
2007-01-22 5:48 am
回答 (2)
2007-01-22 6:06 am
✔ 最佳答案
a1)F'(x) = 2x-6 = 0
x = 3
F(3) = (3)^2-6(3)+k = k-9
The coordinates of V is (3, k-9)
a2)
x1 and x2 are the roots of F(x) = 0
Therefore, x1+x2 = 6 and x1x2 = k
Length of AB = |x1-x2| = sq root[(x1+x2)^2- 4x1x2]
= sq root (6^2-4k) = sq root (36-4k) = 2 * sq root (9-k)
a3)
The base of triangle VAB = length of AB = 2 * sq root (9-k)
height of triangle VAB = |k-9|
Area of triangle VAB = 8
ie. 0.5 * 2 * sq root (9-k) |k-9| = 8
(9-k)^3 = 64
9-k = 4
Therefore, k = 5 as required
b1)
Since the curve shifted downwards for m units
Therefore, G(x) = F(x) - m
= x^2 - 6x + (5-m)
Length of A'B' = sq root [6^2 - 4(5-m)]
= 2 sq root (4+m)
b2)
A'B' = 2AB = 2 * 4 = 8
Therefore 2 sq root (4+m) = 8
ie. sq root (4+m) = 4
4+m = 16
m = 12
2007-01-22 6:25 am
a(i)F(x)=x^2-6x+k=x^2-6x+9+(k-9)
=(x-3)^2+(k-9)
V=(3,k-9)
a(ii).AB=difference of root=[(sum of root)^2-4(product of root)]^1/2
=(36-4k)^1/2
a(iii) Simply by applying base*height/2, the result, k=5 can be obtained
=(x-3)^2+(k-9)
V=(3,k-9)
a(ii).AB=difference of root=[(sum of root)^2-4(product of root)]^1/2
=(36-4k)^1/2
a(iii) Simply by applying base*height/2, the result, k=5 can be obtained
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