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有这样一道数学题:如图1:在直角坐标系中,抛物线y=x2+bx+c与轴交于A、B两点,与y轴交于点C,点B的坐标为(3,0)直线l:y=kx+3过C、B两点.⑴求直线l与抛物线所对应的函数关系式;⑵设抛物线的顶点为D,点P在抛物线的对称轴上,且∠APD=∠ACB,求点P的坐标;⑶连结CD,求∠OCA与∠OCD的两角和的度数.初看题目时,觉得图形很复杂,解题的思路会很局限,但是在实际教学过程中,我发现学生还是有一定的建构能力,想出了好几种非常好的解题方法,这里先介绍几种典型的解法,然后谈谈我的个人感受。解法一:构造相似三角形,寻找相
There is such a math problem: Figure 1: In the Cartesian coordinate system, the parabola y = x2 + bx + c and the axis intersect at A, B two points, and the y-axis intersect at point C, the coordinates of point B is (3, 0) straight line l: y = kx + 3 over C, B two points. (1) find the linear l and parabola corresponding function; ⑵ set the vertex of the parabola to D, point P on the parabola axis of symmetry, and ∠ APD = ∠ ACB, find the coordinates of point P; ⑶ link CD, find ∠ OCA and ∠ OCD of the two degrees and degrees. When you first look at the topic, think graphics are complex, problem-solving ideas will be very limited, but in the actual teaching process , I found that students still have a certain ability to construct, came up with several very good solution to the problem, here to introduce several typical solutions, and then talk about my personal feelings. Solution One: Construction of similar triangles, looking for phase