1. **State the problem:** We are given the hyperbola equation $$\frac{(y+2)^2}{36} - \frac{(x+5)^2}{64} = 1$$ and need to analyze its properties including center, vertices, and asymptotes. 2. **Identify the center:** The hyperbola is in the form $$\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$$ which means it opens vertically. Here, $$h = -5$$ and $$k = -2$$ so the center is at $$(-5, -2)$$. 3. **Determine $$a$$ and $$b$$:** From the denominators, $$a^2 = 36$$ so $$a = 6$$, and $$b^2 = 64$$ so $$b = 8$$. 4. **Find the vertices:** Since the hyperbola opens vertically, vertices are $$a$$ units above and below the center: $$(-5, -2 + 6) = (-5, 4)$$ and $$(-5, -2 - 6) = (-5, -8)$$. 5. **Write the equations of the asymptotes:** For vertical hyperbolas, asymptotes are given by: $$y - k = \pm \frac{a}{b} (x - h)$$ Substitute values: $$y + 2 = \pm \frac{6}{8} (x + 5) = \pm \frac{3}{4} (x + 5)$$ 6. **Check the alternate asymptote given:** The problem also mentions $$y + 2 = \pm \frac{4}{3} (x + 5)$$ which corresponds to $$\frac{b}{a}$$ instead of $$\frac{a}{b}$$. This is the asymptote form for a horizontal hyperbola, so it is not correct for this vertical hyperbola. 7. **Summary:** - Center: $$(-5, -2)$$ - Vertices: $$(-5, 4)$$ and $$(-5, -8)$$ - Asymptotes: $$y + 2 = \pm \frac{3}{4} (x + 5)$$ These asymptotes are lines the hyperbola approaches but never touches. **Final answer:** The hyperbola centered at $$(-5, -2)$$ with vertical transverse axis has vertices at $$(-5, 4)$$ and $$(-5, -8)$$ and asymptotes: $$y + 2 = \pm \frac{3}{4} (x + 5)$$.