Question

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

Answer 1

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->-3^+)f(x)=lim_(x->-3^+)-2/(x+3)=-oo`

`lim_(x->-3^-)f(x)=lim_(x->-3^+)-2/(x+3)=+oo`

The vertical asymptote is `x=-3`

To calculate the horizontal asymptote, we perform

`lim_(x->+oo)f(x)=lim_(x->+oo)-2/(x+3)=0^-`

`lim_(x->-oo)f(x)=lim_(x->-oo)-2/(x+3)=0^+`

The horizontal asymptote is `y=0`

The general form of the graph is

`color(white)(aaaa)` `x` `color(white)(aaaaaaa)` `-oo` `color(white)(aaaaaaaa)` `-3` `color(white)(aaaaaaa)` `+oo`

`color(white)(aaaa)` `-(x+3)` `color(white)(aaaaa)` `+` `color(white)(aaaaaa)` `||` `color(white)(aaa)` `-`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaaaa)` `↗` `color(white)(aaaaaa)` `||` `color(white)(aaa)` `↗`

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

`f(0)=-2/3`

So the intercept is `(0,-2/3)`

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