Yours Truly answers: The first and second derivative tables provide precise geometrical information.
 
Step 1 (Domain)
Clearly, the function is defined for any value of  x. So, the domain is  R.
 
Step 2 (Intercepts)
Since  f (0) = 0  and  f (x) = 0  ⇔  x = 0, the graph meets the x-axis only once at the origin.
 
Step 3 (Symmetry)
None, because  f (–1) ≠ f (1)  and  f (–1) ≠ – f (1).
 
Step 4 (Asymptotes: Vertical, horizontal, and oblique)
Vertical Asymptotes: Since the domain is  R, there is no vertical asymptote.
 
Horizontal Asymptotes: Observe that
Hence,  y = 0  (x-axis)  is a horizontal asymptote as  x → ∞. Moreover, the graph approaches this horizontal asymptote from above as  x → ∞. This is a significant piece of geometrical information. On the other hand,   f (x) → – ∞  as  x → – ∞. This means that there is no horizontal asymptote as  x → – ∞. As seen here, the investigation of horizontal asymptotes must be done separately for  x → ∞  and  x → – ∞.
 
Oblique Asymptotes: There is no oblique asymptote as   x → ∞   because there is already a horizontal asymptote as  x → ∞. There is no oblique asymptote as  x → – ∞, either, because  f (x)  does not stabilize toward a non-constant function. As seen here, the investigation of oblique asymptotes must be done separately for  x → ∞  and  x → – ∞.
 
Before we proceed to the next step, note that Steps 1–4 do not use calculus. Precalculus skills or lack thereof either make or break any graphing problem.
 
What calculus does is to extract precise information of the graph through Steps 5 & 6.
 
Step 5 (f ' -Table: Increasing or Decreasing)
Local extrema, if they exist, should clearly be noted in the table. First,

f ' (x) = e^–2x – 2xe^–2x =  e^–2x(1 – 2x).

So, there is only one critical number at  x = ½. A critical number does not automatically give rise to a local extremum. (For example,  x = 0  is a critical number of  y = x³. But, the function of course does not have a local extremum at  x = 0.) That is why the table is needed to investigate the signs of the first derivative.
 
In the last row of the table, be sure to supply information on  f (x)  as  x → ∞  and  x → – ∞. It will help you when you transfer the table information to the graph.
 
 
Step 6 (f " -Table: Concave Up or Down)
Inflection points, if they exist, should clearly be noted in the table.
As we compute f ", differentiate  e^–2x – 2xe^–2x  instead of  e^–2x(1 – 2x). Now,

f "(x) = –2e^–2x – 2(e^–2x – 2xe^–2x) =  – 4e^–2x + 4xe^–2x = 4e^–2x(x – 1).

So, x = 1  could give rise to an inflection point. But, we do not know until we investigate the signs of the second derivative in the table.
 
 
Step 7 (Graphing)
Transfer the information obtained in Steps 1–6 to the graph. On the graph itself, all relevant information should directly be labeled. Compare the colors on the graph with the tables above.

 

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The example above is more difficult than polynomial functions because it has a horizontal asymptote. However, rational functions are even more challenging because they often have multiple vertical, horizontal, and oblique asymptotes.
 
What we often forget to investigate is vertical tangent lines. A vertical tangent line could occur where  f '  is undefined but  f   is defined.