Question

How to find the asymptotes of `F(x)=(x^2+x-12) /( x^2-4)`?

Answer 1

Vertical asymptotes: `x=-2color(white)("XXX")andcolor(white)("XXX")x=+2`
Horizontal asymptote: `F(x)= 1`

Explanation:

For an expression composed of a fraction with a polynomial numerator `N(x)` and a polynomial denominator `D(x)`

Vertical asymptotes exist as `x=a`
for any and all point for which `D(a)=0` and `N(a)!=0`

With `F(x)=(N(x))/(D(x)) =(x^2+x-12)/(x^2-4)`

`color(white)("XXX")D(x) = 0` for `x=+-2`
`color(white)("XXXXXXX")`hopefully this is obvious
and
`color(white)("XXX")N(x)!=0` for either `x=-2` or `x=2`

Therefore both `x=-2` and `x=2` are vertical asymptotes.

Horizontal asymptotes exist as `F(x)=b`
if `lim_(xrarroo) F(x) = b`

`color(white)("XXX")=lim_(xrarroo)F(x)`

`color(white)("XXX")lim_(xrarroo) (x^2+x-12)/(x^2-4)`

`color(white)("XXX")= lim_(xrarroo) 1 + (x-8)/(x^2-4)`

`color(white)("XXX")= 1 + lim_(xrarroo)(x-8)/(x^2-4)`

`color(white)("XXX")= 1 + 0 = 1`
graph{(x^2+x-12)/(x^2-4) [-5.45, 7.033, -1.27, 4.98]}