f(x)=1/(x+2) Graph the function. State the domain and range.

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) .