# How do you find all the asymptotes for function y = x/(x-6) ?

Jul 6, 2015

$x = 6$ and $y = 1$

#### Explanation:

3 types of asymptotes : Vertical, horizontal and oblique.

Vertical :

The straight line $x = a$ is a vertical asymptote for f(x) if ${\lim}_{x \to a} f \left(x\right) = \pm \infty$

Let $f \left(x\right) = \frac{x}{x - 6}$

The domain of f(x) is ]-oo;6[uu]6;+oo[ <=> 6 is a forbidden value

We generally find a vertical asymptote in a rational function, when the value of denominator is equal to 0 (except if numerator is zero in the same time)

Look at $f \left(x\right)$ when we are close to $6$.

${\lim}_{x \to {6}^{-}} f \left(x\right) = - \infty$ and ${\lim}_{x \to {6}^{+}} f \left(x\right) = + \infty$

We comply with the starting conditions : $x = 6$ is a vertical asymptote.


Horizontal :

The straight line $y = b$ is a horizontal asymptote for $f \left(x\right)$ if ${\lim}_{x \to \pm \infty} f \left(x\right) = b$, ($b \in \mathbb{R}$)

To find easily the limit when x go to $\pm \infty$, we will modify $f \left(x\right)$ :

$\forall x \in \mathbb{R} - \left\{6\right\} , f \left(x\right) = \frac{x}{x - 6} = \frac{x - 6 + 6}{x - 6} = 1 \cdot \cancel{\frac{x - 6}{x - 6}} + \frac{6}{x - 6} = 1 + \frac{6}{x - 6}$

Limit calculation :

${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} 1 + \frac{6}{x - 6} = 1 + {\lim}_{x \to + \infty} \frac{6}{x - 6}$

And ${\lim}_{x \to + \infty} \frac{6}{x - 6} = 6 \times {\lim}_{x \to + \infty} \frac{1}{x - 6} = 6 \times {\lim}_{x \to + \infty} \frac{1}{x} = 0$

Then :
${\lim}_{x \to + \infty} f \left(x\right) = 1 + 0 = 1$

Do the same reasoning when $x \to - \infty$

Therefore : $y = 1$ is a horizontal asymptote for $f \left(x\right)$

Oblique :

No oblique asymptote because ${\lim}_{x \to \pm \infty} f \left(x\right) = 1$

To sum up, look at the graph below :

Note : $y = 1$ is in green.