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

##### 1 Answer

May 6, 2016

vertical asymptote

horizontal asymptote

#### 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]}