Question

Determine any asymptotes (Horizontal, vertical or oblique). Find holes, intercepts and state it's domain.


`g(x) = (x^(2) -4)/(x^(3) +x^(2) -4x - 4)`

Answer 1

Factorize the denominator:

`(x^2-4)/(x^3+x^2-4x-4)=(x^2-4)/(x^2(x+1)-4(x+1))=(x^2-4)/((x^2-4)(x+1))`

If
`x\\eq\pm2`, we can cancel the factors of
`x^2-4`, which makes
`x=-2` and
`x=2` removable discontinuities that appear as holes in the plot of
`g(x)`.

We're then left with

`\frac1{x+1}`

which is undefined when
`x=-1`, so this is the site of a vertical asymptote.

As
`x` gets arbitrarily large in magnitude, we find

`\displaystyle\lim_(x\to-\infty)g(x)=\lim_(x\to+\infty)g(x)=0`

since the degree of the denominator (3) is greater than the degree of the numerator (2). So
`y=0` is a horizontal asymptote.

Intercepts occur where
`g(x)=0` (
`x`-intercepts) and the value of
`g(x)` when
`x=0` (
`y`-intercept). There are no
`x`-intercepts because
`\frac1{x+1}` is never 0. On the other hand,

`g(0)=(0-4)/(0+0-0-4)=1`

so there is one
`y`-intercept at (0, 1).

The domain of
`g(x)` is the set of values that
`x` can take on for which
`g(x)` exists. We've already shown that
`x` can't be -2, 2, or -1, so the domain is the set

`\{x\in\mathbb R\mid x\\eq-2,x\\eq-1,x\\eq2\}`