1. **State the problem:** We are given the hyperbola equation $$\frac{(x-3)^2}{36} - \frac{(y+2)^2}{9} = 1$$ and need to verify its properties and help graph it. 2. **Identify the center:** The center is at $$C = (3, -2)$$ from the terms $$x-3$$ and $$y+2$$. 3. **Determine $a$ and $b$:** From the denominators, $$a^2 = 36 \Rightarrow a = 6$$ and $$b^2 = 9 \Rightarrow b = 3$$. 4. **Vertices:** Vertices lie along the transverse axis (horizontal here) at $$x = 3 \pm 6$$, so vertices are $$ (9, -2) $$ and $$ (-3, -2) $$. 5. **Foci:** Calculate $$c$$ using $$c^2 = a^2 + b^2 = 36 + 9 = 45$$, so $$c = 3\sqrt{5}$$. Foci are at $$ (3 \pm 3\sqrt{5}, -2) $$. 6. **Eccentricity:** $$e = \frac{c}{a} = \frac{3\sqrt{5}}{6} = \frac{\sqrt{5}}{2}$$. 7. **Asymptotes:** Equations are $$y + 2 = \pm \frac{b}{a} (x - 3) = \pm \frac{3}{6} (x - 3) = \pm \frac{1}{2} (x - 3)$$. 8. **Axes lengths:** Transverse axis length $$= 2a = 12$$. Conjugate axis length $$= 2b = 6$$. 9. **Graphing tips:** - Plot center at (3, -2). - Mark vertices at (9, -2) and (-3, -2). - Draw rectangle centered at (3, -2) with width 12 and height 6 to guide asymptotes. - Draw asymptotes through corners of this rectangle. - Plot foci at (3 ± 3√5, -2). - Sketch hyperbola branches opening left and right along the transverse axis. This confirms your calculations and provides a clear guide to graph the hyperbola.