# How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x)=(5x-15) /( 2x+4)?

Jul 20, 2016

vertical asymptote x = -2
horizontal asymptote $y = \frac{5}{2}$

#### Explanation:

The denominator of f(x) cannot be zero as this is 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: 2x + 4 = 0 → x = -2 is the asymptote

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

$\frac{\frac{5 x}{x} - \frac{15}{x}}{\frac{2 x}{x} + \frac{4}{x}} = \frac{5 - \frac{15}{x}}{2 + \frac{4}{x}}$

as $x \to \pm \infty , f \left(x\right) \to \frac{5 - 0}{2 + 0}$

$\Rightarrow y = \frac{5}{2} \text{ is the asymptote}$

Oblique 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 oblique asymptotes.
graph{(5x-15)/(2x+4) [-40, 40, -20, 20]}