(1) f（1）=3,f（2）=9/2
3=a*1²+1+1/b*1 a=2-1/b
9/2=a*(9/2)²+1/b*(9/2) 9/2a+1/b=1
9/2(2-1/b)+1/b=1
9-9/(2b)+1/b=1
-7/(2b)=-8
b=7/16
a=2-1/(7/16)=-2/7
f(x)=(-2/7+1)x²+1/(7/16)x
=5/7x²+16/7x
(2) f(x)=5/7(x²+16/5x)
=5/7(x+8/5)²-64/35
对称轴x=-8/5的右边为增函数,即单增区间：x>-8/5
∴f(x)在[√2/2,+∞)上是增函数