Question

如图,已知抛物线y=k(x+1)(x-3k)(且k>0)与x轴分别交于A,B两点,A点在B点左边,与y轴交于C点,连接B

Answer 1

y=k(x+1)*(x-3k),k>0

A(-1,0),B(3k,0)

x=0,y=-3k^2

(1)

C(0,-3k^2)

(2)

k=1,B(3,0)

y=x^2-2x-3......(1)

k(AE)=k(BC)=1

AE:y=x+1......(2)

(1),(2):

E(4,5)

k(CP)=k(BE)=5

CP:y=5x-3

y=0,x=0.6

P(0.6,0)

(3)

AE:y=x+1

Q(q,1+q),B(3,0),O(0,0)

k(OQ)*k(BQ)=-1

[(1+q)/q]*[(1+q)/(q-3)]=-1

2q^2-q+1=0

(-1)^2-4*2*1