如图,在平面直角坐标系$xOy$中,已知抛物线$y=ax^{2}+bx(a>0)$的顶点为$C$,与$x$轴的正半轴交于点$A$,它的对称轴与抛物线$y=ax^{2}(a>0)$交于点$B$.若四边形$ABOC$是正方形,则$b$的值是<ImageHere>
由抛物线$y=ax^{2}+bx$可知,点$C$的横坐标为$-frac{b}{2a}$,纵坐标为$-frac{b^{2}}{4a}$.$because$四边形$ABOC$是正方形,$therefore-frac{b}{2a}=-left(-frac{b^{2}}{4a}right)$.$thereforeb=-2$.