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