How do you find vertical, horizontal and oblique asymptotes for #(3x-12)/(4x-2)#?

1 Answer
May 6, 2016

Answer:

vertical asymptote #x=1/2#
horizontal asymptote #y=3/4#

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero.

solve : 4x - 2 = 0 → 4x = 2

#rArr x=1/2" is the asymptote "#

Horizontal asymptotes occur as #lim_(x to +- oo) , f(x) to 0 #

divide terms on numerator/denominator by x

#((3x)/x-12/x)/((4x)/x-2/x)=(3-12/x)/(4-2/x)#

as # x to +- oo , y to (3-0)/(4-0)#

#rArr y=3/4" is the asymptote "#

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1 and denominator-degree 1)hence there are no oblique asymptotes.
graph{(3x-12)/(4x-2) [-10, 10, -5, 5]}