解:设$A(x_{1},$$y)、$$B(x_{2},$$y)$
由题意可得$\begin{cases}{x_{1}+x_{2}=0}\\{x_{2}-x_{1}=1}\end{cases},$解得$\begin{cases}{x_{1}=-\dfrac 12}\\{x_{2}=\dfrac 12}\end{cases}$
$y=-2×(\frac 12)^2=-\frac 12$
∴$A(-\frac 12,$$-\frac 12)、$$B(\frac 12,$$-\frac 12)$