jmc

The vertex of the parabola appears to be at the value $x=1$, where $y=3$. Therefore we should have $$y=a(x-1{)}^2+3$$ for some integer $a$. We also know that $(0,1)$ is on the graph of the equation, so $$1=a(0-1{)}^2+3=a+3.$$ Therefore $$a=1-3=\boxed{-2}.$$