Question

Describe the vertical asymptote(s) and hole(s) for the graph of y = (x - 5)/(x^2 + 4x + 3)

asymptotes: x = –3, –1 and no holes.
asymptote: x = –3 and hole: x = –5
asymptotes: x = –3, –1 and hole: x = –5
asymptote: x = –5 and hole: x = –3

Answer 1

\[y = \frac{ x - 5 }{ x^2 + 4x + 3 }\]

Answer 2

@BAdhi so the answer is A ?

Answer 3

Sorry I've got it wrong previously,

asymptote is the points where $$(x^2+4x+3)$$ and holes are where both denominator and devisor equal to zero. http://www.sparknotes.com/math/algebra2/specialgraphs/section2.rhtml

Answer 4

@BAdhi Ok So Now Im Confused

Answer 5

Ok let's do this again,

First by just looking at the function we see that there is no hole in the graph, since there doesn't exist at least one value for x such that we get 0/0 for the function.

Now lets consider asymptotes,

the given function is $$f(x)=\frac{(x-5)}{(x+3)(x+1)}$$ when $$x\to-3\; \mathbb{or}\;x\to-1$$ we can see that f(x) tends to infinity according to definition these points (x=-3 and x=-1) are asymptotes.

Further if we consider the limit of f when x reaches plus or minus infinity,
$$\lim\limits_{x\to \infty}\frac{x+5}{(x+3)(x+1)}=\lim\limits_{x\to \infty}\frac{\left(\frac{1}{x}+\frac{5}{x^2}\right)}{\left(1+\frac{3}{x}\right)\left(1+\frac{1}{x}\right)}=0$$
similarly
$$\lim\limits_{x\to -\infty}\frac{x+5}{(x+3)(x+1)}=0$$
So, according to the definition in the resource y=0 is an asymptote

But I think they don't consider horizontal asymptotes, thus the answer will be A