M 2-2m-3 < 0, and because the images are distributed in one or two quadrants, so

Is even, that is, negative even,

m^2-2m-3=(m- 1)^2-4≥-4

So-4 ≤ m 2-2m-3

When m 2-2m-3 =-4, = = & gtm= 1

When m 2-2m-3 =-2, = = & gtm= 1 √2.