Final answer:
The center of the hyperbola is (-12, 4), and the vertices are (-12, 9) and (-12, -1). The asymptotes are given by the equations y-4 = \(\frac{5}{3}(x+12)\) and y-4 = -\(\frac{5}{3}(x+12)\).
Step-by-step explanation:
To find the center, vertices, and the asymptotes of the hyperbola described by the equation \((\frac{y-4}{25}) - (\frac{x+12}{9}) = 1\), we first need to recognize that this equation is in the standard form of a hyperbola with the y-term first. The standard form of a hyperbola with a vertical transverse axis is \((\frac{y-k}^2{a^2}) - (\frac{x-h}^2{b^2}) = 1\) where \((h,k)\) is the center, \(a\) is the distance to the vertices along the vertical axis, and \(b\) is related to the asymptotes.
The center of the hyperbola is found by setting the terms \((y-4)\) and \((x+12)\) equal to zero, yielding the point \((-12, 4)\). The vertices are found by adding and subtracting the value of \(a\), the square root of the denominator under the y-term, from the y-coordinate of the center: \((4 \pm 5)\), resulting in the points \((-12, 9)\) and \((-12, -1)\).
To find the asymptotes of the hyperbola, we use the slopes \(\pm \frac{a}{b}\) from the denominators of the hyperbola equation. Here \(a\) is \(\sqrt{25}=5\) and \(b\) is \(\sqrt{9}=3\). Therefore, the slopes are \(\pm\frac{5}{3}\). The asymptotes are linear equations that pass through the hyperbola's center, so they can be written as \(y-4 = \pm\frac{5}{3}(x+12)\).