1. **State the problem:** Find the co-vertices of the ellipse given by the equation $$\frac{(x+3)^2}{4} + \frac{(y+1)^2}{9} = 1$$ rounded to the nearest tenth if necessary. 2. **Identify the ellipse parameters:** The ellipse is centered at $(-3, -1)$. 3. **Determine the major and minor axes:** The denominators under the squared terms indicate the lengths of the axes. - The larger denominator is 9 under $(y+1)^2$, so the major axis is vertical. - The smaller denominator is 4 under $(x+3)^2$, so the minor axis is horizontal. 4. **Calculate the lengths of the semi-major and semi-minor axes:** - Semi-major axis length $a = \sqrt{9} = 3$ - Semi-minor axis length $b = \sqrt{4} = 2$ 5. **Find the co-vertices:** Co-vertices lie along the minor axis (horizontal axis here), so their $x$-coordinates vary by $\pm b$ from the center, and $y$-coordinate remains the same. - Center: $(-3, -1)$ - Co-vertices: $(-3 \pm 0, -1 \pm 2)$ since minor axis is horizontal, co-vertices are on vertical axis. Actually, since the major axis is vertical, co-vertices are on the horizontal axis: - Co-vertices: $( -3 \pm 2, -1 )$ 6. **Calculate co-vertices coordinates:** - $(-3 + 2, -1) = (-1, -1)$ - $(-3 - 2, -1) = (-5, -1)$ 7. **Final answer:** The co-vertices are at $(-1, -1)$ and $(-5, -1)$. Note: The user mentioned the co-vertices at $(-3, 2)$ and $(-3, -4)$ which correspond to the vertices (major axis endpoints), not co-vertices. Hence, the co-vertices are $\boxed{(-1, -1) \text{ and } (-5, -1)}$.