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

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 $4 x - 12$

For horizontal asymptotes $4 x - 12 = 0$ $\implies x - 3 = 0 \implies x = 3$

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$= \frac{3}{4}$

Jan 23, 2016

vertical asymptote at $x = \frac{1}{2}$
horizontal asymptote at y = $\frac{3}{4}$

#### 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 $= \frac{1}{2}$

[ 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 = $\frac{3}{4}$

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