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

1 Answer
Jan 30, 2016

Answer:

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]}