Question

若已知二次函数y=f(x)的图象过原点,且1小于等于f(-1)小于等于2,3小于等于f(1)小于等于4,求f(-2)的范围

Answer 1

设f(x)=ax"2+bx+c.则f(0)=c=0。1<=f(-1)=a-b<=2,3<=f(1)=a+b<=4。所以3<=3(a-b)<=6。所以f(-2)=4a-b=3(a-b)+(a+b),故6<=f(-2)<=10。