How do you find vertical, horizontal and oblique asymptotes for $\frac{1}{x^{2} - 2 x - 8}$ $1/(x^2-2x-8)$ ?

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Answer 1 · 2017-02-26T14:43:51

The vertical asymptotes are $x = 4$ $x=4$ and $x = - 2$ $x=-2$
The horizontal asymptote is $y = 0$ $y=0$
No oblique asymptote

Let's factorise the denominator

$x^{2} - 2 x - 8 = (x - 4) (x + 2)$ $x^2-2x-8=(x-4)(x+2)$

As you canot divide by $0$ $0$ , $\Rightarrow$ $=>$ , $x \neq 4$ $x!=4$ and $x \neq - 2$ $x!=-2$

The vertical asymptotes are $x = 4$ $x=4$ and $x = - 2$ $x=-2$

As the degree of the numerator is $<$ $<$ than the degree of the denominator, there is no oblique asymptote.

$lim_(x->+-oo)1/(x^2-2x-8)=lim_(x->+-oo)1/x^2=0^+$

The horizontal asymptote is $y=0$

graph{(y-1/(x^2-2x-8))(y-10000(x-4))(y-1000(x+2))(y)=0 [-3.85, 7.25, -2.893, 2.66]}

How do you find vertical, horizontal and oblique asymptotes for 1/(x^2-2x-8)1x2−2x−81/(x^2-2x-8)?