Answer:
x=-2, 2
Step-by-step explanation:
Aighty check this out
[tex]f(x) = \frac{ {x}^{2} + 3x + 6 }{ {x}^{2} - 4 } = = = > \\(x + 2)(x - 2) \\ so \: x = - 2 \: and \: x = 2 \: are \: not \: defined \: \\ and \: are \: the \: candidates [/tex]
if non of the root give out 0/0 both can be the asymptotes:
[tex]f(2) = \frac{{2}^{2} + 3(2) + 6 }{ { - 2}^{2} + 4 } = = = > \\ f(2) = \frac{16}{0} [/tex]
this one is OK
[tex]f( - 2) = \frac{ { - 2}^{2} + 3( - 2) + 6}{ { - 2}^{2} + 4 } = = = > \\ f( - 2) = \frac{4}{0} [/tex]
so is this one
so the asymptotes are x=-2, 2
