Question

若关于x的函数y=-x·2+2mx-4的顶点在直线y=2x-1上,则m=

Answer 1

y=-x^2+2mx-4

=-(x-m)^2-4+m^2

顶点(m,m^2-4)

m^2-4=2m-1

m^2-2m-3=0

m=3 or =-1