# How do you graph f(x)=x/(x-2) using holes, vertical and horizontal asymptotes, x and y intercepts?

Jul 26, 2018

Below

#### Explanation:

$f \left(x\right) = \frac{x}{x - 2}$
$f \left(x\right) = \frac{\left(x - 2\right) + 2}{x - 2}$
$f \left(x\right) = 1 + \frac{2}{x - 2}$

For vertical asymptotes, $x - 2 \ne 0$ since the denominator cannot equal to 0 as the function will be undefined at that point. To find at what point the function is undefined, we can go $x - 2 = 0$ so $x = 2$ is our vertical asymptote.

For horizontal asymptotes, we let $x \to \infty$. As $x \to \infty$, $\frac{2}{x - 2} \to 0$. Hence, $f \left(\infty\right) = 1 + 0 = 1$. Therefore, there is an horizontal asymptote at $y = 1$

For intercepts,
When $y = 0$, $x = 0$
When $x = 0$, $y = 0$

Plotting your intercepts and drawing in your asymptotes (remember that the asymptotes influence the endpoints of your graph ONLY and nothing else)

graph{x/(x-2) [-10, 10, -5, 5]}