Sean M. answered • 04/23/20
PhD Student in Mathematics
When doing this kind of problem, I find it helpful to run through a bit of a "checklist" and use all gathered information to sketch the graph.
Asymptotes: In this case, you can see you'll have a vertical asymptote at x=2 by setting the denominator of f(x)=0 and solving for x. Horizontal asymptotes appear when the degree of the numerator is less than or equal to the degree of the denominator. If the degree of the denominator is greater than that of the numerator, the horizontal asymptote can be found by dividing the lead coefficients. For f(x), this yields 3/1 or y=3. The final thing to look out for here are holes. Holes are created when you are able to manipulate the expression to cancel out a term or "factor" in the denominator. There are no holes present in this function.
Critical Points/Behavior: To find critical points, consider f'(x). For any x such that f'(x)=0, (x,y) is considered to be a "critical" or "stationary" point--typically a point where the graph changes from increasing/decreasing. Additionally, if f'(x) does not exist at a point, that too is a critical point. At no point does f'(x)=0, however it is undefined at x=2, which gives us our first critical point. The next step is to examine the behavior to the left and right of each critical point found to determine upon which intervals f is increasing, decreasing or constant, as well as any relative maxima/minima. Plugging in x=0,1 will determine the behavior to the left of the asymptote at x=2. Plugging in x=3,4 will do the same thing on the right side.
f'(0)=-3/2 f'(1)=-3
So we can see as f(x) approaches x=2 from the left side, the function is decreasing on the interval (-inf,2). Now for the right side:
f'(3)=-6 f'(4)=-3/2
Again, f(x) is decreasing on the interval (2,+inf)
Concavity/Points of Inflection:
Here, consider f''(x). When f''(x)=0 or does not exist, (x,y) is considered a point of inflection. Similarly to critical points, we simply point these out for now and examine behavior surrounding them. Once again, your only point of interest here is x=2. As before, we will test arbitrary points on either side of the asymptote to determine concavity. We need only to test one point on each side. If f''(x)>0, the function is "concave up" along that interval. Conversely, a function is "concave down" when f''(x)<0.
f''(0)=-4/3 therefore f is concave down along (-inf,2)
f''(3)=12 therefore f is concave up along (2,+inf).
As the concavity changed "around" x=2, we say x=2 is a point of inflection. From here, you should have enough information to get an idea of the visual of your graph. It should look something like an "L" in Quadrant I and a backwards, upside-down "L" that mirrors the other L about the line x=2. If you're ever in doubt, plugging in some points to get an idea of how the graph looks is always helpful.
