Question

一道数学题，请详细解释一下，谢谢

已知二次函数y=(m-2)x^2+2mx-(3-m)的图像的开口向上，顶点在第三象限，且交于y轴的负半轴，则m的取值范围是

Answer 1

开口向上则m-2>0
顶点在第三象限则-(2m)/2(m-2)<0且[4(m-2)(m-3)-(2m)＾2]/4(m-2)<0
交于y轴的负半轴则-(3-m)<0
解上述四个不等式求交集得2<m<3

Answer 2

开口向上 => m-2>0 => m>2
顶点在第三象限，顶点是(m/(2-m),(6-5m)/(m-2)^2) =>
m/(2-m)<0, (6-5m)/(m-2)^2<0 => m>6/5
交于y轴的负半轴 => x=0时,y<0 => -(3-m)<0 => m<3
=> 2<m<3

Answer 3

m-2>0
-(2m)/2(m-2)<0且[4(m-2)(m-3)-(2m)＾2]/4(m-2)<0
-(3-m)<0
2<m<3