Question

又是关于二次函数...已知二次函数Y=X²-kX+k-5(1)求证：无论k为何实数时,此二次函数与X轴都有两个

Answer 1

(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))