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幼苗
共回答了21个问题采纳率:85.7% 举报
1)由题意,-2和0是方程ax^2 + bx + c = 0的两根,即得c = 0、b = 2a
∵函数有最小值,∴f(x)开口向上,∴a>0,f(x) = a(x+1)^2 - a最小值为-a = -1,∴a = 1,b = 2
∴y = f(x) = x^2 + 2x
2)F(x) = t*x^2 + 2tx - x - 3 = t*x^2 + (2t-1)x - 3 = t* {x - [(2t-1)/(2t)]}^2 - [(2t-1)^2 /(4t)] - 3
当对称轴x = (2t-1)/(2t) 在[-3/2,2]之内时,H(t) = F(2) = 8t - 5
由不等式组:-3t《2t-1 ,2t-1《4t,求得t范围:t》1/5;
当对称轴x = (2t-1)/(2t) 在[-3/2,2]右边时,H(t) = F(-3/2) = (9t/4) - 3t - (3/2) = -(3t/4) - (3/2)
解不等式(2t-1)/(2t)>2,得到t范围0>t>-1/2,与t》0矛盾;
当对称轴x = (2t-1)/(2t) 在[-3/2,2]左边时,H(t) = F(2) = 8t - 5,解不等式(2t-1)/(2t)
1年前
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