Drawing Graphs of Functions

Learning Outcomes

Analyze a function and its derivatives to draw its graph

Guidelines for Graphing 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.

Problem-Solving 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 $\underset{x\to \infty }{\lim}f(x)$ and $\underset{x\to −\infty }{\lim}f(x)$ 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 $\infty$ or $−\infty$ , determine whether $f$ has an oblique asymptote. If $f$ is 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
$f(x)=\dfrac{p(x)}{q(x)}=g(x)+\dfrac{r(x)}{q(x)}$ ,

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 $x\to \pm \infty$ . If $g(x)$ is a linear function, it is known as an oblique asymptote.
Determine whether $f$ has any vertical asymptotes.
Calculate $f^{\prime}$ . 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 $f^{\prime \prime}$ . 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: Sketching a Graph of a Polynomial

Sketch a graph of $f(x)=(x-1)^2 (x+2)$

Show 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 $y$ -intercept is $(0,2)$ . To find the $x$ -intercepts, we need to solve the equation $(x-1)^2 (x+2)=0$ , which gives us the $x$ -intercepts $(1,0)$ and $(-2,0)$

Step 3. We need to evaluate the end behavior of $f$ . As $x\to \infty$ , $(x-1)^2 \to \infty$ and $(x+2)\to \infty$ . Therefore, $\underset{x\to \infty }{\lim}f(x)=\infty$ . As $x\to −\infty$ , $(x-1)^2 \to \infty$ and $(x+2) \to −\infty$ . Therefore, $\underset{x\to -\infty }{\lim}f(x)=−\infty$ . 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} - 3 x + 2

$f(x)=(x-1)^2 (x+2)=x^3-3x+2$

Since the leading term of $f$ is $x^3$ , we conclude that $f$ behaves like $y=x^3$ as $x\to \pm \infty$ .

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) = 3 x^{2} - 3

$f^{\prime}(x)=3x^2-3$

Therefore, $f$ has two critical points: $x=1,-1$ . Divide the interval $(−\infty ,\infty)$ into the three smaller intervals: $(−\infty ,-1)$ , $(-1,1)$ , and $(1,\infty )$ . Then, choose test points $x=-2$ , $x=0$ , and $x=2$ from these intervals and evaluate the sign of $f^{\prime}(x)$ at each of these test points, as shown in the following table.

Interval	Test Point	Sign of Derivative $f^{\prime}(x)=3x^2-3=3(x-1)(x+1)$	Conclusion
$(−\infty ,-1)$	$x=-2$	$(+)(−)(−)=+$	$f$ is increasing.
$(-1,1)$	$x=0$	$(+)(−)(+)=−$	$f$ is decreasing.
$(1,\infty )$	$x=2$	$(+)(+)(+)=+$	$f$ 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) = 6 x

$f^{\prime \prime}(x)=6x$

The second derivative is zero at $x=0$ . Therefore, to determine the concavity of $f$ , divide the interval $(−\infty ,\infty)$ into the smaller intervals $(−\infty ,0)$ and $(0,\infty )$ , 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 $f^{\prime \prime}(x)=6x$	Conclusion
$(−\infty ,0)$	$x=-1$	$-$	$f$ is concave down.
$(0,\infty )$	$x=1$	$+$	$f$ 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 $f^{\prime}(x)=0$ at those points—combined with the fact that $f^{\prime \prime}$ 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.

The function f(x) = (x −1)2 (x + 2) is graphed. It crosses the x axis at x = −2 and touches the x axis at x = 1.

Figure 23.

Watch the following video to see the worked solution to Example: Sketching a Graph of a Polynomial.

Closed Captioning and Transcript Information for Video

Try It

Sketch a graph of $f(x)=(x-1)^3 (x+2)$

Hint

$f$ is a fourth-degree polynomial.

Show Solution

Example: Sketching a Rational Function

Sketch the graph of $f(x)=\dfrac{x^2}{1-x^2}$

Show 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 $x=\pm 1$ .

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 \to \pm \infty} \frac{x^{2}}{1 - x^{2}} = lim_{x \to \pm \infty} \frac{1}{\frac{1}{x^{2}} - 1} = - 1

$\underset{x\to \pm \infty }{\lim}\frac{x^2}{1-x^2}=\underset{x\to \pm \infty }{\lim}\frac{1}{\frac{1}{x^2}-1}=-1$ .

Therefore, $f$ has a horizontal asymptote of $y=-1$ as $x\to \infty$ and $x\to −\infty$ .

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 $x=\pm 1$ . To determine whether the lines $x=1$ or $x=-1$ are vertical asymptotes of $f$ , evaluate $\underset{x\to 1}{\lim}f(x)$ and $\underset{x\to −1}{\lim}f(x)$ . By looking at each one-sided limit as $x\to 1$ , we see that

lim_{x \to 1^{+}} \frac{x^{2}}{1 - x^{2}} = - \infty

$\underset{x\to 1^+}{\lim}\frac{x^2}{1-x^2}=−\infty$ and

lim_{x \to 1^{-}} \frac{x^{2}}{1 - x^{2}} = \infty

$\underset{x\to 1^-}{\lim}\frac{x^2}{1-x^2}=\infty$ .

In addition, by looking at each one-sided limit as $x\to −1$ , we find that

lim_{x \to - 1^{+}} \frac{x^{2}}{1 - x^{2}} = \infty

$\underset{x\to −1^+}{\lim}\frac{x^2}{1-x^2}=\infty$ and

lim_{x \to - 1^{-}} \frac{x^{2}}{1 - x^{2}} = - \infty

$\underset{x\to −1^-}{\lim}\frac{x^2}{1-x^2}=−\infty$ .

Step 5. Calculate the first derivative:

f^{'} (x) = \frac{(1 - x^{2}) (2 x) - x^{2} (- 2 x)}{(1 - x^{2})^{2}} = \frac{2 x}{(1 - x^{2})^{2}}

$f^{\prime}(x)=\frac{(1-x^2)(2x)-x^2(-2x)}{(1-x^2)^2}=\frac{2x}{(1-x^2)^2}$ .

Critical points occur at points $x$ where $f^{\prime}(x)=0$ or $f^{\prime}(x)$ is undefined. We see that $f^{\prime}(x)=0$ when $x=0$ . The derivative $f^{\prime}$ is not undefined at any point in the domain of $f$ . However, $x=\pm 1$ are not in the domain of $f$ . Therefore, to determine where $f$ is increasing and where $f$ is decreasing, divide the interval $(−\infty ,\infty )$ into four smaller intervals: $(−\infty ,-1)$ , $(-1,0)$ , $(0,1)$ , and $(1,\infty )$ , and choose a test point in each interval to determine the sign of $f^{\prime}(x)$ 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 $f^{\prime}(x)=\frac{2x}{(1-x^2)^2}$	Conclusion
$(−\infty ,-1)$	$x=-2$	$−/+=−$	$f$ is decreasing.
$(-1,0)$	$x=-1/2$	$−/+=−$	$f$ is decreasing.
$(0,1)$	$x=1/2$	$+/+=+$	$f$ is increasing.
$(1,\infty )$	$x=2$	$+/+=+$	$f$ 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:

\begin{array}{ll} f^{''} (x) & = \frac{(1 - x^{2})^{2} (2) - 2 x (2 (1 - x^{2}) (- 2 x))}{(1 - x^{2})^{4}} \\ = \frac{(1 - x^{2}) [2 (1 - x^{2}) + 8 x^{2}]}{(1 - x^{2})^{4}} \\ = \frac{2 (1 - x^{2}) + 8 x^{2}}{(1 - x^{2})^{3}} \\ = \frac{6 x^{2} + 2}{(1 - x^{2})^{3}} \end{array}

$\begin{array}{ll} f^{\prime \prime}(x) & =\frac{(1-x^2)^2(2)-2x(2(1-x^2)(-2x))}{(1-x^2)^4} \\ & =\frac{(1-x^2)[2(1-x^2)+8x^2]}{(1-x^2)^4} \\ & =\frac{2(1-x^2)+8x^2}{(1-x^2)^3} \\ & =\frac{6x^2+2}{(1-x^2)^3} \end{array}$

To determine the intervals where $f$ is concave up and where $f$ is concave down, we first need to find all points $x$ where $f^{\prime \prime}(x)=0$ or $f^{\prime \prime}(x)$ is undefined. Since the numerator $6x^2+2 \ne 0$ for any $x$ , $f^{\prime \prime}(x)$ is never zero. Furthermore, $f^{\prime \prime}$ is not undefined for any $x$ in the domain of $f$ . However, as discussed earlier, $x=\pm 1$ are not in the domain of $f$ . Therefore, to determine the concavity of $f$ , we divide the interval $(−\infty ,\infty )$ into the three smaller intervals $(−\infty ,-1)$ , $(-1,-1)$ , and $(1,\infty )$ , and choose a test point in each of these intervals to evaluate the sign of $f^{\prime \prime}(x)$ 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 $f^{\prime \prime}(x)=\frac{6x^2+2}{(1-x^2)^3}$	Conclusion
$(−\infty ,-1)$	$x=-2$	$+/-=−$	$f$ is concave down.
$(-1,-1)$	$x=0$	$+/+=+$	$f$ is concave up.
$(1,\infty )$	$x=2$	$+/-=−$	$f$ 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 $x=1$ .

The function f(x) = x2/(1 − x2) is graphed. It has asymptotes y = −1, x = −1, and x = 1.

Figure 25.

Try It

Sketch a graph of $f(x)=\dfrac{3x+5}{8+4x}$

Hint

A line $y=L$ is a horizontal asymptote of $f$ if the limit as $x\to \infty$ or the limit as $x\to −\infty$ of $f(x)$ is $L$ . A line $x=a$ is a vertical asymptote if at least one of the one-sided limits of $f$ as $x\to a$ is $\infty$ or $−\infty$ .

Show Solution

Example: Sketching a Rational Function with an Oblique Asymptote

Sketch the graph of $f(x)=\dfrac{x^2}{x-1}$

Show 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}

$f(x)=\frac{x^2}{x-1}=x+1+\frac{1}{x-1}$ .

Since $1/(x-1)\to 0$ as $x\to \pm \infty$ , $f(x)$ approaches the line $y=x+1$ as $x\to \pm \infty$ . 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 one-sided limits as $x\to 1$ , we find

lim_{x \to 1^{+}} \frac{x^{2}}{x - 1} = \infty

$\underset{x\to 1^+}{\lim}\frac{x^2}{x-1}=\infty$ and

lim_{x \to 1^{-}} \frac{x^{2}}{x - 1} = - \infty

$\underset{x\to 1^-}{\lim}\frac{x^2}{x-1}=−\infty$ .

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) (2 x) - x^{2} (1)}{(x - 1)^{2}} = \frac{x^{2} - 2 x}{(x - 1)^{2}}

$f^{\prime}(x)=\frac{(x-1)(2x)-x^2(1)}{(x-1)^2}=\frac{x^2-2x}{(x-1)^2}$ .

We have $f^{\prime}(x)=0$ 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 $(−\infty ,\infty )$ into the smaller intervals $(−\infty ,0)$ , $(0,1)$ , $(1,2)$ , and $(2,\infty )$ , and choose a test point from each interval to evaluate the sign of $f^{\prime}(x)$ 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 $f^{\prime}(x)=\frac{x^2-2x}{(x-1)^2}=\frac{x(x-2)}{(x-1)^2}$	Conclusion
$(−\infty ,0)$	$x=-1$	$(−)(−)/+=+$	$f$ is increasing.
$(0,1)$	$x=1/2$	$(+)(−)/+=−$	$f$ is decreasing.
$(1,2)$	$x=3/2$	$(+)(−)/+=−$	$f$ is decreasing.
$(2,\infty )$	$x=3$	$(+)(+)/+=+$	$f$ is increasing.

From this table, we see that $f$ has a local maximum at $x=0$ and a local minimum at $x=2$ . The value of $f$ at the local maximum is $f(0)=0$ and the value of $f$ at the local minimum is $f(2)=4$ . Therefore, $(0,0)$ and $(2,4)$ are important points on the graph.

Step 6. Calculate the second derivative:

\begin{array}{ll} f^{''} (x) & = \frac{(x - 1)^{2} (2 x - 2) - (x^{2} - 2 x) (2 (x - 1))}{(x - 1)^{4}} \\ = \frac{(x - 1) [(x - 1) (2 x - 2) - 2 (x^{2} - 2 x)]}{(x - 1)^{4}} \\ = \frac{(x - 1) (2 x - 2) - 2 (x^{2} - 2 x)}{(x - 1)^{3}} \\ = \frac{2 x^{2} - 4 x + 2 - (2 x^{2} - 4 x)}{(x - 1)^{3}} \\ = \frac{2}{(x - 1)^{3}} \end{array}

$\begin{array}{ll} f^{\prime \prime}(x) & =\frac{(x-1)^2(2x-2)-(x^2-2x)(2(x-1))}{(x-1)^4} \\ & =\frac{(x-1)[(x-1)(2x-2)-2(x^2-2x)]}{(x-1)^4} \\ & =\frac{(x-1)(2x-2)-2(x^2-2x)}{(x-1)^3} \\ & =\frac{2x^2-4x+2-(2x^2-4x)}{(x-1)^3} \\ & =\frac{2}{(x-1)^3} \end{array}$

We see that $f^{\prime \prime}(x)$ is never zero or undefined for $x$ in the domain of $f$ . Since $f$ is undefined at $x=1$ , to check concavity we just divide the interval $(−\infty ,\infty )$ into the two smaller intervals $(−\infty ,1)$ and $(1,\infty )$ , and choose a test point from each interval to evaluate the sign of $f^{\prime \prime}(x)$ 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 $f^{\prime \prime}(x)=\frac{2}{(x-1)^3}$	Conclusion
$(−\infty ,1)$	$x=0$	$+/-=−$	$f$ is concave down.
$(1,\infty )$	$x=2$	$+/+=+$	$f$ is concave up.

From the information gathered, we arrive at the following graph for $f.$

The function f(x) = x2/(x − 1) is graphed. It has asymptotes y = x + 1 and x = 1.

Figure 27.

Watch the following video to see the worked solution to Example: Sketching a Rational Function with an Oblique Asymptote.

Closed Captioning and Transcript Information for Video

Try It

Find the oblique asymptote for $f(x)=\dfrac{3x^3-2x+1}{2x^2-4}$

Hint

Use long division of polynomials.

Show Solution

$y=\frac{3}{2}x$

Example: Sketching the Graph of a Function with a Cusp

Sketch a graph of $f(x)=(x-1)^{\frac{2}{3}}$

Show Solution

Step 1. Since the cube-root 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 $\underset{x\to \pm \infty }{\lim}(x-1)^{2/3}=\infty$ , the function continues to grow without bound as $x\to \infty$ and $x\to −\infty$ .

Step 4: The function has no vertical asymptotes.

Step 5: To determine where $f$ is increasing or decreasing, calculate $f^{\prime}$ . We find

f^{'} (x) = \frac{2}{3} (x - 1)^{- 1 / 3} = \frac{2}{3 (x - 1)^{1 / 3}}

$f^{\prime}(x)=\frac{2}{3}(x-1)^{-1/3}=\frac{2}{3(x-1)^{1/3}}$ .

This function is not zero anywhere, but it is undefined when $x=1$ . Therefore, the only critical point is $x=1$ . Divide the interval $(−\infty ,\infty )$ into the smaller intervals $(−\infty ,1)$ and $(1,\infty )$ , and choose test points in each of these intervals to determine the sign of $f^{\prime}(x)$ 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 $f^{\prime}(x)=\frac{2}{3(x-1)^{1/3}}$	Conclusion
$(−\infty ,1)$	$x=0$	$+/-=−$	$f$ is decreasing.
$(1,\infty )$	$x=2$	$+/+=+$	$f$ is increasing.

We conclude that $f$ has 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 $f^{\prime}(1)$ is undefined, so to determine the behavior of the function at this critical point, we need to examine $\underset{x\to 1}{\lim}f^{\prime}(x)$ . Looking at the one-sided limits, we have

lim_{x \to 1^{+}} \frac{2}{3 (x - 1)^{1 / 3}} = \infty

$\underset{x\to 1^+}{\lim}\frac{2}{3(x-1)^{1/3}}=\infty$ and

lim_{x \to 1^{-}} \frac{2}{3 (x - 1)^{1 / 3}} = - \infty

$\underset{x\to 1^-}{\lim}\frac{2}{3(x-1)^{1/3}}=−\infty$ .

Therefore, $f$ has a cusp at $x=1$ .

Step 6: To determine concavity, we calculate the second derivative of $f$ :

f^{''} (x) = - \frac{2}{9} (x - 1)^{- 4 / 3} = \frac{- 2}{9 (x - 1)^{4 / 3}}

$f^{\prime \prime}(x)=-\frac{2}{9}(x-1)^{-4/3}=\frac{-2}{9(x-1)^{4/3}}$ .

We find that $f^{\prime \prime}(x)$ is defined for all $x$ , but is undefined when $x=1$ . Therefore, divide the interval $(−\infty ,\infty )$ into the smaller intervals $(−\infty ,1)$ and $(1,\infty )$ , and choose test points to evaluate the sign of $f^{\prime \prime}(x)$ 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 $f^{\prime \prime}(x)=\frac{-2}{9(x-1)^{4/3}}$	Conclusion
$(−\infty ,1)$	$x=0$	$−/+=−$	$f$ is concave down.
$(1,\infty )$	$x=2$	$−/+=−$	$f$ is concave down.

From this table, we conclude that $f$ is concave down everywhere. Combining all of this information, we arrive at the following graph for $f$ .

The function f(x) = (x − 1)2/3 is graphed. It touches the x axis at x = 1, where it comes to something of a sharp point and then flairs out on either side.

Figure 28.

Watch the following video to see the worked solution to Example: Sketching the Graph of a Function with a Cusp.

Closed Captioning and Transcript Information for Video

Try It

Consider the function $f(x)=5-x^{\frac{2}{3}}$ . 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, $f^{\prime}(a)$ is undefined, one of the one-sided limits as $x\to a$ of $f^{\prime}(x)$ is $+\infty$ , and the other one-sided limit is $−\infty$ .

Show Solution

The function $f$ has a cusp at $(0,5)$ : $\underset{x\to 0^-}{\lim}f^{\prime}(x)=\infty$ , $\underset{x\to 0^+}{\lim}f^{\prime}(x)=−\infty$ . For end behavior, $\underset{x\to \pm \infty }{\lim}f(x)=−\infty$ .

Module 4: Applications of Derivatives