How do you find the asymptotes for #Y=(3x)/(x^2-x-6) + 3#?

1 Answer
Mar 8, 2016

Answer:

Horizontal asymptote is #y=3#
I have taken you to a point where you should be able to take over for the vertical ones.

Explanation:

As #abs(x)# increases then #abs(x^2)#becomes significantly larger and the other values of the fraction become more an more insignificant

#color(blue)("Horizontal asymptote")#

#lim_(abs(x)->oo) [(3x)/(x^2-x-6) + 3] =lim_(abs(x)->oo) [(3x)/(x^2-x-6)] +3 =3#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The equation becomes undefined as the fractions denominator approaches 0. So it is a matter of solving
#" "x^2-x-6=0#

#" "(x+2)(x-3)#

#color(blue)("Vertical asymptotes at "x=-2" and "x=+3)#

Tony B