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Mathematics

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melomori [17]3 years ago

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It is D Because I'm soo Smart And the Question is easy

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

vesna_86 [32]

Factorize the denominator:

$\dfrac{x^2-4}{x^3+x^2-4x-4}=\dfrac{x^2-4}{x^2(x+1)-4(x+1)}=\dfrac{x^2-4}{(x^2-4)(x+1)}$

If $x\neq\pm2$ , 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

$\dfrac1{x+1}$

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 $\dfrac1{x+1}$ is never 0. On the other hand,

$g(0)=\dfrac{0-4}{0+0-0-4}=1$

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

$\{x\in\mathbb R\mid x\neq-2,x\neq-1,x\neq2\}$