To find the vertices of the hyperbola given by the equation
[tex]\[
\frac{(y+6)^2}{81} - \frac{(x+8)^2}{49} = 1,
\][/tex]
we need to recognize that it is in the standard form for a hyperbola with a vertical transverse axis. The general equation of a hyperbola with a vertical transverse axis is
[tex]\[
\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1,
\][/tex]
where \((h, k)\) is the center of the hyperbola.
1. Identify the center \((h, k)\) of the hyperbola:
- The given equation is written as \(\left(y + 6\right)^2\) and \(\left(x + 8\right)^2\), so:
- \(h = -8\),
- \(k = -6\).
2. Determine the value of \(a\):
- The value of \(a^2\) is the denominator of the first term, which is 81.
- Therefore, \(a = \sqrt{81} = 9\).
3. Calculate the vertices:
- Since the transverse axis is vertical, the vertices are located \(a\) units above and below the center.
- Starting from the center \((-8, -6)\), we add and subtract the value of \(a\):
- Vertex 1: \((h, k + a) = (-8, -6 + 9) = (-8, 3)\).
- Vertex 2: \((h, k - a) = (-8, -6 - 9) = (-8, -15)\).
Therefore, the coordinates of the vertices are:
[tex]\[
\boxed{(-8, -15) \text{ and } (-8, 3)}.
\][/tex]
Hence, the correct answer is:
[tex]\[
(-8, -15) \text{ and } (-8, 3).
\][/tex]
