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=______.

∵抛物线的顶点横坐标是3,
∴-

b
2a
=-
12
2a
=3,解得,a=-2.