How do you find vertical, horizontal and oblique asymptotes for (5x-15)/(2x+4)?

May 11, 2016

Vertical asymptote: $x = - 2$
Horizontal asymptote: $y = \frac{5}{2}$
There is not oblique asymptote.

Explanation:

Given $y = \frac{5 x - 15}{2 x + 4}$ (after converting to an equation)

As $\left(2 x + 4\right) \rightarrow 0$
$\textcolor{w h i t e}{\text{XXX}} y \rightarrow \pm \infty$ (whether it is plus or minus depends upon from which side $\left(2 x + 4\right)$ approaches zero.
$\textcolor{w h i t e}{\text{XXX}} 2 x + 4 = 0$ implies $x = - 2$ the vertical asymptote.

The horizontal asymptote is given as the limit value of $y$ as $x \rightarrow \infty$
$\textcolor{w h i t e}{\text{XXX}} {\lim}_{x \rightarrow \infty} y$
$\textcolor{w h i t e}{\text{XXX}} = \lim \left(x \rightarrow \infty\right) \frac{5 x - 15}{2 x + 4}$

$\textcolor{w h i t e}{\text{XXX}} = \lim \left(x \rightarrow \infty\right) \frac{5 - \frac{15}{x}}{2 + \frac{4}{x}}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{5}{2}$

An oblique asymptote only exists if the degree of the numerator is greater than the degree of the denominator. $\left(5 x - 15\right)$ and $\left(2 x + 4\right)$ are both of degree $1$; so there is no oblique asymptote.

graph{(5x-15)/(2x+4) [-16.86, 15.2, -7.67, 8.37]}