抛物线ax²+bx+c过点(0,-1)和点(3,2).a＜0,顶点在直线y=3x–3上.求函数解析式

顶点在直线y=3x–3设顶点是(m,3m-3)则y=a(x-m)^2+3m-3所以-1=a(0-m)^2+3m-33=a(2-m)^2+3m-3相减4=a(4-4m)a=1/(1-m)所以-1=m^2/(1-m)+3m-3所以m-1=m^2-3m^2+6m-32m^2-5m+2=0(2m-1)(m-2)=0a1所以m=2所以y=-x^2+4x-1