Final answer:
The hyperbola's vertices are at (-1, 12) and (-1, 0), the foci are approximately at (-1, 13.21) and (-1, -1.21), and the equations for the asymptotes are y = 1.5x + 7.5 and y = -1.5x + 4.5.
Step-by-step explanation:
The equation provided is for a hyperbola in the form f(y-6)²/36 - (x+1)²/16=1. The center of this hyperbola is at the point (-1, 6). The denominators in the equation give us the squares of the semi-axes lengths of the hyperbola: 36 for the y-axis part (which is 'a' squared) and 16 for the x-axis part (which is 'b' squared).
Vertices of the hyperbola are determined by adding and subtracting the length of the semi-major axis 'a' to the y-coordinate of the center, since the y-term is positive and in the numerator. In this case, a = √36 = 6, so the vertices are (center.x, center.y ± a) = (-1, 6 ± 6), giving us the points (-1, 12) and (-1, 0).
Foci of the hyperbola can be found using the formula c = √(a² + b²), where c is the distance from the center to each focus along the axis of the hyperbola. Since b² = 16, we find c = √(36 + 16) = √52 ≈ 7.21. The foci are thus (center.x, center.y ± c) = (-1, 6 ± 7.21), resulting in approximate coordinates (-1, 13.21) and (-1, -1.21).
The equations for the asymptotes are found using the slope 'a/b' of the asymptotes and the center of the hyperbola. For this hyperbola, 'a' = 6 and 'b' = 4, so the slope of the asymptotes is ± a/b = ± 6/4 = ± 1.5. The equations of the asymptotes are y - center.y = ±(slope)(x - center.x), which results in y - 6 = ± 1.5(x + 1). Simplifying these gives the asymptote equations y = 1.5x + 7.5 and y = -1.5x + 4.5.