SOLUTION: Graph the Rational Function & give the equations of its asymptotes. F(x)= 1/x+2 +3 Vertical Asymptote Horizontal Asymptote

Question 250662: Graph the Rational Function & give the equations of its asymptotes.

F(x)= 1/x+2 +3

Vertical Asymptote
Horizontal Asymptote
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!

Vertical asymptotes. For rational functions like this the vertical asymptotes will occur for any values of x, if any, that makes a denominator zero. There is only one denominator and there is only one value for x that will make it zero: -2. (If you don't see this, then set the denominator equal to zero and solve for x.) So there is one vertical asymptote:
x = -2

Horizontal asymptotes. If there are any horizontal asymptotes, they occur for large positive and/or negative values. So we have to figure out what happens to f(x) when x is large positive or negative numbers. When x is large and positive the denominator of the fraction in f(x) also becomes large and positive. The larger x gets the larger the denominator gets. And the larger the denominator gets (while the numerator stays at 1) the smaller the fraction as a whole gets. In fact, for very large values of x the fraction approaches 0 in value. And since f(x) is this fraction plus 3, f(x) approaches 3 in value for very large positive values of x. So y = 3 is a horizontal asymptote, at least for large positive values of x.

For very large negative values of x, the denominator becomes very large and negative, too. This means that the fraction as a whole approaches 0 and f(x) approaches 3 for very large negative values of x. So y = 3 is also a horizontal asymptote for very large negative values of x.

We can take our analysis of the horizontal asymptote a step further. For large positive x's the fraction is a very small positive number near zero. So the graph of f(x) approaches 3 from above for large positive x's. And for very large negative x's the fraction is a very small negative number near zero. So the graph of f(x) will approach y = 3 from below for large negative x's.

Having the asymptotes is useful but we need some points, too. You choose some numbers for x and find what f(x) is for each x. You can choose any numbers you like for x (except -2, of course). I suggest that you pick x's centered on the vertical asymptote. For example, -1, 0 and 1 would be 1, 2 and 3 units to the right of the vertical asymptote and -3, -4 and -5 would be 1, 2 and 3 units to the left of the vertical asymptote. (If there were more than one vertical asymptote, then you would pick x values on either side of each one.)

I'll get you started. Let's try x = -1:

So (-1, 4) will be a point on the graph of f(x).
I will leave it up to you to pick other x's, find their y's and graph the points. Below is a graph of f(x). (Note Algebra.com's graphing facility does not draw the asymptotes (which are usually drawn as dotted lines).)

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