Question

已知二次函数y=mx²+（m-3)x+2m+3与x轴交于M、N两点,与y轴交于H,若△MNH是RT△,求m

Answer 1

H(0,2m+3) M(a,0) N(b,0)

HM⊥HN

[(2m+3)/a][(2m+3)/b]=-1

ab=-(2m+3)²

韦达定理

ab=(2m+3)/m

-(2m+3)²= (2m+3)/m

m(2m+3)²+ (2m+3)=0

(2m+3)[m(2m+3)+1]=0

(2m+3)[2m²+3m+1]=0

(2m+3)≠0

2m²+3m+1=0

(2m+1)(m+1)=0

m=-1/2 m=-1