Answer
See graph and explanations.
Work Step by Step
1. Rewrite the equation as $x=-4(y^2+y)-3=-4(y+\frac{1}{2})^2-2$ or $(y+\frac{1}{2})^2=-\frac{1}{4}(x+2)$,
2. thus we can identify it as a parabola opens left with $p=\frac{1}{16}$,
3. vertex $(-2,-\frac{1}{2})$,
4. focus $(-2-\frac{1}{16},-\frac{1}{2})$ or $(-\frac{33}{16},-\frac{1}{2})$,
5. directrix $x=-2+\frac{1}{16}=-\frac{31}{16}$,
6. axis of symmetry $y=-\frac{1}{2}$.
7. See graph.
