设该二次函数解析式为：y = ax^2 + bx + c
由二次函数y=f(x)的两零点为x1=0,x2=1得到对称轴为：x = (0+1)/2 = 1/2 ,
∵顶点在对称轴上 ,∴定点横坐标 = 1/2 ,又∵顶点也在y = log2(x)上 ,∴顶点纵坐标 = log2(1/2) = -1
配方得：y = a[x + (b/2a)]^2 + (4ac - b^2)/4a
∴-b/2a = 1/2 ,(4ac - b^2)/4a = -1
将两个零点(0 ,0)、(1 ,0)代入解析式 ,列方程组解得：
a = 4 ,b = -4 ,c = 0
∴f(x) = 4x^2 - 4x
f(x) = 4[x - (1/2)]^2 - 1
当t > 1/2 时 ,g(t) = f(t) = 4t^2 - 4t
当t + 1《 1/2 ,即t《 -1/2时 ,g(t) = f(t+1) = 4t^2 + 4t
当 t《 1/2 < t+1 时 ,g(t) = f(1/2) = -1