1.b^2- 4ac = (2m+1)^2 - 4(2m-2) = 4m^2 + 4m +1 - 8m + 8
= 4m^2 - 4m + 9 = (2m - 1)^2 + 8 > 0
所以,該方程有兩個(gè)不相等的實(shí)數(shù)根．
2．根據(jù)韋達(dá)定理,有：X1＋X2＝2m+1,X1X2＝2m-2
則 y=1/2(X1^2X2+X1X2^2)＝X1X2（X1＋X2）／2 ＝ （2m-2）（2m+1）／2
＝（m - 1)(2m +1）= 2m^2 - m - 1
所以,函數(shù)的解析式y(tǒng) = 2m^2 - m - 1
3．沒法畫圖,只能用數(shù)學(xué)解法．
2m^2 - m - 1 ＜＝ 2m+1
2m^2 - 3m - 2＜＝0
（2m + 1)(m - 2)＜＝0
-1/2