How do you find the vertical, horizontal and slant asymptotes of: $f (x) = \frac{1}{x^{2} - 2 x - 8}$ $f(x)=1/(x^2-2x-8 )$ ?

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Answer 1 · 2017-01-21T12:44:39

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

Let's factorise the denominator

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

So,

$f (x) = \frac{1}{x^{2} - 2 x - 8} = \frac{1}{(x + 2) (x - 4)}$ $f(x)=1/(x^2-2x-8)=1/((x+2)(x-4))$

As you cannot divide by $0$ $0$ , $x \neq - 2$ $x!=-2$ and $x \neq 4$ $x!=4$

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

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

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

The horizontal asymptote is $y=0$

graph{(y-1/(x^2-2x-8))(y)(y-50x-100)(y-50x+200)=0 [-9.84, 10.18, -5.175, 4.825]}

How do you find the vertical, horizontal and slant asymptotes of: f(x)=1/(x^2-2x-8 )f(x)=1x2−2x−8f(x)=1/(x^2-2x-8 )?