抛物线y=ax²-4ax+4a+c与x轴交于A,B与y轴正半轴交于C,点A(1,0),OB=OC,求顶点D,求解析式(特别急!迅速呀!跪求!)
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![抛物线y=ax²-4ax+4a+c与x轴交于A,B与y轴正半轴交于C,点A(1,0),OB=OC,求顶点D,求解析式(特别急!迅速呀!跪求!)](/uploads/image/z/8670298-58-8.jpg?t=%E6%8A%9B%E7%89%A9%E7%BA%BFy%3Dax%26%23178%3B-4ax%2B4a%2Bc%E4%B8%8Ex%E8%BD%B4%E4%BA%A4%E4%BA%8EA%2CB%E4%B8%8Ey%E8%BD%B4%E6%AD%A3%E5%8D%8A%E8%BD%B4%E4%BA%A4%E4%BA%8EC%2C%E7%82%B9A%EF%BC%881%2C0%EF%BC%89%2COB%3DOC%2C%E6%B1%82%E9%A1%B6%E7%82%B9D%2C%E6%B1%82%E8%A7%A3%E6%9E%90%E5%BC%8F%EF%BC%88%E7%89%B9%E5%88%AB%E6%80%A5%21%E8%BF%85%E9%80%9F%E5%91%80%21%E8%B7%AA%E6%B1%82%21%EF%BC%89)
抛物线y=ax²-4ax+4a+c与x轴交于A,B与y轴正半轴交于C,点A(1,0),OB=OC,求顶点D,求解析式(特别急!迅速呀!跪求!)
抛物线y=ax²-4ax+4a+c与x轴交于A,B与y轴正半轴交于C,点A(1,0),OB=OC,求顶点D,求解析式
(特别急!迅速呀!跪求!)
抛物线y=ax²-4ax+4a+c与x轴交于A,B与y轴正半轴交于C,点A(1,0),OB=OC,求顶点D,求解析式(特别急!迅速呀!跪求!)
抛物线y=ax²-4ax+4a+c与x轴交于A(1,0)所以a-4a+4a+c=0得a+c=0,所以c=-a,所以抛物线y=ax²-4ax+4a+c即y=ax^2-4a+3a=a(x-1)(x-3)所以抛物线交x轴于A(1,0),B(3,0)又OC=OB,点C在y轴正半轴上,所以C(0,3)所以得c=3,所以a=-3,所以解析式为y=-3x^2+12x-9,由配方法可得D(2,-1)