二次函数y1=ax^2-2bx+c和y=(a+1)*x^2-2(b+2)x+c+3在同一坐标系中的图像如图所示,若OB=OA,BC=DC,且点B,C的横坐标分别为1,3,求着两个函数的解析式.
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![二次函数y1=ax^2-2bx+c和y=(a+1)*x^2-2(b+2)x+c+3在同一坐标系中的图像如图所示,若OB=OA,BC=DC,且点B,C的横坐标分别为1,3,求着两个函数的解析式.](/uploads/image/z/5560662-30-2.jpg?t=%E4%BA%8C%E6%AC%A1%E5%87%BD%E6%95%B0y1%3Dax%5E2-2bx%2Bc%E5%92%8Cy%3D%28a%2B1%29%2Ax%5E2-2%28b%2B2%29x%2Bc%2B3%E5%9C%A8%E5%90%8C%E4%B8%80%E5%9D%90%E6%A0%87%E7%B3%BB%E4%B8%AD%E7%9A%84%E5%9B%BE%E5%83%8F%E5%A6%82%E5%9B%BE%E6%89%80%E7%A4%BA%2C%E8%8B%A5OB%3DOA%2CBC%3DDC%2C%E4%B8%94%E7%82%B9B%2CC%E7%9A%84%E6%A8%AA%E5%9D%90%E6%A0%87%E5%88%86%E5%88%AB%E4%B8%BA1%2C3%2C%E6%B1%82%E7%9D%80%E4%B8%A4%E4%B8%AA%E5%87%BD%E6%95%B0%E7%9A%84%E8%A7%A3%E6%9E%90%E5%BC%8F.)
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