find center, foci, vertices, and asymptotes of this hyperbola?

Question

(x-8)^2 - (y+6)^2 = 1

Answer 1

(x-8)²/1 - (y+6)²/1 = 1

center (8,-6)
vertices (7,-6) and (9, -6)........(at 8 ± 1)
foci (8 ± √2, -6)
asymptotes with slope ± 1 passing through (8, -6), so
y + 6 = ±(x - 8)

Answer 2

The formula for a pair of hyperbolas that open sideways with center (h,k) in standard form is:

(x - h)²/a² + (y - k)²/b² = 1

The center (h,k) = (8,-6).

a = 1
b = 1
c² = a² + b² = 1 + 1 = 2
c = √2

The foci are (h-c, k) and (h+c, k).
(8-√2, -6) and (8+√2, -6)

The vertices are (h-a, k) and (h+a, k).
(7, -6) and (9, -6)

The slopes of the asymptotes is m = ∆y/∆x = ±b/a.
m = ±1/1 = ±1

The asymptotes go thru the center of the hyperbola so their equations are:

y + 6 = ±(x - 8)

y = x - 14
and
y = -x + 2