For the following function: g(x) = 2x + 2 / x^2 - 6x - 7 (a) Find the domain. Answer .................. (b) horizontal asymptotes ? (c) vertical asymptotes ? (d) is there discontinuity?

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For the following function: g(x) = 2x + 2 / x^2 - 6x - 7 (a) Find the domain. Answer .................. (b) horizontal asymptotes ? (c) vertical asymptotes ? (d) is there discontinuity?

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Answer 1 · 2015-02-27T11:41:40

The only "forbidden" values for $x$ $x$ are the ones that make the denominator equal to $0$ $0$ .

$\frac{2 x + 2}{x^{2} - 6 x - 7} = \frac{2 (x + 1)}{(x + 1) (x - 7)}$ $(2x+2)/(x^2-6x-7)=(2(x+1))/((x+1)(x-7)$

The domain is now limited to $x \neq - 1 and x \neq 7$ $x!=-1 and x!=7$
These also give the vertical asymptotes $x = - 1 and x = 7$ $x=-1 and x=7$
At these $x$ $x$ -values there are also discontinuities , where $y$ $y$ goes from $+ \infty$ $+oo$ to $- \infty$ $-oo$ and back.

As for the horizontal asymptote we look at what happens when $x$ $x$ gets larger and larger.
In this case we may cancel out the $(x + 1)$ $(x+1)$ 's to get:

$lim_(x->oo)2/(x-7)=0$

Because as $x$ gets larger, the fraction will get smaller.
So the horizontal asymptote has the equation $y=0$

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For the following function: g(x) = 2x + 2 / x^2 - 6x - 7 (a) Find the domain. Answer .................. (b) horizontal asymptotes ? (c) vertical asymptotes ? (d) is there discontinuity?

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