# How do you graph f(x)=-2/(x+3) using holes, vertical and horizontal asymptotes, x and y intercepts?

Jul 30, 2017

The vertical asymptote is $x = - 3$
The horizontal asymptote is $y = 0$
See the graph below

#### Explanation:

To calculate the vertical asymptote, we perform

${\lim}_{x \to - {3}^{+}} f \left(x\right) = {\lim}_{x \to - {3}^{+}} - \frac{2}{x + 3} = - \infty$

${\lim}_{x \to - {3}^{-}} f \left(x\right) = {\lim}_{x \to - {3}^{+}} - \frac{2}{x + 3} = + \infty$

The vertical asymptote is $x = - 3$

To calculate the horizontal asymptote, we perform

${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} - \frac{2}{x + 3} = {0}^{-}$

${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} - \frac{2}{x + 3} = {0}^{+}$

The horizontal asymptote is $y = 0$

The general form of the graph is

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a a a}$$- 3$$\textcolor{w h i t e}{a a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$- \left(x + 3\right)$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a a a}$$| |$$\textcolor{w h i t e}{a a a}$$-$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a a a}$↗$\textcolor{w h i t e}{a a a a a a}$$| |$$\textcolor{w h i t e}{a a a}$↗

The intercept with the y-axis is when $x = 0$

$f \left(0\right) = - \frac{2}{3}$

So the intercept is $\left(0 , - \frac{2}{3}\right)$

graph{-2/(x+3) [-18.02, 18.03, -9.01, 9.01]}