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

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Answer 1 · 2016-02-25T12:49:15

There is a horizontal asymptote $y = 2$ $y = 2$ .

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

This is the graph of $f (x)$ $f(x)$ .

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

First rewrite $f (x)$ $f(x)$ by simplifying it.

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

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

You can see that

$lim_{x->oo} f(x) = lim_{x->oo} (2-frac{4}{2x-1})$

$= 2-0$

$= 2$

Similarly,

$lim_{x->-oo} f(x) = 2$

There is a horizontal asymptote $y = 2$ .

$f(x)$ is undefined at $x = 1/2$ .

There is a vertical asymptote $x = 1/2$ .

How do you find the vertical, horizontal or slant asymptotes for f(x) = (8x-12)/( 4x-2)f(x)=8x−124x−2f(x) = (8x-12)/( 4x-2)?