Part 1 of 4 \( 1 \leftarrow \quad \) Find the center, foci, and vertices of the ellipse. Graph the equation. \( (x-2)^{2}+9(y+3)^{2}=36 \)

To find the center, foci, and vertices of the given ellipse, we start by rewriting it in standard form: \[ \frac{(x-2)^2}{36} + \frac{(y+3)^2}{4} = 1 \] Here, we identify that \( a^2 = 36 \) (thus \( a = 6 \)) and \( b^2 = 4 \) (thus \( b = 2 \)). The center of the ellipse is at the point \( (2, -3) \). The vertices are located along the major axis, which is horizontal since \( a > b \). Therefore, the vertices are \( (2 \pm 6, -3) \), resulting in \( (8, -3) \) and \( (-4, -3) \). To find the foci, we calculate \( c \) using the formula \( c = \sqrt{a^2 - b^2} \): \[ c = \sqrt{36 - 4} = \sqrt{32} = 4\sqrt{2} \] The foci are at \( (2 \pm 4\sqrt{2}, -3) \), which approximately gives us \( (2 + 4\sqrt{2}, -3) \approx (8.656, -3) \) and \( (2 - 4\sqrt{2}, -3) \approx (-4.656, -3) \). For graphing, draw the center at \( (2, -3) \), plot the vertices, and mark the foci along the major axis. The ellipse will appear wider along the x-axis due to the horizontal stretch of 6 units and slightly compressed along the y-axis by 2 units.