1. **State the problem:** We are given the equation of an ellipse: $$\frac{(x+1)^2}{9} + \frac{(y+3)^2}{4} = 1$$ and we want to understand its properties and graph. 2. **Formula and rules:** This is the standard form of an ellipse centered at $(-1, -3)$ with horizontal radius $a = 3$ (since $9 = 3^2$) and vertical radius $b = 2$ (since $4 = 2^2$). 3. **Find the center:** The center is at $(-1, -3)$. 4. **Find the vertices:** - Horizontal vertices are at $(h \pm a, k) = (-1 \pm 3, -3)$, which are $(-4, -3)$ and $(2, -3)$. - Vertical vertices are at $(h, k \pm b) = (-1, -3 \pm 2)$, which are $(-1, -1)$ and $(-1, -5)$. 5. **Plot points:** The points given approximately $(-5, -4)$, $(-3, -4)$, and $(1, -4)$ lie near the ellipse but are not vertices; they help visualize the shape. 6. **Summary:** The ellipse is centered at $(-1, -3)$, stretches 3 units left and right, and 2 units up and down. **Final answer:** The ellipse has center $(-1, -3)$, horizontal radius 3, vertical radius 2, and equation $$\frac{(x+1)^2}{9} + \frac{(y+3)^2}{4} = 1$$.