3.3 Graphs of Rational Functions

Vanessa Coffelt

3.3 Graphs of Rational Functions

In Section 3.2, we learned about the types of behaviors to expect from graphs of rational functions: vertical asymptotes, holes in graph, horizontal and slant asymptotes. Moreover, Theorems 3.2, 3.3, and 3.4 tell us exactly when and where these behaviors will occur. We used rough sketches extensively in the last section to help us verify results. In this section, we delve more deeply into graphing rational functions with the goal of sketching relatively accurate graphs without the aid of a graphing utility. Your instructor will ultimately communicate the level of detail expected out of you when it comes to producing graphs of rational functions; what we provide here is an attempt to glean as much information about the graph as possible given the analytical tools at our disposal.

One of the standard tools we will use is the sign diagram which was first introduced in Section 1.5.1, and then revisited in Section 2.1. In these sections, to construct a sign diagram for a function $f$ , we first found the zeros of $f$ . The zeros broke the domain of $f$ into a series of intervals. We determined the sign of $f(x)$ over the entire interval by finding the sign of $f(x)$ for just one test value per interval. The theorem that justified this approach was the Intermediate Value Theorem, which says that continuous functions cannot change their sign between two values unless there is a zero between those two values.

This strategy fails in general with rational functions. Indeed, the very first function we studied in Section 3.2, $r(x) = \frac{1}{x}$ changes sign between $x=-1$ and $x=1$ , but there is no zero between these two values – instead, the graph changes sign across a vertical asymptote. We could also well imagine the graph of a rational function having a hole where an $x$ -intercept should be.^[1] It turns out that with Calculus we can show rational functions are continuous on their domains. What this means for us is when we construct sign diagrams, we need to choose test values on either side of values excluded from the domain in addition to checking around zeros.^[2]

Steps for Constructing a Sign Diagram for a Rational Function

Suppose $f$ is a rational function.

Determine the domain of $f$ .
Identify holes and vertical asymptotes for the graph of $f$ . Indicate any holes in the graph by placing them on the number line with a capital H above them and any vertical asymptotes with a vertical dashed line above them.
Determine the zeros of $f$ and place them on the number line with the number $0$ above them.
Choose a test value in each of the intervals determined in steps 1 and 2.
Determine and record the sign of $f(x)$ for each test value in step 3.

We now present our procedure for graphing rational functions and apply it to a few exhaustive examples. Please note that we decrease the amount of detail given in the explanations as we move through the examples. The reader should be able to fill in any details in those steps which we have abbreviated.

Steps for Graphing Rational Functions

Suppose $r$ is a rational function.

Determine the domain of $r$ .
Reduce $r(x)$ to lowest terms, if applicable.
Determine the location of any vertical asymptotes or holes in the graph, if they exist.
Identify the axis intercepts, if they exist.
Analyze the end behavior of $r$ . Find the horizontal or slant asymptote, if one exists.
Use a sign diagram and plot additional points, as needed, to sketch the graph.^[3]

Example 3.3.1

Sketch a detailed graph of $f(x) = \dfrac{3x}{x^2-4}$ .

Solution:

We follow the six step procedure outlined above.

Determine the domain.
To find the domain, we first find the excluded values. To that end, we solve $x^2 - 4 = 0$ and find $x = \pm 2$ .
Our domain is $\{ x \in \mathbb{R} \, | \, x \neq \pm 2\}$ , or, using interval notation, $(-\infty, -2) \cup (-2,2) \cup (2,\infty)$ .
Reduce $f(x)$ to lowest terms.
We check if $f(x)$ is in lowest terms by factoring:
$f(x) = \frac{3x}{(x-2)(x+2)}$

There are no common factors which means $f(x)$ is already in lowest terms.
Determine the location of any vertical asymptotes or holes in the graph, if they exist.
Per Theorem 3.2, vertical asymptotes and holes in the graph come from values excluded from the domain of .
The two numbers excluded from the domain of are and , didn’t reduce, thus Theorem 3.2 tells us and are vertical asymptotes of the graph.
We can actually go a step further at this point and determine exactly how the graph approaches the asymptote near each of these values. Though not absolutely necessary,^[4] it is good practice for those heading off to Calculus. For the discussion that follows, we use the factored form of .
- The behavior of $y=f(x)$ as $x \rightarrow -2$ : Suppose $x \rightarrow -2^{-}$ . If we were to build a table of values, we’d use $x$ -values a little less than $-2$ , say $-2.1$ , $-2.01$ and $-2.001$ . While there is no harm in actually building a table like we did in Section 3.2, we want to develop a `number sense’ here. Let’s think about each factor in the formula of $f(x)$ as we imagine substituting a number like $x=-2.000001$ into $f(x)$ . The quantity $3x$ would be very close to $-6$ , the quantity $(x-2)$ would be very close to $-4$ , and the factor $(x+2)$ would be very close to $0$ . More specifically, $(x+2)$ would be a little less than $0$ , in this case, $-0.000001.$ We will call such a number a `very small $(-)$ ‘, `very small’ meaning close to zero in absolute value. So, mentally, as $x \rightarrow -2^{-}$ ,
  $\begin{array}{rcl} f(x) &=& \dfrac{3x}{(x-2)(x+2)} \\[8pt] &\approx& \dfrac{-6}{(-4)\left( \text{very small } (-)\right)} \\[8pt] &=& \dfrac{3}{2 \left( \text{very small } (-)\right)} \end{array}$
  
  Now, the closer $x$ gets to $-2$ , the smaller $(x+2)$ will become, so even though we are multiplying our `very small $(-)$ ‘ by $2$ , the denominator will continue to get smaller and smaller, and remain negative. The result is a fraction whose numerator is positive, but whose denominator is very small and negative. Mentally,
  
  $\begin{array}{rcl} f(x) &\approx& \dfrac{3}{2 \left( \text{very small } (-)\right)} \\[8pt] &\approx& \dfrac{3}{\text{very small } (-)} \\[8pt] &\approx& \text{very big } (-) \end{array}$
  
  The term `very big $(-)$ ‘ means a number with a large absolute value which is negative.^[5]
  What all of this means is that as $x \rightarrow -2^{-}$ , $f(x) \rightarrow -\infty$ .
  Now suppose we wanted to determine the behavior of $f(x)$ as $x \rightarrow -2^{+}$ . If we imagine substituting something a little larger than $-2$ in for $x$ , say $-1.999999$ , we mentally estimate
  
  $\begin{array}{rcl} f(x) &\approx& \dfrac{-6}{(-4)\left( \text{very small } (+) \right)} \\[8pt] &=& \dfrac{3}{2 \left( \text{very small } (+) \right)} \\[8pt] &\approx& \dfrac{3}{\text{very small } (+)} \\[8pt] &\approx & \text{very big } (+) \end{array}$
  
  We conclude that as $x \rightarrow -2^{+}$ , $f(x) \rightarrow \infty$ .
- The behavior of $y=f(x)$ as $x \rightarrow 2$ : Consider $x \rightarrow 2^{-}$ . We imagine substituting $x = 1.999999$ . Approximating $f(x)$ as we did above, we get
  $\begin{array}{rcl} f(x) & \approx & \dfrac{6}{\left( \text{very small } (-) \right)(4)} \\[8pt] &=& \dfrac{3}{2 \left( \text{very small } (-) \right)} \\[8pt] &\approx & \dfrac{3}{\text{very small } (-)} \\[8pt] &\approx & \text{very big } (-) \end{array}$
  
  We conclude that as $x \rightarrow 2^{-}$ , $f(x) \rightarrow -\infty$ .
  Similarly, as $x \rightarrow 2^{+}$ , we imagine substituting $x = 2.000001$ to get
  
  $\begin{array}{rcl} f(x) & \approx & \frac{3}{\text{\scriptsize very small } (+)} \\[8pt] &\approx & \text{very big } (+) \end{array}$
  
  So as $x \rightarrow 2^{+}, f(x) \rightarrow \infty$ .
We interpret this graphically below.
Identify the axis intercepts, if they exist.
To find the $x$ -intercepts of the graph, we set $y=f(x) = 0$ . Solving $\frac{3x}{(x-2)(x+2)} = 0$ results in $3x=0,$ thus $x=0$ . Because $x=0$ is in our domain, $(0,0)$ is the $x$ -intercept.
This is also the $y$ -intercept,^[6] as we can quickly verify with $f(0) = \frac{3(0)}{0^2-4} = 0$ .
Analyze the end behavior of . Find the horizontal or slant asymptote, if one exists.Next, we determine the end behavior of the graph of . The degree of the numerator is , and the degree of the denominator is , and so Theorem 3.3 tells us that is the horizontal asymptote. As with the vertical asymptotes, we can glean more detailed information using `number sense’. For the discussion below, we use the formula .
- The behavior of $y=f(x)$ as $x \rightarrow -\infty$ : If we were to make a table of values to discuss the behavior of $f$ as $x \rightarrow -\infty$ , we would substitute very `large’ negative numbers in for $x$ , say for example, $x = -1 \text{ billion}$ . The numerator $3x$ would then be $-3 \, \text{billion}$ , whereas the denominator $x^2-4$ would be $(-1 \text{ billion})^2 - 4$ , which is pretty much the same as $1(\text{billion})^2$ . Hence,
  $\begin{array}{rcl} f\left( -1 \text{ billion}\right) &\approx& \dfrac{-3 \text{ billion}}{1(\text{billion})^2} \\[8pt] &\approx& - \dfrac{3}{\text{billion}} \\[8pt] &\approx& \text{very small } (-) \end{array}$
  
  Notice that if we substituted in $x = -1 \text{ trillion}$ , essentially the same kind of division would occur, and we would be left with an even `smaller’ negative number. This not only confirms the fact that as $x \rightarrow -\infty$ , $f(x) \rightarrow 0$ , it tells us that $f(x) \rightarrow 0^{-}$ .
  In other words, the graph of $y=f(x)$ is a little bit below the $x$ -axis as we move to the far left.
- The behavior of $y=f(x)$ as $x \rightarrow \infty$ : On the flip side, we can imagine substituting very large positive numbers in for $x$ and looking at the behavior of $f(x)$ . For example, let $x = 1 \text{ billion}$ . Proceeding as before, we get
  $\begin{array}{rcl} f\left(\text{1 billion}\right) &\approx& \dfrac{3 \, \text{billion}}{1(\text{billion})^2} \\[8pt] &\approx & \dfrac{3}{\text{billion}} \\[8pt] &\approx& \text{very small } (+) \end{array}$
  
  The larger the number we put in, the smaller the positive number we would get out. In other words, as $x \rightarrow \infty$ , $f(x) \rightarrow 0^{+}$ , so the graph of $y=f(x)$ is a little bit above the $x$ -axis as we look toward the far right.
We interpret these findings graphically below.
Use a sign diagram and plot additional points, as needed, to sketch the graph.
Lastly, we construct a sign diagram for $f(x)$ . The $x$ -values excluded from the domain of $f$ are $x = \pm 2$ , and the only zero of $f$ is $x=0$ .
Displaying these appropriately on the number line gives us four test intervals, and we choose the test values^[7] $x=-3$ , $x=-1$ , $x=1$ and $x=3$ . We find $f(-3)$ is $(-)$ , $f(-1)$ is $(+)$ , $f(1)$ is $(-)$ and $f(3)$ is $(+)$ .
As we begin our sketch, it certainly appears as if the graph could be symmetric about the origin. Taking a moment to check for symmetry, we find $f(-x) = \frac{3(-x)}{(-x)^2-4} = -\frac{3x}{x^2-4} = -f(x)$ . Hence, $f$ is odd and the graph of $y = f(x)$ is symmetric about the origin.
Putting all of our work together, we get the graph below.

Something important to note about the above example is that while $y=0$ is the horizontal asymptote, the graph of $f$ actually crosses the $x$ -axis at $(0,0)$ . The myth that graphs of rational functions can’t cross their horizontal asymptotes is completely false,^[8] as we shall see again in our next example.

Example 3.3.2

Sketch a detailed graph of $g(t) = \dfrac{2t^2-3t-5}{t^2-t-6}$ .

Solution:

Determine the domain.
To find the values excluded from the domain of $g$ , we solve $t^2-t-6 = 0$ and find $t = -2$ and $t=3$ .
Hence, our domain is $\{ t \in \mathbb{R} \, | \, t \neq -2 3 \}$ , or using interval notation: $(-\infty, -2) \cup (-2,3) \cup (3,\infty)$ .
Reduce $f(x)$ to lowest terms.
To check if $g(t)$ is in lowest terms, we factor:
$g(t) = \frac{(2t-5)(t+1)}{(t-3)(t+2)}$

There is no cancellation, so $g(t)$ is in lowest terms.
Determine the location of any vertical asymptotes or holes in the graph, if they exist.
Due to the fact that was given to us in lowest terms, we have, once again by Theorem 3.2 vertical asymptotes and .
Keeping in mind , we proceed to our analysis near each of these values.
- The behavior of $y=g(t)$ as $t \rightarrow -2$ :} As $t \rightarrow -2^{-}$ , we imagine substituting a number a little bit less than $-2$ . We have
  $\begin{array}{rcl} g(t) &\approx& \frac{(-9)(-1)}{(-5)(\text{very small } (-))} \\[8pt] &\approx& \frac{9}{\text{very small } (+)} \\[8pt] &\approx& \text{very big } (+) \end{array}$
  
  so as $t \rightarrow -2^{-}$ , $g(t) \rightarrow \infty$ .
  On the flip side, as $t \rightarrow -2^{+}$ , we get
  
  $\begin{array}{rcl} g(t) &\approx& \frac{9}{\text{ very small } (-)} \\[8pt] &\approx & \text{very big } (-) \end{array}$
  
  so as $t \rightarrow -2^{+}$ , $g(t) \rightarrow -\infty$ .
- The behavior of $y=g(t)$ as $t \rightarrow 3$ : As $t \rightarrow 3^{-}$ , we imagine plugging in a number just shy of $3$ . We have
  $\begin{array}{rcl} g(t) &\approx & \frac{(1)(4)}{(\text{ very small } (-)) (5)} \\[8pt] &\approx & \frac{4}{\text{very small } (-)} \\[8pt] &\approx & \text{very big } (-) \end{array}$
  
  Hence, as $t \rightarrow 3^{-}$ , $g(t) \rightarrow -\infty$ .
  As $t \rightarrow 3^{+}$ , we get
  
  $\begin{array}{rcl} g(t) &\approx& \frac{4}{\text{ very small } (+)} \\[8pt] &\approx& \text{very big } (+) \end{array}$
  
  so as $t \rightarrow 3^{+}$ , $g(t) \rightarrow \infty$ .
We interpret this analysis graphically below.
Identify the axis intercepts, if they exist.
To find the $t$ -intercepts we set $y = g(t) = 0$ . Using the factored form of $g(t)$ above, we find the zeros to be the solutions of $(2t-5)(t+1)=0$ . We obtain $t = \frac{5}{2}$ and $t=-1$ . Both of these numbers are in the domain of $g$ , therefore we have two $t$ -intercepts, $\left( \frac{5}{2},0\right)$ and $(-1,0)$ .
To find the $y$ -intercept, we find $y = g(0) = \frac{5}{6}$ , so our $y$ -intercept is $\left(0, \frac{5}{6}\right)$ .
Analyze the end behavior of . Find the horizontal or slant asymptote, if one exists.
As the degrees of the numerator and denominator of are the same, we know from Theorem 3.3 that we can find the horizontal asymptote of the graph of by taking the ratio of the leading terms coefficients, .
However, if we take the time to do a more detailed analysis, we will be able to reveal some `hidden’ behavior which would be lost otherwise. Using long division, we may rewrite as We focus our attention on the term .
- The behavior of $y=g(t)$ as $t \rightarrow -\infty$ : If imagine substituting $t = -1 \text{ billion}$ into $\frac{t-7}{t^2-t-6}$ , we estimate
  $\begin{array}{rcl} \frac{t-7}{t^2-t-6} &\approx& \frac{-1 \text{\scriptsize billion}}{1 \text{\scriptsize billion}^2} \\[8pt] &=& \frac{-1}{\text{\scriptsize billion}} \\[8pt] &\approx& \text{very small } (-) \end{array}$
  
  ^[9]
  Hence,
  
  $\begin{array}{rcl} g(t) &=& 2 - \frac{t-7}{t^2-t-6} \\[8pt] &\approx & 2 - \text{very small } (-) \\ &=& 2 + \text{very small } (+) \end{array}$
  
  Hence, as $t \rightarrow -\infty$ , the graph is a little bit above the line $y=2$ .
- The behavior of $y=g(t)$ as $t \rightarrow \infty$ . To consider $\frac{t-7}{t^2-t-6}$ as $t \rightarrow \infty$ , we imagine substituting $t= 1 \text{ billion}$ and, going through the usual mental routine, find
  $\frac{t-7}{t^2-t-6} \approx \text{very small } (+)$
  
  Hence, $g(t) \approx 2 - \ \text{very small } (+)$ , so the graph is just below the line $y=2$ as $t \rightarrow \infty$ .

We sketch the end behavior below.

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Use a sign diagram and plot additional points, as needed, to sketch the graph.
Finally we construct our sign diagram. We place an dashed line above $t=-2$ and $t=3$ , and a ` $0$ ‘ above $t = \frac{5}{2}$ and $t=-1$ .Choosing test values in the test intervals gives us $g(t)$ is $(+)$ on the intervals $(-\infty, -2)$ , $\left(-1, \frac{5}{2}\right)$ and $(3, \infty)$ , and $(-)$ on the intervals $(-2,-1)$ and $\left(\frac{5}{2}, 3\right)$ .As we piece together all of the information, it stands to reason the graph must cross the horizontal asymptote at some point after $t=3$ in order for it to approach $y=2$ from underneath.^[10]To find where $y = g(t)$ intersects $y = 2$ , we solve $g(t) = 2 - \frac{t-7}{t^2-t-6} = 2$ and get $t-7= 0$ , or $t=7$ . Note that $t-7$ is the remainder when $2t^2-3t-5$ is divided by $t^2-t-6$ , so it makes sense that for $g(t)$ to equal the quotient $2$ , the remainder from the division must be $0$ . Sure enough, we find $g(7)=2$ .The location of the $t$ -intercepts alone dashes all hope of the function being even or odd (do you see why?) so we skip the symmetry check in this case.

More can be said about the graph of $y = g(t)$ . It stands to reason that $g$ must attain a local minimum at some point past $t=7$ because the graph of $g$ crosses through $y=2$ at $(2,7)$ but approaches $y=2$ from below as $t \rightarrow \infty$ . Calculus verifies a local minimum at $(13, 1.96)$ . We invite the reader to verify this claim using a graphing utility.

Example 3.3.3

Sketch a detailed graph of $h(x) = \dfrac{2x^3+5x^2+4x+1}{x^2+3x+2}$ .

Solution:

Determine the domain.
Solving $x^2+3x+2 = 0$ gives $x = -2$ and $x=-1$ as our excluded values.
Hence, the domain is $\{ x \in \mathbb{R} \, | \, x \neq -1, -2 \}$ or, using interval notation, $(-\infty, -2) \cup (-2, -1) \cup (-1, \infty)$ .
Reduce $h(x)$ to lowest terms.
To reduce $h(x)$ , we need to factor the numerator and denominator. To factor the numerator, we use the techniques^[11] set forth in Section 0.3 and get
$\begin{array}{rcl} h(x) &=& \dfrac{2x^3+5x^2+4x+1}{x^2+3x+2} \\[10pt] &=& \dfrac{(2x+1)(x+1)^2}{(x+2)(x+1)} \\[8pt] &=& \dfrac{ (2x+1) (x+1)^{\cancelto{1}{2}} }{(x+2)\cancel{(x+1)}} \\[10pt] &=& \dfrac{(2x+1)(x+1)}{x+2}\end{array}$
Note we can use this formula for $h(x)$ in our analysis of the graph of $h$ as long as we are not substituting $x=-1$ . To make this exclusion specific, we write
$h(x) = \dfrac{(2x+1)(x+1)}{x+2}, \; \; x \neq -1$
Determine the location of any vertical asymptotes or holes in the graph, if they exist.
From Theorem 3.2, we know that is a zero of the denominator of the reduced form of , thus we have a vertical asymptote there.
As for , the factor was divided from the denominator when we reduced , so there will be a hole when . To find the -coordinate of the hole, we substitute into , per Theorem 3.2 and get . Hence, we have a hole on the -axis at .It should make you uncomfortable plugging into the reduced formula for , especially because we’ve made such a big deal about the stipulation `‘ that goes along with that formula. What we are really doing is taking a Calculus short-cut to the more detailed kind of analysis near which we will show below.
- The behavior of $y=h(x)$ as $x \rightarrow -2$ : As $x \rightarrow -2^{-}$ , we imagine substituting a number a little bit less than $-2$ . We have
  $\begin{array}{rcl} h(x) &\approx & \frac{(-3)(-1)}{(\text{\scriptsize very small } (-))} \\[8pt] &\approx & \frac{3}{(\text{\scriptsize very small } (-))} \\[8pt] &\approx & \text{very big } (-) \end{array}$
  
  thus as $x \rightarrow -2^{-}$ , $h(x) \rightarrow -\infty$ .
  On the other side of $-2$ , as $x \rightarrow -2^{+}$ , we find that
  
  $\begin{array}{rcl} h(x) &\approx & \frac{3}{\text{ very small } (+)} \\[8pt] &\approx & \text{very big } (+) \end{array}$
  
  so $h(x) \rightarrow \infty$ .
- The behavior of $y=h(x)$ as $x \rightarrow -1$ : As $x \rightarrow -1^{-}$ , we imagine plugging in a number a bit less than $x=-1$ . We have
  $\begin{array}{rcl} h(x) &\approx & \frac{(-1)(\text{ very small } (-))}{1} \\[8pt] &=& \text{very small } (+) \end{array}$
  
  Hence, as $x \rightarrow -1^{-}$ , $h(x) \rightarrow 0^{+}$ . This means that as $x \rightarrow -1^{-}$ , the graph is a bit above the point $(-1,0)$ .As $x \rightarrow -1^{+}$ , we get
  
  $\begin{array}{rcl} h(x) &\approx & \frac{(-1)(\text{\scriptsize very small }(+))}{1} \\[8pt] &=& \text{very small } (-) \end{array}$
  
  This gives us that as $x \rightarrow -1^{+}$ , $h(x) \rightarrow 0^{-}$ , so the graph is a little bit lower than $(-1,0)$ here.
We interpret this graphically below.
Identify the axis intercepts, if they exist.
To find the $x$ -intercepts, as usual, we set $h(x) = 0$ and solve. Solving $\frac{(2x+1)(x+1)}{x+2}=0$ yields $x=-\frac{1}{2}$ and $x=-1$ .The latter isn’t in the domain of $h$ , in fact, we know there is a hole at $(-1,0)$ , so we exclude it. Our only $x$ -intercept is $\left(-\frac{1}{2}, 0\right)$ .To find the $y$ -intercept, we set $x=0$ . Due to the fact that $0 \neq -1$ , we can use the reduced formula for $h(x)$ and we get $h(0) = \frac{1}{2}$ for a $y$ -intercept of $\left(0,\frac{1}{2}\right)$ .
Analyze the end behavior of . Find the horizontal or slant asymptote, if one exists.
For end behavior, we note that the degree of the numerator of , , is and the degree of the denominator, , is so by Theorem 3.4, the graph of has a slant asymptote.
For , we are far enough away from to use the reduced formula, , .To perform long division, we multiply out the numerator and get , , and rewrite , .By Theorem 3.4, the slant asymptote is , and to better see how the graph approaches the asymptote, we focus our attention on the term generated from the remainder, .
- The behavior of $y=h(x)$ as $x \rightarrow -\infty$ : Substituting $x = -1 \text{ billion}$ into $\frac{3}{x+2}$ , we get the estimate
  $\begin{array}{rcl} \frac{3}{-1 \text{\scriptsize billion}} &\approx& \text{very small } (-) \end{array}$
  
  Hence,
  
  $\begin{array}{rcl} h(x) &=& 2x-1+\frac{3}{x+2} \\[8pt] &\approx& 2x-1 + \text{very small } (-) \end{array}$
  
  This means the graph of $y=h(x)$ is a little bit below the line $y=2x-1$ as $x \rightarrow -\infty$ .
- The behavior of $y=h(x)$ as $x \rightarrow \infty$ : If $x \rightarrow \infty$ , then
  $\begin{array}{rcl} \frac{3}{x+2} &\approx& \text{very small } (+) \end{array}$
  
  This means
  
  $\begin{array}{rcl} h(x) &\approx& 2x-1 + \text{very small } (+) \end{array}$
  
  or that the graph of $y=h(x)$ is a little bit above the line $y=2x-1$ as $x \rightarrow \infty$ .
We sketch the end behavior below.
Use a sign diagram and plot additional points, as needed, to sketch the graph.
To make our sign diagram, we place a dashed line above $x=-2$ and an `H’ above $x=-1$ and a ` $0$ ‘ above $x=-\frac{1}{2}$ .On our four test intervals, we find $h(x)$ is $(+)$ on $(-2,-1)$ and $\left(-\frac{1}{2}, \infty\right)$ and $h(x)$ is $(-)$ on $(-\infty, -2)$ and $\left(-1,-\frac{1}{2}\right)$ .Putting all of our work together yields the graph below.

To find if the graph of $h$ ever crosses the slant asymptote, we solve $h(x) = 2x-1+\frac{3}{x+2}= 2x-1$ . This results in $\frac{3}{x+2} = 0$ , which has no solution.^[12] Hence, the graph of $h$ never crosses its slant asymptote.^[13]

Our last graphing example is challenging in that our six step process provides us little information to work with.

Example 3.3.4

Sketch a detailed graph of $r(x) = \dfrac{x^4+1}{x^2+1}$ .

Solution:

Determine the domain.
The denominator $x^2+1$ is never zero which means there are no excluded values.The domain is $\mathbb{R}$ , or using interval notation, $(-\infty, \infty)$ .
Reduce $r(x)$ to lowest terms. With no real zeros in the denominator, $x^2+1$ is an irreducible quadratic. Our only hope of reducing $r(x)$ is if $x^2+1$ is a factor of $x^4+1$ . Performing long division gives us
$\frac{x^4+1}{x^2+1} = x^2-1+\frac{2}{x^2+1}$

The remainder is not zero so $r(x)$ is already reduced.
Determine the location of any vertical asymptotes or holes in the graph, if they exist. There are no numbers excluded from the domain of $r$ , so there are no vertical asymptotes or holes in the graph of $r$ .
Identify the axis intercepts, if they exist. To find the $x$ -intercept, we’d set $r(x) = 0$ . As there are no real solutions to $x^4+1=0$ , we have no $x$ -intercepts. $r(0) = 1$ , thus we do get $(0,1)$ as the $y$ -intercept.
Analyze the end behavior of $f(x)$ . Find the horizontal or slant asymptote, if one exists. For end behavior, we note that because the degree of the numerator is exactly two more than the degree of the denominator, neither Theorems 3.3 nor 3.4 apply. We know from our attempt to reduce $r(x)$ that we can rewrite $r(x) = x^2-1+\frac{2}{x^2+1}$ , so we focus our attention on the term corresponding to the remainder, $\frac{2}{x^2+1}$ It should be clear that as $x \rightarrow \pm \infty$ ,
$\frac{2}{x^2+1} \approx \text{very small } (+)$

which means

$r(x) \approx x^2-1 + \text{very small } (+)$

So the graph $y=r(x)$ is a little bit above the graph of the parabola $y=x^2-1$ as $x \rightarrow \pm \infty$ .
Use a sign diagram and plot additional points, as needed, to sketch the graph. There isn’t much work to do for a sign diagram for $r(x)$ , because its domain is all real numbers and it has no zeros. Our sole test interval is $(-\infty, \infty)$ , and we know $r(0) = 1$ , so we conclude $r(x)$ is $(+)$ for all real numbers. We check for symmetry, and find $r(-x) = \frac{(-x)^4+1}{(-x)^2+1} = \frac{x^4+1}{x^2+1} = r(x)$ , so $r$ is even and, hence, the graph is symmetric about the $y$ -axis. It may be tempting at this point to call it quits, reach for a graphing utility, or ask someone who knows Calculus.^[14] It turns out, we can do a little bit better. Recall from Section 2.2.1, that when $|x| <1$ but $x \neq 0$ , $x^4 < x^2$ , hence $x^4+1 < x^2+1$ . This means for $-1<x<0$ and $0<x<1$ , $r(x) = \frac{x^4+1}{x^2+1} < 1$ . Because $r(0) = 1$ , the graph of $y = r(x)$ must fall to either side before heading off to $\infty$ . This means $(0,1)$ is a local maximum and, moreover, there are at least two local minimums, one on either side of $(0,1)$ . We invite the reader to confirm this using a graphing utility.

Our last example turns the tables and invites us to write formulas for rational function given their graphs.

Example 3.3.5

Write formulas for rational functions $r(x)$ and $F(x)$ given their graphs below:

Rendered by QuickLaTeX.com

Solution:

The good news is the graph of $r$ closely resembles the graph of $F$ , so once we know an expression for $r(x)$ , we should be able to modify it to obtain $F(x)$ .

We are told $r$ is a rational function, so we know there are polynomial functions $p$ and $q$ so that $r(x) = \frac{p(x)}{q(x)}$ . We can factor $p(x)$ and $q(x)$ completely in terms of their leading coefficients and their zeros. For simplicity’s sake, we assume neither $p$ nor $q$ has any non-real zeros.

We focus our attention first on finding an expression for $p(x)$ . When finding the $x$ -intercepts, we look for the zeros of $r$ by solving $r(x) = \frac{p(x)}{q(x)} = 0$ . This equation quickly reduces to solving $p(x) =0$ .

As $\left(\frac{5}{3}, 0 \right)$ is an $x$ -intercept of the graph, we know $x = \frac{5}{3}$ is a zero of $r$ , and, hence, a zero of $p$ . Due to the fact that we are shown no other $x$ -intercepts, we assume $r$ , hence $p$ , have no other real zeros (and no non-real zeros by our assumption.) Definition 2.9 gives $p(x) = a\left(x - \frac{5}{3}\right)^m$ where $a$ is the leading coefficient of $p(x)$ and $m$ is the multiplicity of the zero $x = \frac{5}{3}$ . The graph of $y = r(x)$ crosses through the $x$ -axis in what appears to be a fairly linear fashion at $\left(\frac{5}{3}, 0 \right)$ , so it seems reasonable to set $m=1$ . Hence, $p(x) = a \left(x - \frac{5}{3}\right)$ .

Next, we focus our attention on finding $q(x)$ . Theorem 3.2 says $x=1$ comes from a factor of $(x-1)$ in the denominator of $r(x)$ . This means $(x-1)$ is a factor of $q(x)$ . Because there are no other vertical asymptotes or holes in the graph, $x=1$ is the only real zero, hence (per our assumption) the only zero of $q$ . At this point, we have $q(x) = b(x-1)^m$ where $b$ is the leading coefficient of $q(x)$ and $m$ is the multiplicity of the zero $x=1$ . As the graph of $r$ has the horizontal asymptote $y = 3$ ,Theorem 3.4 tells us two things: first, degree of $q$ must match the degree of $p$ ; second, the ratio $\frac{a}{b} = 3$ . Hence, the degree of $q$ is $1$ so that:

$\begin{array}{rcl} r(x) & = & \dfrac{a \left(x - \frac{5}{3}\right)}{b(x-1)} \\ & = & \dfrac{a}{b} \left(\dfrac{x - \frac{5}{3}}{x-1}\right) \\ & = & 3 \left(\dfrac{x - \frac{5}{3}}{x-1}\right) \\ & = & \dfrac{3x-5}{x-1}. \end{array}$

We have yet to use the $y$ -intercept, $(0,5)$ . In this case, we use it as a partial check: $r(0) = \frac{3(0)-5}{0-1} = 5$ , as required. We can sketch $y=r(x)$ by hand, to give a better check of our work.

Now it is time to find a formula for $F(x)$ . The graphs of $r$ and $F$ look identical except the graph of $F$ has a hole at $\left(\frac{5}{3}, 0 \right)$ instead of an $x$ -intercept. Theorem 3.2 tells us this happens because a factor of $\left(x - \frac{5}{3} \right)$ divides out from the denominator when the formula for $F(x)$ is reduced. Hence, we reverse this process and multiply the numerator and denominator of our expression for $r(x)$ by $\left(x - \frac{5}{3} \right)$ :

$\begin{array}{rclr} F(x) & = & r(x) \cdot \dfrac{\left(x - \frac{5}{3} \right)}{\left(x - \frac{5}{3} \right)} & \\[8pt] & = & \dfrac{3x-5}{x-1} \cdot \dfrac{\left(x - \frac{5}{3} \right)}{\left(x - \frac{5}{3} \right)} & \\[10pt] & = & \dfrac{3x^2-10x+\frac{25}{3}}{x^2 - \frac{8}{3} x + \frac{5}{3}} & \text{expand}\\[10pt] & = & \dfrac{9x^2-30x+25}{3x^2-8x+5} & \text{multiply by } 1 = \frac{3}{3} \text{ to reduce complex fractions.}\\ \end{array}$

Again, we can check our answer by applying the six step method to this function or, for a quick verification, we can use a graphing utility.^[15]

Another way to approach Example 3.3.5 is to take a cue from Theorem 3.1. The graph of $y=r(x)$ certainly appears to be the result of moving around the graph of $f(x) = \frac{1}{x}$ . To that end, suppose $r(x) = \frac{a}{x-k} + k$ . As the vertical asymptote is $x=1$ and the horizontal asymptote is $y=3$ , we get $h=1$ and $k=3$ . At this point, we have $r(x) = \frac{a}{x-1}+3$ . We can determine $a$ by using the $y$ -intercept, $(0,5)$ : $r(0) =5$ gives us $-a+3 = 5$ so $a = -2$ . Hence, $r(x) = \frac{-2}{x-1}+3$ . At this point we could check the $x$ -intercept $\left(\frac{5}{3}, 0 \right)$ is on the graph, check our answer using a graphing utility, or even better, get common denominators and write $r(x)$ as a single rational expression to compare with our answer in the above example.

As usual, the authors offer no apologies for what may be construed as `pedantry’ in this section. We feel that the detail presented in this section is necessary to obtain a firm grasp of the concepts presented here and it also serves as an introduction to the methods employed in Calculus. In the end, your instructor will decide how much, if any, of the kinds of details presented here are `mission critical’ to your understanding of Precalculus. Without further delay, we present you with this section’s Exercises.

3.3.1 Section Exercises

In Exercises 1 – 16, use the six-step procedure to graph the rational function. Be sure to draw any asymptotes as dashed lines.

$f(x) = \dfrac{4}{x + 2}$
$f(x) = 5x(6-2x)^{-1}$
$g(t) = t^{-2}$
$g(t) = \dfrac{1}{t^{2} + t - 12}$
$r(z) = \dfrac{2z - 1}{-2z^{2} - 5z + 3}$
$r(z) = \dfrac{z}{z^{2} + z - 12}$
$f(x) = 4x(x^2+4)^{-1}$
$f(x) = 4x(x^2-4)^{-1}$
$g(t) = \dfrac{t^2-t-12}{t^2+t-6}$
$g(t) = 3- \dfrac{5t-25}{t^2-9}$
$r(z) = \dfrac{z^2-z-6}{z+1}$
$r(z) =-z-2+\dfrac{6}{3-z}$
$f(x) = \dfrac{x^3+2x^2+x}{x^2-x-2}$
$f(x) = \dfrac{5x}{9-x^2} - x$
$g(t) =\dfrac{1}{2}t-1 + \dfrac{t+1}{t^2+1}$
$g(t) = \dfrac{t^2-2t+1}{t^{3}+t^{2}-2t}$

In Exercises 17 – 20, write a possible formula for the function whose graph is given.

Section 3.3 Exercise Answers can be found in the Appendix … Coming soon

Take $f(x) = \frac{x^2}{x}$ , for instance. ↵
As excluded values are zeros of the denominator, we can think of this as really just generalizing what we already do. ↵
It doesn't hurt to check for symmetry at this point, if convenient. ↵
The sign diagram in step 6 will also determine the behavior near the vertical asymptotes. ↵
The actual retail value of $f(-2.000001)$ is approximately $-1,500,000$ . ↵
Per Exercise 77, functions can have at most one $y$ -intercept. $(0,0)$ is on the graph, thus it is the $y$ -intercept. ↵
In this particular case, we don't need test values because our analysis of the behavior of $f$ near the vertical asymptotes and our end behavior analysis have given us the signs on each of the test intervals. In general, however, this won't always be the case, so for demonstration purposes, we continue with our usual construction. ↵
That's why we called it a MYTH! ↵
We are once again using the fact that for polynomials, end behavior is determined by the leading term, so in the denominator, the $t^2$ term dominates the $t$ and constant terms. ↵
This subtlety would have been missed had we skipped the long division and subsequent end behavior analysis. ↵
Bet you never thought you'd never see that stuff again before the Final Exam! ↵
Alternatively, the remainder after the long division was $r=3$ which is never $0$ . ↵
But rest assured, some graphs do! ↵
This is exactly what the authors did in the Third Edition. Special thanks go to Erik Boczko from Ohio University for showing us that, in fact, we could do more with this example algebraically. ↵
Be warned, however, a graphing utility may not show the hole at $\left(\frac{5}{3}, 0 \right)$ . ↵