抛物线y=3-2x-x2,开口向_,顶点坐标为_.对称轴是直线:_.
问题描述:
抛物线y=3-2x-x2,开口向______,顶点坐标为______.对称轴是直线:______.
答
y=3-2x-x2,
=-(x2+2x+1)+1+3,
=-(x+1)2+4,
∵a=-1<0,
∴开口向下,顶点坐标是(-1,4),对称轴是直线x=-1.
故答案为:向下,(-1,4),x=-1.