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Yuki888 [10]
3 years ago
13

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

Mathematics
1 answer:
vesna_86 [32]3 years ago
8 0

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\}

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