我们在课堂教学中,如果注意对某些典型题目进行一题多解的讲授和训练,不仅可以提高课堂教学的功效,同时也可以培养学生发散性思维的能力。 题目:二次函数y=ax~2+bx+c(a≠0)在x=-1时有最小值-4,它的图象与x轴交点的横坐标分别为x_1、x_2,且(x_1)~c+(x_2)~2=10。求函数解析式。 解法1:设x_1、x_2是方程ax~2+bx+c=0(a≠0)的解。 In classroom teaching, if we pay attention to the teaching and training of some solutions to some typical topics, we can not only improve the effectiveness of classroom teaching, but also develop the ability of students to develop divergent thinking. Title: The quadratic function y=ax~2+bx+c(a≠0) has a minimum value of -4 at x=-1, and its abscissa at the intersection of the image and the x-axis is x_1, x_2, respectively. X_1)~c+(x_2)~2=10. Find the function analytical formula. Solution 1: Let x_1, x_2 be the solution of the equation ax~2+bx+c=0(a≠0).