# How do you find the vertical, horizontal or slant asymptotes for f(x)=(x-3 )/ (2x-1 )?

##### 1 Answer
Apr 8, 2016

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

#### Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation let the denominator equal zero.

solve: 2x - 1 = 0 $\Rightarrow x = \frac{1}{2} \text{ is the asymptote }$

Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} f \left(x\right) \to 0$

divide all terms on numerator/denominator by x

$\frac{\frac{x}{x} - \frac{3}{x}}{\frac{2 x}{x} - \frac{1}{x}} = \frac{1 - \frac{3}{x}}{2 - \frac{1}{x}}$

as $x \to \pm \infty , \frac{3}{x} \text{ and } \frac{1}{x} \to 0$

$\Rightarrow y = \frac{1}{2} \text{ is the asymptote }$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no slant asymptotes.

Here is the graph of f(x).
graph{(x-3)/(2x-1) [-10, 10, -5, 5]}