`f(x) = 1/(x+2)`

First, determine the vertical asymptote of the rational function. Take note that vertical asymptote refers to the values of x that make the function undefined. Since it is undefined when the denominator is zero, to find the VA, set the denominator equal to zero.

`x+2=0`

`x=-2`

Graph this vertical asymptote on the grid. Its graph should be a dashed line. (See attachment.)

Next, determine the horizontal or slant asymptote. To do so, compare degree of the numerator and denominator.

degree of numerator = 0

degree of the denominator = 1

Since the degree of the numerator is less than the degree of the denominator, the asymptote is horizontal, not slant. And its horizontal asymptote is:

`y=0`

Graph this horizontal asymptote on the grid. Its graph should be a dashed line.(See attachment.)

Next, find the intercepts.

y-intercept:

`y=1/(0+2)=1/2`

So the y-intercept is `(0,1/2)` .

x-intercept:

`0=1/(x+2)`

`(x+2)*0=1/(x+2)*(x+2)`

`0=1`

So, the function has no x-intercept.

Also, determine the other points of the function. To do so, assign any values to x, except -2. And solve for the y values.

`x=-10` , `y=1/(-10+2) = -1/8`

`x=-7` , `y=1/(-7+2)=-1/5`

`x=-4` ,`y=1/(-4+2)=-1/2`

`x=-1` , `y=1/(-1+2)=1`

`x=1` , `y=1/(1+2)=1/3`

`x=3` , `y=1/(3+2)=1/5`

`x=8` , `y=1/(8+2)=1/10`

Then, plot the points `(-10,-1/8)` , `(-7,-1/5)` , `(-4,-1/2)` ,`(-1,1)` , `(0,1/2)` , `(1,1/3)` , `(3,1/5)` , and `(8,1/10)`.

And connect them.

**Therefore, the graph of the function is:**

Base on the graph,** the domain of the function is `(-oo, -2) uu (2,oo)` . And its range is `(-oo, 0) uu (0, oo)` **.