作业帮函数 如图所示问题
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函数 如图所示问题
函数 如图所示问题
函数 如图所示问题
1,很简单,就不写过程了
A(1,0) B(3,0) C(2,√3)
2,把A,B,C三点坐标分别带入解析式y=ax²+bx+c
解得解析式：y=—√3（x²—4x+3）
3,因为过D点,所以也过D点关于直线x=2的对称点F（如同）所以可知F点为（4.0）
所以过D,F.. a是已知值所以可得出解析式：y=-√3x²+4√3x+√3
1)
Let (x1, 0) and (x2, 0) be the coordinates of A and B, respectively.
So, c is at [(x1+x2)/2, sqrt(3)]
Since ABCD is a rhombus, x2 - x1 = (x1+x2)/2
=> x2 = 3x1
CB^2 = [(x1+x2)/2 ...
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1)
Let (x1, 0) and (x2, 0) be the coordinates of A and B, respectively.
So, c is at [(x1+x2)/2, sqrt(3)]
Since ABCD is a rhombus, x2 - x1 = (x1+x2)/2
=> x2 = 3x1
CB^2 = [(x1+x2)/2 - x2]^2 + 3 = (x2-x1)^2
Plug in x2 = 3x1,
x1^2 + 3 = 4x1^2
x1 = 1
x2 = 3
A: (1, 0)
B: (3, 0)
C: (2, sqrt(3))
2)
x1+x2 = 4
x1*x2 = 3
So, the equation can be written as
y = a(x^2 - 4x + 3)
Plug in the vertex at C,
sqrt(3) = a[4 - 8 + 3)
a = -sqrt(3)
y = -sqrt(3)[x^2 - 4x + 3]
3)
Let y = -sqrt(3)[x^2 - 4x + 3] + k
Plug in (0, sqrt(3)),
sqrt(3) = -3sqrt(3) + k
k = 4sqrt(3)
y = -sqrt(3)[x^2 - 4x + 3] + 4sqrt(3) = -sqrt(3)[x^2 - 4x - 1]
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