How do you find the asymptotes for #(3x-12)/( 4x-2)#?

2 Answers
Jan 23, 2016

Horizontal asymptotes of a rational function occurs when the function in the denominator becomes zero.
In this case the function in denominator is #4x-12#

For horizontal asymptotes #4x-12=0# #implies x-3=0 implies x=3#

Hence horizontal asymptote is #3#

Vertical asymptotes accurs when the degree of numerator and denominator is equal. In this case both numerator and denominator have a degree #1# i.e, the degree is equal. The asymptote is found out by the ratio of leading coefficients.
In this case the leading coefficients of numerator and denominator are #3# and #4# respectively.
#implies# vertical asymptote#=3/4#

Jan 23, 2016

Answer:

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

Explanation:

Vertical asymptotes can be found when the denominator of

the rational function is zero.

This will be when : 4x - 2 =0 hence 4x = 2 so x # = 1/2 #

[ Horizontal asymptotes can be found when the degree of the

numerator and the degree of the denominator are equal ]

In this question they are both of degree 1 and so equal.

The asymptote can be found by taking the ratio of leading

coefficients hence y = # 3/4 #

graph{(3x-12)/(4x - 2) [-22.5, 22.5, -11.25, 11.25]}