Answer
There is no symmetry about the x-axis, y-axis nor the origin.
See graph
The intercepts: $(-2,0)$, $(0,0)$, $(0,0)$.
Work Step by Step
$$y=-x^2-2x$$
Testing for x-symmetry:
$$y=-(-x)^2-2(-x)$$ $$y=-x^2+2x$$
Since the resulting equation is not the same as the original equation, there is no x-symmetry.
Testing for y-symmetry:
$$-y=-x^2-2x$$ $$y=x^2+2x$$
Since the resulting equation is not the same as the original equation, there is no y-symmetry.
Testing for origin-symmetry:
$$-y=-(-x)^2-2(-x)$$ $$-y=-x^2+2x$$ $$y=x^2-2x$$
Since the resulting equation is not the same as the original equation, there is no origin-symmetry.
Rewriting the equation in vertex form:
$$-y=x^2+2x$$
$$-y+(\frac{-2}{2})^2=x^2+2x+(\frac{-2}{2})^2$$
$$-y+1=x^2+2x+1$$
$$-(y-1)=(x+1)^2$$
$$y-1=-(x+1)^2$$
Thus, the vertex is at $(-1,1)$.
Finding the x-intercepts:
$$0=-x^2-2x$$
$$0=-x(x+2)$$
$$0=x(x+2)$$
$$x=0$$
$$x+2=0$$
$$x=-2$$
Thus, the x-intercepts are $(-2,0)$ and $(0,0)$.
Using the points, the graph is as shown.
The intercepts are as for x-intercepts $(-2,0)$ and $(0,0)$, and for y-intercept $(0,0)$.
