1. **State the problem:** Given the parabola equation $$(x + 2)^2 = -12(y - 5),$$ find the vertex, focus, endpoints of the latus rectum, and the equation of the directrix. Then sketch the parabola. 2. **Identify the form:** This is a parabola in the form $$(x - h)^2 = 4p(y - k),$$ where the vertex is at $$(h, k).$$ Here, $$h = -2$$ and $$k = 5$$. 3. **Find $$p$$:** Compare $$-12$$ with $$4p$$, so $$4p = -12 \\ p = -3.$$ Since $$p < 0,$$ the parabola opens downward. 4. **Vertex:** The vertex is at $$(-2, 5).$$ 5. **Focus:** The focus lies $$p$$ units from the vertex along the axis of symmetry (vertical line $$x = -2$$). Since $$p = -3,$$ the focus is at: $$ (-2, 5 + p) = (-2, 5 - 3) = (-2, 2). $$ 6. **Directrix:** The directrix is a horizontal line $$p$$ units opposite the focus from the vertex: $$ y = k - p = 5 - (-3) = 8. $$ 7. **Latus rectum:** The length of the latus rectum is $$|4p| = 12.$$ The endpoints are horizontally aligned with the focus, so their coordinates are: $$ \left(-2 - \frac{12}{2}, 2\right) = (-8, 2), \quad \left(-2 + \frac{12}{2}, 2\right) = (4, 2). $$ **Summary:** - Vertex: $$(-2, 5)$$ - Focus: $$(-2, 2)$$ - Directrix: $$y = 8$$ - Endpoints of latus rectum: $$(-8, 2)$$ and $$(4, 2)$$ This parabola opens downward with vertex at $$(-2, 5)$$.