Question 250662
{{{f(x) = 1/(x+2) + 3}}}
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<br>
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 <i>smaller</i> 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.<br>
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.<br>
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.<br>
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.)<br>
I'll get you started. Let's try x = -1:
{{{f(-1) = 1/((-1) + 2) + 3 = 1/1 + 3 = 1 + 3 = 4}}}
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).)
{{{graph(400, 400, -7, 5, -1, 11, 1/(x+2)+3)}}}