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

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`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.



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:


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

Next, find the intercepts.



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





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)` .

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