已知抛物线y=ax^2+bx+c经过点(4,-6),(-2,0),a＞0,与x轴的交于A,B两点,与y轴交于点C,求△ABC面积S的最小值

因抛物线y=ax^2+bx+c经过点(4,-6),(-2,0),故16a+4b+c=-6,4a-2b+c=0,解得b=-(2a+1),c=-(8a+2) ①S△ABC=(1/2)|AB||c|=(4a+1)[√(b²-4ac)]/a,将①代入得,S△ABC=(4a+1)(6a+1)/a=24a+1/a +10≥4√6+10,当且仅当a=...