# How do you find the vertical, horizontal and slant asymptotes of: f(x)=(3x-12) /( 4x-2)?

Oct 20, 2016

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

#### Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: $4 x - 2 = 0 \Rightarrow x = \frac{1}{2} \text{ is the asymptote}$

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$

divide all terms on numerator/denominator by x

$f \left(x\right) = \frac{\frac{3 x}{x} - \frac{12}{x}}{\frac{4 x}{x} - \frac{2}{x}} = \frac{3 - \frac{12}{x}}{4 - \frac{2}{x}}$

as $x \to \pm \infty , f \left(x\right) \to \frac{3 - 0}{4 - 0}$

$\Rightarrow y = \frac{3}{4} \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 1 ) Hence there are no slant asymptotes.
graph{(3x-12)/(4x-2) [-10, 10, -5, 5]}