Scott M. answered 09/05/17
Tutor
4.9
(118)
Math and Physics Teacher
you can factor the bottom first on its own:
(2x2 - 8x - 10)
2(x2 - 4x - 5)
2(x - 5)(x+1)
now, factor the top:
(x2 - 25)
(x + 5)(x - 5)
therefore, because of the (x - 5) on both top and bottom, x = 5 is a removable discontinuity (hole), and (x + 1) on the bottom makes an asymptote at x = -1. both numbers are excluded from the domain.
domain is (-infinity, -1) U (-1,5) U (5,+infinity)
(2x2 - 8x - 10)
2(x2 - 4x - 5)
2(x - 5)(x+1)
now, factor the top:
(x2 - 25)
(x + 5)(x - 5)
therefore, because of the (x - 5) on both top and bottom, x = 5 is a removable discontinuity (hole), and (x + 1) on the bottom makes an asymptote at x = -1. both numbers are excluded from the domain.
domain is (-infinity, -1) U (-1,5) U (5,+infinity)