3 The following figure [ http://webdisk2.cliffhk.net/10140/2.gif ]
shows the graph of y = ax² + bx + c
(a) find the values of a,b and c
(b)find the coordinates of vertex P
4 The following figure [ http://webdisk2.cliffhk.net/10140/4.gif ]
shows the graph of y = a(x - 2)(x + 3) with the vertex at ( h,8)
(a)find the value of h
(b)find the value of a
(c)find the y-intercept of the graph
F4 MATHS ( 20點)
2006-11-10 6:26 am
回答 (2)
2006-11-10 9:55 pm
✔ 最佳答案
The following is more quicker than the steps above. I'm using the symmetry here.3a. sum of roots = 2, product of roots = -8 when y = 0.
The equation is
x^2 - 2x - 8 = 0
Since it is the curve that the coeff of x^2 is negative,
so, -x^2 + 2x + 8 = 0.
k(-x^2 + 2x + 8) = 0
There are many other curves that satisfy the condition.
So, the curve is y = k(-x^2 + 2x + 8)
Therefore, a = -k, b = 2k, c = 8k where k is a constant and k GT 0.
(This question has 3 unknowns but only has 2 equations, GT: is greater than)
b. Since -2 and 4 are the roots for y = 0. Then, the x-coordinate of the vertex must be between 2 roots (i.e. half of the sum of the roots).
x = sum of roots/2 = 1
Put x = 1 back in the curve,
y = k(-1^2 + 2*1 + 8) = k(-1 + 2 + 8) = 9k
Therefore, vertex P is (1, 9k)
4a. The x-coordinate of the vertex is equal to the half of the sum of roots for y = 0.
x = [2 + (-3)]/2 = -0.5
So, h = -0.5.
b. Put the vertex into the equation of the curve,
8 = a(h - 2)(h + 3)
8 = a(-2.5)(2.5)
a = -8/6.25
a = -32/25
a = -1.28
c. Put x=0 into the curve,
y = a(0 - 2)(0 + 3)
= -6a
= 7.68
參考: my mathematical knowledge
2006-11-10 7:06 am
3 according to where the curve cuts x-axis, it is y = (x+2)(x-4)
y=x^2-2x-8
a=1
b=-2
c=-8
2006-11-09 23:17:17 補充：
It should beaccording to where the curve cuts x-axis, it is y = (x+2)(x-4)but as the curve goes downward as x inrease or decrease, it should be y = -(x+2)(x-4)y = -x^2 + 2x + 8a = -1b = +2c = +8
2006-11-09 23:39:16 補充：
y= -x^2 + 2x + 8 = -(x^2 - 2x +1 -1 -8) = -(x-1)^2 + 9when x = 1, the curve reached maximun height, 9 , point P is (1,9)
2006-11-09 23:39:46 補充：
4 the curve is y = a(x+3)(x-2) = -ax^2 -ax + 6ay = -a(x^2 + x - 6) = -a( x^2 + x + 1/4 - 1/4 + 6 )= -a(x^2 + x + 1/4) + a/4 - 6a = -(x + 1/2)^2 - 23a/4when x = -1/2, -23a/4 = 8, a = -(32/23)
2006-11-09 23:42:05 補充：
h = -1/2when x = 0, y= -6a = (-6)(-32/23) = 186/23y intercept = 186/23
2006-11-09 23:53:27 補充：
Sorry it should be 4 the curve is y = a(x+3)(x-2) = ax^2 -ax - 6ay = a(x^2 + x - 6) = a( x^2 + x + 1/4 - 1/4 - 6 )= a(x^2 + x + 1/4) - a/4 - 6a = a(x + 1/2)^2 - 25a/4when x = -1/2, -25a/4 = 8, a = -(32/25) h = -1/2when x = 0, y= -6a = (-6)(-32/25) = 186/25y intercept = 186/25
y=x^2-2x-8
a=1
b=-2
c=-8
2006-11-09 23:17:17 補充：
It should beaccording to where the curve cuts x-axis, it is y = (x+2)(x-4)but as the curve goes downward as x inrease or decrease, it should be y = -(x+2)(x-4)y = -x^2 + 2x + 8a = -1b = +2c = +8
2006-11-09 23:39:16 補充：
y= -x^2 + 2x + 8 = -(x^2 - 2x +1 -1 -8) = -(x-1)^2 + 9when x = 1, the curve reached maximun height, 9 , point P is (1,9)
2006-11-09 23:39:46 補充：
4 the curve is y = a(x+3)(x-2) = -ax^2 -ax + 6ay = -a(x^2 + x - 6) = -a( x^2 + x + 1/4 - 1/4 + 6 )= -a(x^2 + x + 1/4) + a/4 - 6a = -(x + 1/2)^2 - 23a/4when x = -1/2, -23a/4 = 8, a = -(32/23)
2006-11-09 23:42:05 補充：
h = -1/2when x = 0, y= -6a = (-6)(-32/23) = 186/23y intercept = 186/23
2006-11-09 23:53:27 補充：
Sorry it should be 4 the curve is y = a(x+3)(x-2) = ax^2 -ax - 6ay = a(x^2 + x - 6) = a( x^2 + x + 1/4 - 1/4 - 6 )= a(x^2 + x + 1/4) - a/4 - 6a = a(x + 1/2)^2 - 25a/4when x = -1/2, -25a/4 = 8, a = -(32/25) h = -1/2when x = 0, y= -6a = (-6)(-32/25) = 186/25y intercept = 186/25
收錄日期: 2021-04-25 16:36:20
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061109000051KK04744