The domain of f is all real numbers except 2.

This equals zero when x = 0 or x = 4 . The direction of f (x) based on the sign of the derivative is depicted below:

This never equals zero. The concavity of the graph is depicted below: To find the vertical asymptotes, take limits near the suspected asymptote:

So, x = 2 is a vertical asymptote of the graph To find the horizontal asymptotes, take limits at infinity:

Thus, f grows without bound, and there are no horizontal asymptotes Now, find the exact coordinates of the intercepts and critical points: y -intercept: (0, 0) x -intercept: (0, 0) Critical Point: (0, 0) and (4, 8) Combining the information from the first derivative work and the second derivative work generates this single chart: Based on this information, the graph may now be sketched:

To find an appropriate equation, begin by pulling out relevant features of the graph. f has vertical asymptotes at x = 3 and x = - 3 . Vertical asymptotes can occur where the denominator of a rational function is equal to zero. Thus, the function in question should look something like:

f has a horizontal asymptote at y = 2 , which means that as x approaches infinity (or negative infinity), the function approaches 2. Thus, the function must look something like:

f (x) =

Finally, f (0) = - 3 , so the equation must be modified to:

f (x) =

As in the last problem, start with the vertical asymptotes:

Because of the horizontal asymptotes,

f (x) =

f (0) = 0 , but this does not modify the equation of the function.

The information contained in the graph of f'(x) is enough to determine the sign and concavity of f (x) . It is possible to clearly pick out intervals where f'(x) is positive or negative. Also, f''(x) is positive wherever f'(x) in increasing, and f''(x) is negative wherever f'(x) is decreasing. Putting this information together generates the following chart: This can be graphed in the following way: