已知抛物线y=ax²+bx与x轴的交点A(1,0).B(0,3),且过点C(0,-3) (1)求抛已知抛物线y=ax²+bx与x轴的交点A(1,0).B(0,3),且过点C(0,-3)(1)求抛物线的解析式和顶点坐标(2)请你写出一
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![已知抛物线y=ax²+bx与x轴的交点A(1,0).B(0,3),且过点C(0,-3) (1)求抛已知抛物线y=ax²+bx与x轴的交点A(1,0).B(0,3),且过点C(0,-3)(1)求抛物线的解析式和顶点坐标(2)请你写出一](/uploads/image/z/3770027-35-7.jpg?t=%E5%B7%B2%E7%9F%A5%E6%8A%9B%E7%89%A9%E7%BA%BFy%EF%BC%9Dax%26%23178%3B%2Bbx%E4%B8%8Ex%E8%BD%B4%E7%9A%84%E4%BA%A4%E7%82%B9A%EF%BC%881%2C0%EF%BC%89.B%280%2C3%29%2C%E4%B8%94%E8%BF%87%E7%82%B9C%EF%BC%880%2C-3%EF%BC%89+%EF%BC%881%EF%BC%89%E6%B1%82%E6%8A%9B%E5%B7%B2%E7%9F%A5%E6%8A%9B%E7%89%A9%E7%BA%BFy%EF%BC%9Dax%26%23178%3B%2Bbx%E4%B8%8Ex%E8%BD%B4%E7%9A%84%E4%BA%A4%E7%82%B9A%EF%BC%881%2C0%EF%BC%89.B%280%2C3%29%2C%E4%B8%94%E8%BF%87%E7%82%B9C%EF%BC%880%2C-3%EF%BC%89%EF%BC%881%EF%BC%89%E6%B1%82%E6%8A%9B%E7%89%A9%E7%BA%BF%E7%9A%84%E8%A7%A3%E6%9E%90%E5%BC%8F%E5%92%8C%E9%A1%B6%E7%82%B9%E5%9D%90%E6%A0%87%EF%BC%882%EF%BC%89%E8%AF%B7%E4%BD%A0%E5%86%99%E5%87%BA%E4%B8%80)
已知抛物线y=ax²+bx与x轴的交点A(1,0).B(0,3),且过点C(0,-3) (1)求抛已知抛物线y=ax²+bx与x轴的交点A(1,0).B(0,3),且过点C(0,-3)(1)求抛物线的解析式和顶点坐标(2)请你写出一
已知抛物线y=ax²+bx与x轴的交点A(1,0).B(0,3),且过点C(0,-3) (1)求抛
已知抛物线y=ax²+bx与x轴的交点A(1,0).B(0,3),且过点C(0,-3)
(1)求抛物线的解析式和顶点坐标
(2)请你写出一种平移的方法,使平移后抛物线的顶点落在直线y=-x上,并写出平移后抛物线的解析式
已知抛物线y=ax²+bx与x轴的交点A(1,0).B(0,3),且过点C(0,-3) (1)求抛已知抛物线y=ax²+bx与x轴的交点A(1,0).B(0,3),且过点C(0,-3)(1)求抛物线的解析式和顶点坐标(2)请你写出一
抛物线应该是y=ax²+bx+c吧?B坐标应该是(3,0)吧?我就按我的理解做了
(1)将A,B,C坐标带入方程得
a+b+c=0
9a+3b+c=0
c=-3
解得a=-1,b=4,c=-3
则抛物线方程为y=-x²+4x-3
配方得y=-(x-2)²+1
所以定点坐标(2,1)
(2)将抛物线y向下平移3个单位,使得顶点为(2,-2) (这是最简便的,因为只需调整常数就行)
此时y=-(x-2)²-2=-x²+4x-6