How do you find the vertical, horizontal or slant asymptotes for f(x) = (8x-12)/( 4x-2)?

Feb 25, 2016

There is a horizontal asymptote $y = 2$.

There is a vertical asymptote $x = \frac{1}{2}$.

Explanation:

This is the graph of $f \left(x\right)$.

graph{(8x-12)/(4x-2) [-20, 20, -10, 10]}

First rewrite $f \left(x\right)$ by simplifying it.

$f \left(x\right) = \frac{8 x - 12}{4 x - 2}$

$= 2 - \frac{4}{2 x - 1}$

You can see that

${\lim}_{x \to \infty} f \left(x\right) = {\lim}_{x \to \infty} \left(2 - \frac{4}{2 x - 1}\right)$

$= 2 - 0$

$= 2$

Similarly,

${\lim}_{x \to - \infty} f \left(x\right) = 2$

There is a horizontal asymptote $y = 2$.

$f \left(x\right)$ is undefined at $x = \frac{1}{2}$.

There is a vertical asymptote $x = \frac{1}{2}$.