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一、二次函数与一元二次方程例1已知二次函数y=-x~2+2x+m的部分图象如图1所示.求关于x的一元二次方程-x~2+2x+m=0的解.解如图1,该二次函数的图象经过点(3、0),∴y=-x~2+2x+m可化为0=-3~2+2×3+m,即m=3,所以有-x~2+2x+3=0,即x~2-2x-3=0.解之得x_1=-1,x_2=3,故一元二次方程-x~2+2x+m=0的解是-1和3.二、二次函数与不等式例2二次函数y=ax~2+bc+c(a≠0)的图象如图2所示,则下列结论:①4a-2b+
First, the quadratic function and a quadratic equation Example 1 known quadratic function y = -x ~ 2 +2 x + m part of the image shown in Figure 1. Find the quadratic equation on x -x ~ 2 + 2x + m = 0. The solution is shown in Figure 1. The image of the quadratic function passes through point (3,0) and ∴y = -x ~ 2 + 2x + m can be 0 = -3 ~ 2 + 2 X 3 + m, ie m = 3, so there is -x ~ 2 + 2x + 3 = 0, that is, x ~ 2-2x-3 = 0. Solution of x_1 = -1, x_2 = 3, The solutions of the equations -x ~ 2 + 2x + m = 0 are -1 and 3. Second, Quadratic Functions and Inequalities Example 2 The image of quadratic function y = ax ~ 2 + bc + c (a ≠ 0) 2, the following conclusions: ① 4a-2b +