Determine any asymptotes (Horizontal, vertical or oblique). Find holes, intercepts and state it's domain.[tex]g(x) = \frac{x^{2} -4}{x^{3} +x^{2} -4x - 4}[/tex]

Factorize the denominator:[tex]\dfrac{x^2-4}{x^3+x^2-4x-4}=\dfrac{x^2-4}{x^2(x+1)-4(x+1)}=\dfrac{x^2-4}{(x^2-4)(x+1)}[/tex]If [tex]x\neq\pm2[/tex], we can cancel the factors of [tex]x^2-4[/tex], which makes [tex]x=-2[/tex] and [tex]x=2[/tex] removable discontinuities that appear as holes in the plot of [tex]g(x)[/tex].We're then left with[tex]\dfrac1{x+1}[/tex]which is undefined when [tex]x=-1[/tex], so this is the site of a vertical asymptote.As [tex]x[/tex] gets arbitrarily large in magnitude, we find[tex]\displaystyle\lim_{x\to-\infty}g(x)=\lim_{x\to+\infty}g(x)=0[/tex]since the degree of the denominator (3) is greater than the degree of the numerator (2). So [tex]y=0[/tex] is a horizontal asymptote.Intercepts occur where [tex]g(x)=0[/tex] ([tex]x[/tex]-intercepts) and the value of [tex]g(x)[/tex] when [tex]x=0[/tex] ([tex]y[/tex]-intercept). There are no [tex]x[/tex]-intercepts because [tex]\dfrac1{x+1}[/tex] is never 0. On the other hand,[tex]g(0)=\dfrac{0-4}{0+0-0-4}=1[/tex]so there is one [tex]y[/tex]-intercept at (0, 1).The domain of [tex]g(x)[/tex] is the set of values that [tex]x[/tex] can take on for which [tex]g(x)[/tex] exists. We've already shown that [tex]x[/tex] can't be -2, 2, or -1, so the domain is the set[tex]\{x\in\mathbb R\mid x\neq-2,x\neq-1,x\neq2\}[/tex]