抛物线y=ax2+12x-19顶点横坐标是3,则a=_.
问题描述:
抛物线y=ax2+12x-19顶点横坐标是3,则a=______.
答
∵抛物线的顶点横坐标是3,
∴-
=-b 2a
=3,解得,a=-2.12 2a