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

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]}