又是关于二次函数。。。

发布网友发布时间：2022-04-30 17:31

共3个回答

热心网友时间：2022-06-28 11:35

(1)
x^2-kx+k-5=0
△ = k^2 - 4(1)(k-5)
= k^2-4k+20
= (k-2)^2 + 16 > 0

=>无论k为何实数时，此二次函数与X轴都有两个交点
(2)
称轴为X=1
y=x^2-kx+k-5
y' = 2x - k =0
x = k/2 = 1
=> k = 2
(3)
y=x^2-2x-3
=(x-3)(x+1)
A(-1,0),B=(3,0)
x= 0 , y= -3
C(0,-3)
let D(d,d^2-2d-3)
slope of OD = (d^2-2d-3)/d
slope of CB = (-3-0)/(0-3) = 1
slope of OD x slope of CB = -1
=> (d^2-2d-3)/d = -1
d^2-d-3 = 0
d = (1+√13)/2

when d=(1+√13)/2
y = [(1+√13)/2]^2 -2[(1+√13)/2]-3
= (1+√13)^2/4 - (1+√13) -3
= (1+2√13 + 13)/4 -√13 - 4
= -1/2(1+ √13)
D( (1+√13)/2,-1/2(1+ √13))

热心网友时间：2022-06-28 11:36

1. 得塔>0
2. k/2 = 1 k = 2
3. 求出交点，求BC斜率，进而求出OD斜率，得D点坐标

不会hi我

热心网友时间：2022-06-28 11:36

与x轴的交点即是Y=0时的情况，给y=0带进去用求根公式解答如一楼。