From 11:00PM PDT on Friday, July 1 until 5:00AM PDT on Saturday, July 2, the Shmoop engineering elves will be making tweaks and improvements to the site. That means Shmoop will be unavailable for use during that time. Thanks for your patience!
Vertical asymptotes / holes: f has no vertical asymptotes. It has a hole at (2,1).
Horizontal/etc. asymptotes: f doesn't have any of these, since it's 2 everywhere, as opposed to "approaching'' 2.
The domain of f: f is undefined at x = 2, and 1 everywhere else. .
Taking this information, we draw the graph:
Graph the function .
f has a vertical asymptote at x = 0, and no holes. We can start drawing:
Since the degree of the numerator and denominator are the same, f has a horizontal asymptote at y = 3:
f is 0 when its numerator is 0 and its denominator is not. This occurs when .
The only thing left is to find where f is negative and positive. When the numerator and denominator are both negative, therefore f is positive. When , the numerator is positive and the denominator is negative, so f is negative.
We now have enough information to get a rough sketch of the piece of f that lies to the left of the vertical asymptote at y = 0.
When x > 0, the numerator and denominator are both positive and f is positive. Since f must approach its asymptotes, f looks like this:
We didn't need to worry about f(0) for this function, since f is undefined at 0.
A number line is a useful tool for figuring out where a function is negative and positive, and we'll use this tool in one of the examples.
Wherever f is 0 or undefined, its sign has the ability to change. Draw a number line and mark all the values of x where f is 0 or undefined. From this, find the sign of f in between those marked values.
Graph the function
First we need to factor.
There is no way to simplify this function; no terms will cancel. We find no holes, but vertical asymptotes at x = -3 and x = -1:
Since the degree of the numerator and degree of the denominator are the same, we have a horizontal asymptote at :
The function is 0 when the numerator is 0, which occurs at x = -2 and x = 1:
Now we need to figure out where the function is negative and where it is positive, which we can do with a number line:
We can add this point to the graph:
We know some points of f, we know where it is 0, negative, and positive, and we know it must approach its asymptotes, so f must look like this:
We haven't explained how we know that f has the shape it has, and we won't get to that until we start talking about concavity. For now, think about drawing the functions to make them "smooth," like MJ's criminal.
Remember that intercepts are where a function crosses the axes. The y-intercept is f(0), if that exists. The x-intercepts are all values of x where the function is 0.
Graph the function
Label all asymptotes, intercepts, and holes.
First we factor:
We have a vertical asymptote at x = -1 and a hole at (1,8):
We have no horizontal asymptote, but since the degree of the numerator is one greater than the degree of the denominator we do have a slant asymptote, which we can find via long division:
We have a slant asymptote at y = 4x + 8. We can draw that in too:
Now we need function values. f is 0 when x = -3 and when x = 0:
Finally, we need to figure out where f is positive and negative: