解:设抛物线为y=a(x-h)²+k,则h=2,k=h-1解得h=2,k=1∴y=a(x-2)²+1∴-8=a(3-2)²+1∴a=-9因此所求表达式为y=-9(x-2)²+1即y=-9x²+36x-35。当x=0时y=-35;当y=0时x1=5/3,x2=7/3.因此抛物线与y轴交于点(0,-35),与x轴交于点(5/3,0),(7/3,0,).
