4.4: Graphing
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We have shown how to use the first and second derivatives of a function to describe the shape of a graph. We also know the behavior of \(f\) as \(x→±∞\). In this section, we outline a strategy for graphing an arbitrary function \(f\).
4.4 Graphing
Guidelines for Drawing the Graph of a Function
We now have enough analytical tools to draw graphs of a wide variety of algebraic and transcendental functions. Before showing how to graph specific functions, let’s look at a general strategy to use when graphing any function.
ProblemSolving Strategy: Drawing the Graph of a Function
Given a function \(f\), use the following steps to sketch a graph of \(f\):
 Determine the domain of the function.
 Locate the \(x\) and \(y\)intercepts.
 Check for Even or Odd Symmetry.
 If \(f(x) = f(x)\) then this is an Even function (symmetric about the yaxis).
 If \(f(x) = f(x)\) then this is an Odd function (symmetric about the origin).
 Consider Asymptotes.
 Evaluate \(\lim_{x→∞}f(x)\) and \(\lim_{x→−∞}f(x)\) to determine the end behavior. If either of these limits is a finite number \(L\), then \(y=L\) is a horizontal asymptote.
 Determine whether \(f\) has any vertical asymptotes.
 Calculate \(f′.\) Find all critical points and determine the intervals where \(f\) is increasing and where \(f\) is decreasing. (First Derivative Test)
 Determine whether \(f\) has any local extrema (minimums or maximums). If so, find both x & y values for those points.
 Calculate \(f''.\) Determine the intervals where \(f\) is concave up and where \(f\) is concave down. (Concavity Test)
 Use concavity information to determine whether \(f\) has any inflection points. If so, find both x & y values for those points.
 Sketch the graph. (Use the points and information found from steps 1  8; find additional points if needed.)
Now let’s use this strategy to graph several different functions. We start by graphing a polynomial function.
Example \(\PageIndex{8}\): Sketching a Graph of a Polynomial
Sketch a graph of \(f(x)=(x−1)^2(x+2).\)
Solution
Step 1: Since \(f\) is a polynomial, the domain is the set of all real numbers.
Step 2: When \(x=0,f(x)=2.\) Therefore, the yintercept is \((0,2)\). To find the \(x\)intercepts, we need to solve the equation \((x−1)^2(x+2)=0\), gives us the \(x\)intercepts \((1,0)\) and \((−2,0)\)
Step 3: \(f(−x)=(−x−1)^2(−x+2) =((−1)(x+1))^2(−1)(x2)=(x+1)^2(x−2)\). This is not \(f(x)\) nor \(f(x)\), so this function is not Even & not Odd.
Step 4(a): We need to evaluate the end behavior of \(f.\)As \(x→∞, (x−1)^2→∞\) and \((x+2)→∞\). Therefore, \(\lim_{x→∞}f(x)=∞\) As \(x→−∞, (x−1)^2→∞\) and \((x+2)→−∞\). Therefore, \(\lim_{x→∞}f(x)=−∞\). To get even more information about the end behavior of \(f\), we can multiply the factors of \(f\). When doing so, we see that
\[f(x)=(x−1)^2(x+2)=x^3−3x+2.\]
Since the leading term of \(f\) is \(x^3\), we conclude that \(f\) behaves like \(y=x^3\) as \(x→±∞.\)
Step 4(b): Since \(f\) is a polynomial function, it does not have any vertical asymptotes.
Step 5: The first derivative of \(f\) is
\[f′(x)=3x^2−3.\]
\(Set f′(x)=0\) to find \(f\) has two critical points: \(x=1,−1.\) Divide the interval \((−∞,∞)\) into the three smaller intervals: \((−∞,−1), (−1,1)\), and \((1,∞)\). Then, choose test points \(x=−2, x=0\), and \(x=2\) from these intervals and evaluate the sign of \(f′(x)\) at each of these test points, as shown in the following table.
Interval  Test point  Sign of Derivative \(f'(x)=3x^2−3=3(x−1)(x+1)\)  Conclusion 

\((−∞,−1)\)  \(x=−2\)  \((+)(−)(−)=+\)  \(f\) is increasing 
\((−1,1)\)  \(x=0\)  \((+)(−)(+)=−\)  \(f\) decreasing 
\((1,∞)\)  (x=2\)  \((+)(+)(+)=+\)  \(f\) is increasing 
NOTE: this table is equivalent to the line with asterisks on \(−1\) and \(1\) and the Signs of the Derivative on the line using the technique shown above .
Be sure to write \(f′\) along the line to indicate this is the First Derivative Test.
Step 6: From the table, we see that \(f\) has a local maximum at \(x=−1\) and a local minimum at \(x=1\). Evaluating \(f(x)\) at those two points, we find that the local maximum value is \(f(−1)=4\) and the local minimum value is \(f(1)=0.\) So the local maximum is \((−1,4)\) and the local minimum is \((1,0)\).
Step 7: The second derivative of \(f\) is
\[f''(x)=6x.\]
The second derivative is zero at \(x=0.\) Therefore, to determine the concavity of \(f\), divide the interval \((−∞,∞)\) into the smaller intervals \((−∞,0)\) and \((0,∞)\), and choose test points \(x=−1\) and \(x=1\) to determine the concavity of \(f\) on each of these smaller intervals as shown in the following table.
Interval  Test Point  Sign of \(f''(x)=6x\)  Conclusion 

\((−∞,0)\)  \(x=−1\)  \(−\)  \(f\) is concave down.. 
\((0,∞)\)  \(x=1\)  \(+\)  \(f\) is concave up. 
NOTE: this table is equivalent to the line with a triangle on \(0\) and the Signs of the Second Derivative on the line using the technique shown above .
Be sure to write \(f''\)along the line to indicate this is the Concavity Test.
Step 8: From the table, we see that \(f\) has an inflection point at \(x=0\). \(f(0)=2\) so \((0,2)\) is a point of inflection.
We note that the information in the preceding table confirms the fact, found in step \(6\), that f has a local maximum at \(x=−1\) and a local minimum at \(x=1\). In addition, the information found in step \(6\)—namely, \(f\) has a local maximum at \(x=−1\) and a local minimum at \(x=1\), and \(f′(x)=0\) at those points—combined with the fact that \(f''\) changes sign only at \(x=0\) confirms the results found in step \(6\) on the concavity of \(f\).
Step 9: Combining the information from all our steps, we arrive at the graph of \(f(x)=(x−1)^2(x+2)\) shown in the following graph.
Exercise \(\PageIndex{8}\)
Sketch a graph of \(f(x)=(x−1)^3(x+2).\)
 Hint

\(f\) is a fourthdegree polynomial.
 Answer for the graph
Example \(\PageIndex{9}\): Sketching a Rational Function
Sketch the graph of \(f(x)=\frac{x^2}{1−x^2}\).
Solution
Step 1: The function \(f\) is defined as long as the denominator is not zero. Therefore, the domain is the set of all real numbers \(x\) except \(x=±1.\)
Step 2: Find the intercepts. If \(x=0,\) then \(f(x)=0\), so \(0\) is an intercept. If \(y=0\), then \(\frac{x^2}{(1−x^2)}=0,\) which implies x=0. Therefore, (0,0) is the only intercept.
Step 3: \(f(−x)=\frac{(−x)^2}{1−(−x)^2} = \frac{x^2}{1−x^2}\) This IS \(f(x)\), so this function is Even. It will be symmetric with the yaxis.
Step 4(a): Evaluate the limits at infinity. Since \(f\) is a rational function, divide the numerator and denominator by the highest power in the denominator: \(x^2\). We obtain
\(\lim_{x→±∞}\frac{x^2}{1−x^2}=\lim_{x→±∞}\frac{1}{\frac{1}{x^2}−1}=−1.\)
Therefore, \(f\) has a horizontal asymptote of \(y=−1\) as \(x→∞\) and \(x→−∞.\)
Step 4(b): To determine whether \(f\) has any vertical asymptotes, first check to see whether the denominator has any zeroes. We find the denominator is zero when \(x=±1\). To determine whether the lines \(x=1\) or \(x=−1\) are vertical asymptotes of \(f\), evaluate \(\lim_{x→1}f(x)\) and \(\lim_{x→−1}f(x)\). By looking at each onesided limit as \(x→1,\) we see that
\(\lim_{x→1^+}\frac{x^2}{1−x^2}=−∞\) and \(\lim_{x→1^−}\frac{x^2}{1−x^2}=∞.\)
In addition, by looking at each onesided limit as \(x→−1,\) we find that
\(\lim_{x→−1^+}\frac{x^2}{1−x^2}=∞\) and \(\lim_{x→−1^−}\frac{x^2}{1−x^2}=−∞.\)
Step 5: Calculate the first derivative:
\(f′(x)=\frac{(1−x^2)(2x)−x^2(−2x)}{(1−x^2)^2}=\frac{2x}{(1−x^2)^2}\).
Critical points occur at points \(x\) where \(f′(x)=0\) or \(f′(x)\) is undefined. We see that \(f′(x)=0\) when \(x=0.\) The derivative \(f′\) is not undefined at any point in the domain of \(f\). However, \(x=±1\) are not in the domain of \(f\). Therefore, to determine where \(f\) is increasing and where \(f\) is decreasing, divide the interval \((−∞,∞)\) into four smaller intervals: \((−∞,−1), (−1,0), (0,1),\) and \((1,∞)\), and choose a test point in each interval to determine the sign of \(f′(x)\) in each of these intervals. The values \(x=−2, x=−\frac{1}{2}, x=\frac{1}{2}\), and \(x=2\) are good choices for test points as shown in the following table.
Interval  Test point  Sign of \(f′(x)=\frac{2x}{(1−x^2)^2}\)  Conclusion 

\((−∞,−1)\)  \(x=−2\)  \(−/+=−\)  \(f\) is decreasing. 
\((−1,0)\)  \(x=−/2\)  \(−/+=−\)  \(f\) is decreasing. 
\((0,1)\)  \(x=1/2\)  \(+/+=+\)  \(f\) is increasing. 
\((1,∞)\)  \(x=2\)  \(+/+=+\)  \(f\) is increasing. 
NOTE: this table is equivalent to the line with asterisks on \(−1\), \(0\) and \(1\) and the Signs of the Derivative on the line using the technique shown above .
Be sure to write \(f′\) along the line to indicate this is the First Derivative Test.
Step 6: From the table, we see that \(f\) has a local minimum at \(x=0\) but no local maximum. \(f(0)=0.\) So the local minimum is \((0,0)\).
Step 7: Calculate the second derivative:
\(f''(x)=\frac{(1−x^2)^2(2)−2x(2(1−x^2)(−2x))}{(1−x^2)^4}\)
\(=\frac{(1−x^2)[2(1−x^2)+8x^2]}{(1−x^2)^4}\)
\(=\frac{2(1−x^2)+8x^2}{(1−x^2)^3}\)
\(=\frac{6x^2+2}{(1−x^2)^3}.\)
To determine the intervals where \(f\) is concave up and where \(f\) is concave down, we first need to find all points \(x\) where \(f''(x)=0\) or \(f''(x)\) is undefined. Since the numerator \(6x^2+2≠0\) for any \(x, f''(x)\) is never zero. Furthermore, \(f''\) is not undefined for any \(x\) in the domain of \(f\). However, as discussed earlier, \(x=±1\) are not in the domain of \(f\). Therefore, to determine the concavity of \(f\), we divide the interval \((−∞,∞)\) into the three smaller intervals \((−∞,−1), (−1,−1)\), and \((1,∞)\), and choose a test point in each of these intervals to evaluate the sign of \(f''(x)\). in each of these intervals. The values \(x=−2, x=0\), and \(x=2\) are possible test points as shown in the following table.
Interval  Test Point  Sign of \(f''(x)=\frac{6x^2+2}{(1−x^2)^3}\)  Conclusion 

\((−∞,−1)\)  \(x=−2\)  \(+/−=−\)  \(f\) is concave down. 
\((−1,−1)\)  \(x=0\)  \(+/+=+\)  \(f\) is concave up 
\((1,∞)\)  \(x=2\)  \(+/−=−\)  \(f\) is concave down. 
NOTE: this table is equivalent to the line with triangles on \(−1\) and \(1\) and the Signs of the Second Derivative on the line using the technique shown above .
Be sure to write \(f''\)along the line to indicate this is the Concavity Test.
Step 8: Although \(f\) changes concavity at \(x=−1\) and \(x=1\), there are no inflection points at either of these places because \(f\) is not continuous at \(x=−1\) or \(x=1.\) So, there are no points of inflection.
Step 9: Combining all this information, we arrive at the graph of \(f\) shown below.
\(f(x)=\frac{x^2}{(1−x^2)}\).
Exercise \(\PageIndex{9}\)
Sketch a graph of \(f(x)=\frac{(3x+5)}{(8+4x).}\)
 Hint

A line \(y=L\) is a horizontal asymptote of \(f\) if the limit as \(x→∞\) or the limit as \(x→−∞\) of \(f(x)\) is \(L\). A line \(x=a\) is a vertical asymptote if at least one of the onesided limits of \(f\) as \(x→a\) is \(∞\) or \(−∞.\)
 Answer for the graph
Example \(\PageIndex{10}\): Sketching aNOTHER Rational Function (with an Oblique Asymptote)
Sketch the graph of \(f(x)=\frac{x^2}{x−1}\)
Solution
Step 1: The domain of \(f\) is the set of all real numbers \(x\) except \(x=1.\)
Step 2: Find the intercepts. We can see that when \(x=0, f(x)=0,\) so \((0,0)\) is the only intercept.
Step 3: \(f(−x)=\frac{(−x)^2}{−x−1} = −\frac{x^2}{x+1}\). This is not \(f(x)\) nor \(f(x)\), so this function is not Even & not Odd.
Step 4(a): Evaluate the limits at infinity. Since \(f\) is a rational function, divide the numerator and denominator by the highest power in the denominator: \(x\). We obtain
\(\lim_{x→±∞}\frac{x^2}{(x−1)}=\lim_{x→±∞}\frac{x}{1−\frac{1}{x}}=x.\)
Therefore, \(f\) has no horizontal asymptote, but as \(x→∞\), \(f→∞\) and \(x→−∞\), \(f→−∞\).
*********the following analysis can be used for the oblique asymptote; not required for our course:
Since the degree of the numerator is one more than the degree of the denominator, \(f\) must have an oblique asymptote. To find the oblique asymptote, use long division of polynomials to write
\(f(x)=\frac{x^2}{x−1}=x+1+\frac{1}{x−1}\).
Since \(1/(x−1)→0\) as \(x→±∞, f(x)\) approaches the line \(y=x+1\) as \(x→±∞\). The line \(y=x+1\) is an oblique asymptote for \(f\).
************
Step 4(b): To check for vertical asymptotes, look at where the denominator is zero. Here the denominator is zero at \(x=1.\) Looking at both onesided limits as \(x→1,\) we find
\(\lim_{x→1^+}\frac{x^2}{x−1}=∞\) and \(\lim_{x→1^−}\frac{x^2}{x−1}=−∞.\)
Therefore, \(x=1\) is a vertical asymptote, and we have determined the behavior of \(f\) as \(x\) approaches \(1\) from the right and the left.
Step 5: Calculate the first derivative:
\(f′(x)=\frac{(x−1)(2x)−x^2(1)}{(x−1)^2}=\frac{x^2−2x}{(x−1)^2}.\)
We have \(f′(x)=0\) when \(x^2−2x=x(x−2)=0\). Therefore, \(x=0\) and \(x=2\) are critical points. Since \(f\) is undefined at \(x=1\), we need to divide the interval \((−∞,∞)\) into the smaller intervals \((−∞,0), (0,1), (1,2),\) and \((2,∞)\), and choose a test point from each interval to evaluate the sign of \(f′(x)\) in each of these smaller intervals. For example, let \(x=−1, x=\frac{1}{2}, x=\frac{3}{2}\), and \(x=3\) be the test points as shown in the following table.
Interval  Test point  Sign of \(f′(x)=\frac{x^2−2x}{(x−1)^2}\)  Conclusion 

\((−∞,0)\)  \(x=−1\)  \(+/+=+\)  \(f\) is increasing. 
\((0,1)\)  \(x=1/2\)  \(−/+=−\)  \(f\) is decreasing. 
\((1,2)\)  \(x=3/2\)  \(−/+=−\)  \(f\) is decreasing. 
\((2,∞)\)  \(x=3\)  \(+/+=+\)  \(f\) is increasing. 
NOTE: this table is equivalent to the line with asterisks on \(0\), \(1\) and \(2\) and the Signs of the Derivative on the line using the technique shown above .
Be sure to write \(f′\) along the line to indicate this is the First Derivative Test.
Step 6: We see \(f\) has a local maximum at \(x=0\) and a local minimum at \(x=2\). Evaluating \(f(x)\) at those two points, we find that the local maximum value is \(f(0)=0\) and the local minimum value is \(f(2)=4.\) So the local maximum is \((0,0)\) and the local minimum is \((2,4)\).
Step 7: The second derivative of \(f\) is
\(f''(x)=\frac{2}{(x−1)^3}\)
To determine the intervals where \(f\) is concave up and where \(f\) is concave down, we first need to find all points \(x\) where \(f''(x)=0\) or \(f''(x)\) is undefined. Since the numerator \(2≠0\) for any \(x, f''(x)\) is never zero. Furthermore, \(f''\) is not undefined for any \(x\) in the domain of \(f\). However, as discussed earlier, \(x=1\) is not in the domain of \(f\). Therefore, to determine the concavity of \(f\), we divide the interval \((−∞,∞)\) into the two smaller intervals \((−∞,1), and \((1,∞)\), and choose a test point in each of these intervals to evaluate the sign of \(f''(x)\). in each of these intervals. The values \( x=0\), and \(x=2\) are possible test points as shown in the following table.
Interval  Test Point  Sign of \(f''(x)=\frac{2}{(x−1)^3}\)  Conclusion 

\((−∞,1)\)  \(x=0\)  \(+/−=−\)  \(f\) is concave down. 
\((1,∞)\)  \(x=2\)  \(+/+=+\)  \(f\) is concave up 
NOTE: this table is equivalent to the line with triangles on \(−1\) and \(1\) and the Signs of the Second Derivative on the line using the technique shown above .
Be sure to write \(f''\)along the line to indicate this is the Concavity Test.
Step 8: Although \(f\) changes concavity at \(x=1\), there is no inflection point because \(f\) is not continuous at \(x=1.\) So, there are no points of inflection.
Step 9: From the information gathered, we arrive at the following graph for \(f.\)
Note: drawing in the oblique asymptote is not required
Exercise \(\PageIndex{10}\)
OPTIONAL: Find the oblique asymptote for \(f(x)=\frac{(3x^3−2x+1)}{(2x^2−4)}\).
 Hint

Use long division of polynomials.
 Answer

\(y=\frac{3}{2}x\)
Example \(\PageIndex{11}\): Sketching the Graph of a Function with a Cusp
Sketch a graph of \(f(x)=(x−1)^{2/3}\)
Solution
Step 1: Since the cuberoot function is defined for all real numbers \(x\) and \((x−1)^{2/3}=(\sqrt[3]{x−1})^2\), the domain of \(f\) is all real numbers.
Step 2: To find the \(y\)intercept, evaluate \(f(0)\). Since \(f(0)=1,\) the \(y\)intercept is \((0,1)\). To find the \(x\)intercept, solve \((x−1)^{2/3}=0\). The solution of this equation is \(x=1\), so the \(x\)intercept is \((1,0).\)
Step 3: \(f(−x)=(−x−1)^{2/3} = ((−1)(x+1))^{2/3}\)=(x+1)^{2/3}. This is not \(f(x)\) nor \(f(x)\), so this function is not Even & not Odd.
Step 4(a): Since \(\lim_{x→±∞}(x−1)^{2/3}=∞,\) the function continues to grow without bound as \(x→∞\) and \(x→−∞.\)
Step 4(b): The function has no vertical asymptotes.
Step 5: To determine where \(f\) is increasing or decreasing, calculate \(f′.\) We find
\[f′(x)=\frac{2}{3}(x−1)^{−1/3}=\frac{2}{3(x−1)^{1/3}}\]
This function is not zero anywhere, but it is undefined when \(x=1.\) Therefore, the only critical point is \(x=1.\) Divide the interval \((−∞,∞)\) into the smaller intervals \((−∞,1)\) and \((1,∞)\), and choose test points in each of these intervals to determine the sign of \(f′(x)\) in each of these smaller intervals. Let \(x=0\) and \(x=2\) be the test points as shown in the following table.
Interval  Test Point  Sign of \(f′(x)=\frac{2}{3(x−1)^{1/3}}\)  Conclusion 

\((−∞,1)\)  \(x=0\)  \(+/−=−\)  \(f\) is decreasing 
\((1,∞)\)  \(x=2\)  \(+/+=+\)  \(f\) is increasing 
NOTE: this table is equivalent to the line with an asterisk on \(1\) and the Signs of the Derivative on the line using the technique shown above .
Be sure to write \(f′\) along the line to indicate this is the First Derivative Test.
Step 6: We conclude that \(f\)as a local minimum at \(x=1\). Evaluating \(f\) at \(x=1\), we find that the value of \(f\) at the local minimum is zero. The local minimum is \((1,0)\).Note that \(f′(1)\) is undefined, so to determine the behavior of the function at this critical point, we need to examine \(\lim_{x→1}f′(x).\) Looking at the onesided limits, we have
\[\lim_{x→1^+}\frac{2}{3(x−1)^{1/3}}=∞\] and \[\lim_{x→1^−}\frac{2}{3(x−1)^{1/3}}=−∞.\]
Therefore, \(f\) has a cusp at \(x=1.\)
Step 7: To determine concavity, we calculate the second derivative of \(f:\)
\[f''(x)=−\dfrac{2}{9}(x−1)^{−4/3}=\dfrac{−2}{9(x−1)^{4/3}}.\]
We find that \(f''(x)\) is defined for all \(x\), but is undefined when \(x=1\). Therefore, divide the interval \((−∞,∞)\) into the smaller intervals \((−∞,1)\) and \((1,∞)\), and choose test points to evaluate the sign of \(f''(x)\) in each of these intervals. As we did earlier, let \(x=0\) and \(x=2\) be test points as shown in the following table.
Interval  Test Point  Sign of \(f''(x)=\frac{−2}{3(x−1)^{4/3}}\)  Conclusion 

\((−∞,1)\)  \(x=0\)  \(−/+=−\)  \(f\) is concave down 
\((1,∞)\)  \(x=2\)  \(−/+=−\)  \(f\) is concave down 
NOTE: this table is equivalent to the line with a triangle on \(1\) and the Signs of the Second Derivative on the line using the technique shown above .
Be sure to write \(f''\)along the line to indicate this is the Concavity Test.
From this table, we conclude that \(f\)s concave down everywhere.
Step 8: There are no points of inflection.
Step 9: Combining all of this information, we arrive at the following graph for \(f\).
Exercise \(\PageIndex{11}\)
Consider the function \(f(x)=5−x^{2/3}\). Determine the point on the graph where a cusp is located. Determine the end behavior of \(f\).
 Hint

A function \(f\) has a cusp at a point a if \(f(a)\) exists, \(f'(a)\) is undefined, one of the onesided limits as \(x→a\) of \(f'(x) is +∞\), and the other onesided limit is \(−∞.\)
 Answer

The function \(f\) has a cusp at \((0,5) \lim_{x→0^−}f′(x)=∞\), \(\lim_{x→0^+}f′(x)=−∞\). For end behavior, \(\lim_{x→±∞}f(x)=−∞.\)
DO we want this? is it repeated above? If keep, decide where.
Guidelines for Drawing the Graph of a Function
NOTE: Drawing the Graph will be covered in a later chapter. You do NOT need any of the rest of this section for Unit 1.
We now have enough analytical tools to draw graphs of a wide variety of algebraic and transcendental functions. Before showing how to graph specific functions, let’s look at a general strategy to use when graphing any function.
ProblemSolving Strategy: Drawing the Graph of a Function
Given a function \(f\), use the following steps to sketch a graph of \(f\):
 Determine the domain of the function.
 Locate the \(x\) and \(y\)intercepts.
 Evaluate \(\lim_{x→∞}f(x)\) and \(\lim_{x→−∞}f(x)\) to determine the end behavior. If either of these limits is a finite number \(L\), then \(y=L\) is a horizontal asymptote. If either of these limits is \(∞\) or \(−∞\), determine whether \(f\) has an oblique asymptote. If \(i\)s a rational function such that \(f(x)=\frac{p(x)}{q(x)}\), where the degree of the numerator is greater than the degree of the denominator, then \(f\) can be written as \[f(x)=\frac{p(x)}{q(x)}=g(x)+\frac{r(x)}{q(x),}\] where the degree of \(r(x)\) is less than the degree of \(q(x)\). The values of \(f(x)\) approach the values of \(g(x)\) as \(x→±∞\). If \(g(x)\) is a linear function, it is known as an oblique asymptote.
 Determine whether \(f\) has any vertical asymptotes.
 Calculate \(f′.\) Find all critical points and determine the intervals where \(f\) is increasing and where \(f\) is decreasing. Determine whether \(f\) has any local extrema.
 Calculate \(f''.\) Determine the intervals where \(f\) is concave up and where \(f\) is concave down. Use this information to determine whether \(f\) has any inflection points. The second derivative can also be used as an alternate means to determine or verify that \(f\) has a local extremum at a critical point.
Now let’s use this strategy to graph several different functions. We start by graphing a polynomial function.
Example \(\PageIndex{8}\): Sketching a Graph of a Polynomial
Sketch a graph of \(f(x)=(x−1)^2(x+2).\)
Solution
Step 1: Since \(f\) is a polynomial, the domain is the set of all real numbers.
Step 2: When \(x=0,f(x)=2.\) Therefore, the yintercept is \((0,2)\). To find the \(x\)intercepts, we need to solve the equation \((x−1)^2(x+2)=0\), gives us the \(x\)intercepts \((1,0)\) and \((−2,0)\)
Step 3: We need to evaluate the end behavior of \(f.\)As \(x→∞, (x−1)^2→∞\) and \((x+2)→∞\). Therefore, \(\lim_{x→∞}f(x)=∞\) As \(x→−∞, (x−1)^2→∞\) and \((x+2)→−∞\). Therefore, \(\lim_{x→∞}f(x)=−∞\). To get even more information about the end behavior of \(f\), we can multiply the factors of \(f\).When doing so, we see that
\[f(x)=(x−1)^2(x+2)=x^3−3x+2.\]
Since the leading term of \(f\) is \(x^3\), we conclude that \(f\) behaves like \(y=x^3\) as \(x→±∞.\)
Step 4: Since \(f\) is a polynomial function, it does not have any vertical asymptotes.
Step 5: The first derivative of \(f\) is
\[f′(x)=3x^2−3.\]
Therefore, \(f\) has two critical points: \(x=1,−1.\) Divide the interval \((−∞,∞)\) into the three smaller intervals: \((−∞,−1), (−1,1)\), and \((1,∞)\). Then, choose test points \(x=−2, x=0\), and \(x=2\) from these intervals and evaluate the sign of \(f′(x)\) at each of these test points, as shown in the following table.
Interval  Test point  Sign of Derivative \(f'(x)=3x^2−3=3(x−1)(x+1)\)  Conclusion 

\((−∞,−1)\)  \(x=−2\)  \((+)(−)(−)=+\)  \(f\) is increasing 
\((−1,1)\)  \(x=0\)  \((+)(−)(+)=−\)  \(f\) decreasing 
\((1,∞)\)  (x=2\)  \((+)(+)(+)=+\)  \(f\) is increasing 
From the table, we see that \(f\) has a local maximum at \(x=−1\) and a local minimum at \(x=1\). Evaluating \(f(x)\) at those two points, we find that the local maximum value is \(f(−1)=4\) and the local minimum value is \(f(1)=0.\)
Step 6: The second derivative of \(f\) is
\[f''(x)=6x.\]
The second derivative is zero at \(x=0.\) Therefore, to determine the concavity of \(f\), divide the interval \((−∞,∞)\) into the smaller intervals \((−∞,0)\) and \((0,∞)\), and choose test points \(x=−1\) and \(x=1\) to determine the concavity of \(f\) on each of these smaller intervals as shown in the following table.
Interval  Test Point  Sign of \(''(x)=6x\)  Conclusion 

\((−∞,0)\)  \(x=−1\)  \(−\)  \(f\) is concave down.. 
\((0,∞)\)  \(x=1\)  \(+\)  \(f\) is concave up. 
We note that the information in the preceding table confirms the fact, found in step \(5\), that f has a local maximum at \(x=−1\) and a local minimum at \(x=1\). In addition, the information found in step \(5\)—namely, \(f\) has a local maximum at \(x=−1\) and a local minimum at \(x=1\), and \(f′(x)=0\) at those points—combined with the fact that \(f''\) changes sign only at \(x=0\) confirms the results found in step \(6\) on the concavity of \(f\).
Combining this information, we arrive at the graph of \(f(x)=(x−1)^2(x+2)\) shown in the following graph.
Exercise \(\PageIndex{8}\)
Sketch a graph of \(f(x)=(x−1)^3(x+2).\)
 Answer

\(f\) is a fourthdegree polynomial.
 Answer
Example \(\PageIndex{9}\): Sketching a Rational Function
Sketch the graph of \(f(x)=\frac{x^2}{(1−x^2)}\).
Solution
Step 1: The function \(f\) is defined as long as the denominator is not zero. Therefore, the domain is the set of all real numbers \(x\) except \(x=±1.\)
Step 2: Find the intercepts. If \(x=0,\) then \(f(x)=0\), so \(0\) is an intercept. If \(y=0\), then \(\frac{x^2}{(1−x^2)}=0,\) which implies x=0. Therefore, (0,0) is the only intercept.
Step 3: Evaluate the limits at infinity. Since \(f\)is a rational function, divide the numerator and denominator by the highest power in the denominator: \(x^2\).We obtain
\(\lim_{x→±∞}\frac{x^2}{1−x^2}=\lim_{x→±∞}\frac{1}{\frac{1}{x^2}−1}=−1.\)
Therefore, \(f\) has a horizontal asymptote of \(y=−1\) as \(x→∞\) and \(x→−∞.\)
Step 4: To determine whether \(f\) has any vertical asymptotes, first check to see whether the denominator has any zeroes. We find the denominator is zero when \(x=±1\). To determine whether the lines \(x=1\) or \(x=−1\) are vertical asymptotes of \(f\), evaluate \(\lim_{x→1}f(x)\) and \(\lim_{x→−1}f(x)\). By looking at each onesided limit as \(x→1,\) we see that
\(\lim_{x→1^+}\frac{x^2}{1−x^2}=−∞\) and \(\lim_{x→1^−}\frac{x^2}{1−x^2}=∞.\)
In addition, by looking at each onesided limit as \(x→−1,\) we find that
\(\lim_{x→−1^+}\frac{x^2}{1−x^2}=∞\) and \(\lim_{x→−1^−}\frac{x^2}{1−x^2}=−∞.\)
Step 5: Calculate the first derivative:
\(f′(x)=\frac{(1−x^2)(2x)−x^2(−2x)}{(1−x^2)^2}=\frac{2x}{(1−x^2)^2}\).
Critical points occur at points \(x\) where \(f′(x)=0\) or \(f′(x)\) is undefined. We see that \(f′(x)=0\) when \(x=0.\) The derivative \(f′\) is not undefined at any point in the domain of \(f\). However, \(x=±1\) are not in the domain of \(f\). Therefore, to determine where \(f\) is increasing and where \(f\) is decreasing, divide the interval \((−∞,∞)\) into four smaller intervals: \((−∞,−1), (−1,0), (0,1),\) and \((1,∞)\), and choose a test point in each interval to determine the sign of \(f′(x)\) in each of these intervals. The values \(x=−2, x=−\frac{1}{2}, x=\frac{1}{2}\), and \(x=2\) are good choices for test points as shown in the following table.
Interval  Test point  Sign of \(f′(x)=\frac{2x}{(1−x^2)^2}\)  Conclusion 

\((−∞,−1)\)  \(x=−2\)  \(−/+=−\)  \(f\) is decreasing. 
\((−1,0)\)  \(x=−/2\)  \(−/+=−\)  \(f\) is decreasing. 
\((0,1)\)  \(x=1/2\)  \(+/+=+\)  \(f\) is increasing. 
\((1,∞)\)  \(x=2\)  \(+/+=+\)  \(f\) is increasing. 
From this analysis, we conclude that \(f\) has a local minimum at \(x=0\) but no local maximum.
Step 6: Calculate the second derivative:
\(f''(x)=\frac{(1−x^2)^2(2)−2x(2(1−x^2)(−2x))}{(1−x^2)^4}\)
\(=\frac{(1−x^2)[2(1−x^2)+8x^2]}{(1−x^2)^4}\)
\(=\frac{2(1−x^2)+8x^2}{(1−x^2)^3}\)
\(=\frac{6x^2+2}{(1−x^2)^3}.\)
To determine the intervals where \(f\) is concave up and where \(f\) is concave down, we first need to find all points \(x\) where \(f''(x)=0\) or \(f''(x)\) is undefined. Since the numerator \(6x^2+2≠0\) for any \(x, f''(x)\) is never zero. Furthermore, \(f''\) is not undefined for any \(x\) in the domain of \(f\). However, as discussed earlier, \(x=±1\) are not in the domain of \(f\). Therefore, to determine the concavity of \(f\), we divide the interval \((−∞,∞)\) into the three smaller intervals \((−∞,−1), (−1,−1)\), and \((1,∞)\), and choose a test point in each of these intervals to evaluate the sign of \(f''(x)\). in each of these intervals. The values \(x=−2, x=0\), and \(x=2\) are possible test points as shown in the following table.
Interval  Test Point  Sign of \(f''(x)=\frac{6x^2+2}{(1−x^2)^3}\)  Conclusion 

\((−∞,−1)\)  \(x=−2\)  \(+/−=−\)  \(f\) is concave down. 
\((−1,−1)\)  \(x=0\)  \(+/+=+\)  \(f\) is concave up 
\((1,∞)\)  \(x=2\)  \(+/−=−\)  \(f\) is concave down. 
Combining all this information, we arrive at the graph of \(f\) shown below. Note that, although \(f\) changes concavity at \(x=−1\) and \(x=1\), there are no inflection points at either of these places because \(f\) is not continuous at \(x=−1\) or \(x=1.\)
Exercise \(\PageIndex{9}\)
Sketch a graph of \(f(x)=\frac{(3x+5)}{(8+4x).}\)
 Hint

A line \(y=L\) is a horizontal asymptote of \(f\) if the limit as \(x→∞\) or the limit as \(x→−∞\) of \(f(x)\) is \(L\). A line \(x=a\) is a vertical asymptote if at least one of the onesided limits of \(f\) as \(x→a\) is \(∞\) or \(−∞.\)
 Answer
Example \(\PageIndex{10}\): Sketching a Rational Function with an Oblique Asymptote
Sketch the graph of \(f(x)=\frac{x^2}{(x−1)}\)
Solution
Step 1: The domain of \(f\) is the set of all real numbers \(x\) except \(x=1.\)
Step 2: Find the intercepts. We can see that when \(x=0, f(x)=0,\) so \((0,0)\) is the only intercept.
Step 3: Evaluate the limits at infinity. Since the degree of the numerator is one more than the degree of the denominator, \(f\) must have an oblique asymptote. To find the oblique asymptote, use long division of polynomials to write
\(f(x)=\frac{x^2}{x−1}=x+1+\frac{1}{x−1}\).
Since \(1/(x−1)→0\) as \(x→±∞, f(x)\) approaches the line \(y=x+1\) as \(x→±∞\). The line \(y=x+1\) is an oblique asymptote for \(f\).
Step 4: To check for vertical asymptotes, look at where the denominator is zero. Here the denominator is zero at \(x=1.\) Looking at both onesided limits as \(x→1,\) we find
\(\lim_{x→1^+}\frac{x^2}{x−1}=∞\) and \(\lim_{x→1^−}\frac{x^2}{x−1}=−∞.\)
Therefore, \(x=1\) is a vertical asymptote, and we have determined the behavior of \(f\) as \(x\) approaches \(1\) from the right and the left.
Step 5: Calculate the first derivative:
\(f′(x)=\frac{(x−1)(2x)−x^2(1)}{(x−1)^2}=\frac{x^2−2x}{(x−1)^2}.\)
We have \(f′(x)=0\) when \(x^2−2x=x(x−2)=0\). Therefore, \(x=0\) and \(x=2\) are critical points. Since \(f\) is undefined at \(x=1\), we need to divide the interval \((−∞,∞)\) into the smaller intervals \((−∞,0), (0,1), (1,2),\) and \((2,∞)\), and choose a test point from each interval to evaluate the sign of \(f′(x)\) in each of these smaller intervals. For example, let \(x=−1, x=\frac{1}{2}, x=\frac{3}{2}\), and \(x=3\) be the test points as shown in the following table.
Interval  Test Point  Sign of \(f''(x)=\frac{2}{(x−1)^3}\)  Conclusion 

\((−∞,1)\)  \(x=0\)  \(+/−=−\)  \(f\) is concave down. 
\((1,∞)\)  (x=2\)  \(+/+=+\)  \(f\) is concave up 
From the information gathered, we arrive at the following graph for \(f.\)
Exercise \(\PageIndex{10}\)
Find the oblique asymptote for \(f(x)=\frac{(3x^3−2x+1)}{(2x^2−4)}\).
 Hint

Use long division of polynomials.
 Answer

\(y=\frac{3}{2}x\)
Example \(\PageIndex{11}\): Sketching the Graph of a Function with a Cusp
Sketch a graph of \((x)=(x−1)^{2/3}\)
Solution
Step 1: Since the cuberoot function is defined for all real numbers \(x\) and \((x−1)^{2/3}=(\sqrt[3]{x−1})^2\), the domain of \(f\) is all real numbers.
Step 2: To find the \(y\)intercept, evaluate \(f(0)\). Since \(f(0)=1,\) the \(y\)intercept is \((0,1)\). To find the \(x\)intercept, solve \((x−1)^{2/3}=0\). The solution of this equation is \(x=1\), so the \(x\)intercept is \((1,0).\)
Step 3: Since \(\lim_{x→±∞}(x−1)^{2/3}=∞,\) the function continues to grow without bound as \(x→∞\) and \(x→−∞.\)
Step 4: The function has no vertical asymptotes.
Step 5: To determine where \(f\) is increasing or decreasing, calculate \(f′.\) We find
\[f′(x)=\frac{2}{3}(x−1)^{−1/3}=\frac{2}{3(x−1)^{1/3}}\]
This function is not zero anywhere, but it is undefined when \(x=1.\) Therefore, the only critical point is \(x=1.\) Divide the interval \((−∞,∞)\) into the smaller intervals \((−∞,1)\) and \((1,∞)\), and choose test points in each of these intervals to determine the sign of \(f′(x)\) in each of these smaller intervals. Let \(x=0\) and \(x=2\) be the test points as shown in the following table.
Interval  Test Point  Sign of \(f′(x)=\frac{2}{3(x−1)^{1/3}}\)  Conclusion 

\((−∞,1)\)  \(x=0\)  \(+/−=−\)  \(f\) is decreasing 
\((1,∞)\)  \(x=2\)  \(+/+=+\)  \(f\) is increasing 
We conclude that \(h\)as a local minimum at \(x=1\). Evaluating \(f\) at \(x=1\), we find that the value of \(f\) at the local minimum is zero. Note that \(f′(1)\) is undefined, so to determine the behavior of the function at this critical point, we need to examine \(\lim_{x→1}f′(x).\) Looking at the onesided limits, we have
\[\lim_{x→1^+}\frac{2}{3(x−1)^{1/3}}=∞\) and \(\lim_{x→1^−}\frac{2}{3(x−1)^{1/3}}=−∞.\]
Therefore, \(f\) has a cusp at \(x=1.\)
Step 6: To determine concavity, we calculate the second derivative of \(f:\)
\[f''(x)=−\dfrac{2}{9}(x−1)^{−4/3}=\dfrac{−2}{9(x−1)^{4/3}}.\]
We find that \(f''(x)\) is defined for all \(x\), but is undefined when \(x=1\). Therefore, divide the interval \((−∞,∞)\) into the smaller intervals \((−∞,1)\) and \((1,∞)\), and choose test points to evaluate the sign of \(f''(x)\) in each of these intervals. As we did earlier, let \(x=0\) and \(x=2\) be test points as shown in the following table.
Interval  Test Point  Sign of \(f''(x)=\frac{−2}{3(x−1)^{4/3}}\)  Conclusion 

\((−∞,1)\)  \(x=0\)  \(−/+=−\)  \(f\) is concave down 
\((1,∞)\)  \(x=2\)  \−(+/+=−\)  \(f\) is concave down 
From this table, we conclude that \(i\)s concave down everywhere. Combining all of this information, we arrive at the following graph for \(f\).
Exercise \(\PageIndex{11}\)
Consider the function \(f(x)=5−x^{2/3}\). Determine the point on the graph where a cusp is located. Determine the end behavior of \(f\).
 Hint

A function \(f\) has a cusp at a point a if \(f(a)\) exists, \(f'(a)\) is undefined, one of the onesided limits as \(x→a\) of \(f'(x) is +∞\), and the other onesided limit is \(−∞.\)
 Answer

The function \(f\) has a cusp at \((0,5) \lim_{x→0^−}f′(x)=∞\), \(\lim_{x→0^+}f′(x)=−∞\). For end behavior, \(\lim_{x→±∞}f(x)=−∞.\)
Key Concepts
 The limit of \(f(x)\) is \(L\) as \(x→∞\) (or as \(x→−∞)\) if the values \(f(x)\) become arbitrarily close to \(L\) as \(x\)becomes sufficiently large.
 The limit of \(f(x)\) is \(∞\) as \(x→∞\) if \(f(x)\) becomes arbitrarily large as \(x\) becomes sufficiently large. The limit of \(f(x)\) is \(−∞\) as \(x→∞\) if \(f(x)<0\) and \(f(x)\) becomes arbitrarily large as \(x\) becomes sufficiently large. We can define the limit of \(f(x)\) as \(x\) approaches \(−∞\) similarly.
 For a polynomial function \(p(x)=a_nx^n+a_{n−1}x^{n−1}+…+a_1x+a_0,\) where \(a_n≠0\), the end behavior is determined by the leading term \(a_nx^n\). If \(n≠0, p(x)\) approaches \(∞\) or \(−∞\)at each end.
 For a rational function \(f(x)=\frac{p(x)}{q(x),}\) the end behavior is determined by the relationship between the degree of \(p\) and the degree of \(q\). If the degree of \(p\) is less than the degree of \(q\), the line \(y=0\) is a horizontal asymptote for \(f\). If the degree of \(p\) is equal to the degree of \(q\), then the line \(y=\frac{a_n}{b_n}\) is a horizontal asymptote, where \(a_n\) and \(b_n\) are the leading coefficients of \(p\) and \(q\), respectively. If the degree of \(p\) is greater than the degree of \(q\), then \(f\) approaches \(∞\) or \(−∞\) at each end.
Glossary
 end behavior
 the behavior of a function as \(x→∞\) and \(x→−∞\)
 horizontal asymptote
 if \(\lim_{x→∞}f(x)=L\) or \(\lim_{x→−∞}f(x)=L\), then \(y=L\) is a horizontal asymptote of \(f\)
 infinite limit at infinity
 a function that becomes arbitrarily large as x becomes large
 limit at infinity
 a function that becomes arbitrarily large as \(x\) becomes large
 oblique asymptote
 the line \(y=mx+b\) if \(f(x)\) approaches it as \(x→∞\) or\( x→−∞\)
Contributors
Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CCBYSANC 4.0 license. Download for free at http://cnx.org.