(1)∵抛物线y1=x2+2(1-m)x+n经过点(-1,3m+), ∴3m+=(-1)2+2(1-m)×(-1)+n=1-2+2m+n, 则n-m=;
(2)∵n-m=,即n=m+, ∴y1=x2+2(1-m)x+m+, ∴p=-=m-1, 将p=m-1代入得:q=-m2+3m+, ∵m=p+1, ∴q=-(p+1)2+3(p+1)+, 则q=-p2+p+;
(3)∵y1=x2+2(1-m)x+m+,y2=-2mx-, ∴代入y1≥2y2,得:x2+2(1-m)x+m+≥2(-2mx-), 整理得:x2+2(1+m)x+m+≥0, 由题意得到:△=4(1+m)2-4(m+)=4m2+4m-3≤0, 即(2m-1)(2m+3)≤0, 解得:-≤m≤, 当m=0时,经检验不满足题意, 则m的范围为-≤m≤且m≠0. |