1. Problem statement. The curve shown is symmetric and the choices are $x=0$, $y=0$, $y=-2$, and $x=2$. 2. Observation. The graph resembles a sideways parabola opening rightward and it crosses the y-axis at $y=-1$ and $y=0$. The shape appears mirrored across a vertical line through the origin. 3. Formula and rules. A curve is symmetric about the vertical line $x=a$ if whenever $(x,y)$ is on the curve then the reflected point $(2a-x,y)$ is also on the curve. In equation form, if the curve satisfies $F(x,y)=0$ then symmetry about $x=a$ implies $F(2a-x,y)=0$ whenever $F(x,y)=0$. 4. Apply to the figure. If the symmetry line is $x=0$ then reflections send $(x,y)$ to $(-x,y)$ because $2\cdot 0 - x = -x$. The graph shows matching points on opposite sides of the y-axis and intersections at $y=-1$ and $y=0$ that lie on that axis, which is consistent with symmetry about $x=0$. 5. Evaluate the options. Option (a) $x=0$ is the y-axis and matches the observed vertical mirror symmetry. Options (b) $y=0$ and (c) $y=-2$ are horizontal lines and do not match the observed vertical symmetry. Option (d) $x=2$ is a vertical line shifted right and does not pass through the observed axis of symmetry. 6. Final answer. The curve is symmetric about $x=0$.