4.5: Describing Relationships with Rational Functions
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In this section, we will explore describing relationships with rational functions. A rational functions is a ratio of two polynomials such as \[f(x)=\dfrac{x^2-4x+1}{x^3+5x^2+3} \nonumber \] We will approach rational functions in a way similar to how we approached describing relationships with polynomial functions. We will use the key features of the rational function such as the zeros and the end behavior to graph it. We need to identify other key features of the rational function that are needed to draw its graph and describe the relationship between quantities.
Focus: Determining The End Behavior of Rational Functions
For the end behavior, we need to determine what value(s) \(y=f(x)\) approaches as \(x\) approaches \( \pm \infty \). Let's explore a few different cases.
Case #1: A power in the numerator
A power in the numerator such as \(y=\dfrac{-2x^3}{1}\) simplifies to a single power of \(x\), \(y=-2x^3\). This is an odd power whose coefficient \(a=-2\) causes a reflections about the x-axis. The graph opens down on the right end and up on the left end as shown in the graph below. The end behavior can be describes as \(x \to \infty\), \(y \to -\infty\) and as \(x \to -\infty\), \(y \to \infty\).
Case #2: A power in the denominator
Let's examine the end behavior of a power in the denominator such as \(f(x)=\dfrac{1}{x} \) in the table below.
| x | 1 | 10 | 100 | 1000 | \( \to \infty\) |
|---|---|---|---|---|---|
| \(y=f(x)=\dfrac{1}{x}\) | \(\dfrac{1}{1}=1\) | \(\dfrac{1}{10}=0.1\) | \(\dfrac{1}{100}=0.01\) | \(\dfrac{1}{1000}=0.001\) | \( \to 0\) |
As \(x\) gets larger and larger approaching positive infinity, \(y=f(x) \) approaches zero.
| x | -1 | -10 | -100 | -1000 | \( \to -\infty\) |
|---|---|---|---|---|---|
| \(y=f(x)=\dfrac{1}{x}\) | \(\dfrac{1}{-1}=-1\) | \(\dfrac{1}{-10}=-0.1\) | \(\dfrac{1}{-100}=-0.01\) | \(\dfrac{1}{-1000}=-0.001\) | \( \to 0\) |
As \(x\) approaches negative infinity, \(y=f(x) \) approaches zero. The end behavior of \(f(x)=\dfrac{1}{x} \) is shown on the graph below
For a larger power in the denominator, such as \(f(x)=\dfrac{1}{x^2} \), the end behavior is determined in the table below. Notice, as \(x\) approaches \( \pm \infty\), \(y=f(x) \) approaches zero.
| x | 1 | 10 | 100 | 1000 | \( \to \infty\) |
|---|---|---|---|---|---|
| \(y=f(x)=\dfrac{1}{x^2}\) | \(\dfrac{1}{1}=1\) | \(\dfrac{1}{100}=0.01\) | \(\dfrac{1}{10000}=0.0001\) | \(\dfrac{1}{1000000}=0.000001\) | \( \to 0\) |
| x | -1 | -10 | -100 | -1000 | \( \to -\infty\) |
|---|---|---|---|---|---|
| \(y=f(x)=\dfrac{1}{x^2}\) | \(\dfrac{1}{1}=-1\) | \(\dfrac{1}{100}=0.01\) | \(\dfrac{1}{10000}=0.0001\) | \(\dfrac{1}{1000000}=0.000001\) | \( \to 0\) |
In general, if the denominator of a fraction is getting larger and larger while the numerator is approaching a finite value, the value of the fraction will approach zero. As the whole number power gets larger and larger, the denominator gets larger even faster making the the value of the whole fraction approach zero even faster.
A power in the denominator \(y=\dfrac{1}{x^n} \) has end behavior that is described as \(x \to \pm \infty \), \(y \to 0 \).
Case #3: A rational function with the degree of the numerator > the degree of the denominator
Let \(f(x)=\dfrac{x^4+3x+1}{x^2-3} \). The numerator \(x^4+3x+1\) is a polynomial whose end behavior is the same as the term with the largest power \(x^4\) as we learned in section 4.3. Similarly the denominator \(x^2-3\) is a polynomial whose end behavior is the same as the term with the largest power \(x^2\). Therefore, the rational function \(f(x)=\dfrac{x^4+3x+1}{x^2-3}\) has the same end behavior as the ratio of leading terms \(y=\dfrac{x^4}{x^2} = x^2 \). This is an even power that opens up on both ends (as \(x \to \pm \infty\), \(y \to \infty\)). The end behavior is shown in the graph below.
For the end behavior, we are only talking about the left and right ends of the graph. The functions \(f(x)=\dfrac{x^4+3x+1}{x^2-3} \) and \(y=x^2\) have very different graphs for smaller values of \(x\).
Case #4: A rational function with the degree of the numerator < the degree of the denominator
Let \(f(x)=\dfrac{x^2-3}{x^4+3x+1} \). The polynomial in the numerator, \(x^2-3\), has the same end behavior is the same as the term with the largest power \(x^2\). The polynomial in the denominator\(x^4+3x+1\) has the same end behavior is the same as the term with the largest power \(x^4\). Therefore, the rational function \(f(x)=\dfrac{x^2-3}{x^4+3x+1} \) has the same end behavior as the ratio of leading terms \(y=\dfrac{x^2}{x^4} = \dfrac{1}{x^2} \). This is a power in the denominator that approaches zero on both ends (as \(x \to \pm \infty \), \(y \to 0\) ) as shown in case #2. The possible end behaviors are shown in the graph below.
Notice, the graph of \(f(x)\) can approach \(y=0\) either from above the x-axis or below the x-axis on the right end as \(x \to \infty \). We need more information about the graph to determine from which direction the graph approaches \(y=0\). For example, if we knew that \(y\) was positive on the right end, we could deduce that the graph of \(f(x)\) approaches \(y=0\) from above the x-axis. Similarly, the graph of \(f(x)\) can approach \(y=0\) either from above the x-axis or below the x-axis on the left end as \(x \to -\infty \). If we knew that \(y\) was also positive on the left end, we could deduce that the graph of \(f(x)\) approaches \(y=0\) from above the x-axis.
Case #5: A rational function with the degree of the numerator = the degree of the denominator
Let \(f(x)=\dfrac{2x^3-5x+1}{5x^3+4x} \). The polynomial in the numerator, \(2x^3-5x+1\), has the same end behavior is the same as the term with the largest power \(2x^3\). The polynomial in the denominator \(5x^3+4x\) has the same end behavior is the same as the term with the largest power \(5x^3\). Therefore, the rational function \(f(x)=\dfrac{2x^3-5x+1}{5x^3+4x} \) has the same end behavior as the ratio of leading terms \(y=\dfrac{2x^3}{5x^3} = \dfrac{2}{5} \). On both ends, \(y\) approaches \(y=\dfrac{2}{5} \) (as \(x \to \pm \infty \), \( y \to \dfrac{2}{5}\) ). The possible end behaviors are shown in the graph below (in red and blue).
The end behavior of a rational function \(f(x)=\dfrac{a_{n} x^{n} +\cdots + a_{2} x^{2} +a_{1} x + a_{0}}{b_{m} x^{m} +\cdots + b_{2} x^{2} +b_{1} x + b_{0}}\) is the same as the end behavior of the ratio of leading terms \(y=\dfrac{a_{n}x^n}{b_{n}x^m} \)
For the rational function in case #5, \(f(x)=\dfrac{2x^3-5x+1}{5x^3+4x} \) where as \(x \to \pm \infty \), \(y \to \dfrac{2}{5}\), notice that the graph nears the horizontal line \(y=\dfrac{2}{5} \) shown in green in the graph above. We call this line a horizontal asymptote.
A function \(y=f(x)\) has a horizontal asymptote \(y=a\) if as \(x\) approaches \( \pm \infty \), \(y\) approaches \(a\). The graph of \(y=f(x)\) nears the line \(y=a\) as \( x\) approaches \( \pm \infty \).
A horizontal asymptote represents a type of end behavior. The rational functions in case #2 and #4 have a horizontal asymptote at \(y=0\) since the graphs near the line \(y=0\) since \(y\) approaches 0 as \(x \to \pm \infty \). However, be careful. Not all rational functions have a horizontal asymptote for their end behavior such as the rational functions in cases #1 and #3.
Let \(f(x)=\dfrac{5x^2+2x}{3x^2+3x+1} \). Determine the end behavior. Identify any horizontal asymptotes if they occur. Then sketch a possible graph of the end behavior.
- Answer
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The rational function \(f(x)=\dfrac{5x^2+2x}{3x^2+3x+1} \) has the same end behavior as the ratio of leading terms \9y=\dfrac{5x^2}{3x^2} = \dfrac{5}{3} \). On both ends, \(y\) approaches \(y=\dfrac{5}{3} \) (as \(x \to \pm \infty \), \(y \to \dfrac{5}{3} \) ). A horizontal asymptote occurs at \(y=0\). The possible end behavior is shown in the graph below.
Let \(f(x)=\dfrac{5x+2}{3x^2+3x+1} \). Determine the end behavior. Identify any horizontal asymptotes if they occur. Then sketch a possible graph of the end behavior of the graph.
- Answer
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The rational function \[f(x)=\dfrac{5x+2}{3x^2+3x+1} \) has the same end behavior as the ratio of leading terms \(y=\dfrac{5x}{3x^2} = \dfrac{5}{3x} \). This is a power in the denominator where \(y\) approaches \(y=0 \) as \(x \to \pm \infty \). A horizontal asymptote occurs at \(y=0\). The possible end behavior is shown in a possible graph below.
Write a formula for a rational function with no horizontal asymptote.
- Answer
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Since the end behavior of a rational function is the same as the ratio of leading terms. If this ratio simplifies to a constant such as \(y=k\) or a power of \(x\) in the denominator, a horizontal asymptote would occur. Therefore, the ratio must simplify to a power of \(x\) in the numerator such as \(y=x^3\) which opens up on the right and down on the left. The rational function \(\dfrac{x^5+2x^2+1}{x^2+4x}\) has the same end behavior as the ratio of leading terms \(\dfrac{x^5}{x^2}\) which simplifies to \(y=x^3\)
Focus: Horizontal Asymptotes in Applications
Horizontal Asymptotes describe many natural phenomena. For example, if you bring a hot cup of tea with a temperature of 160 \( \circ \)F into a room with a temperature of 75 \( \circ \)F. The temperature T of the cup of tea will decrease over time to room temperature (as \(t \to \infty, T \to 75\) ). This describes the right end behavior of the graph with a horizontal asymptote occurring at \(T=75\) as shown in the graph below.
Suppose a mass is attached to a spring with a natural length of 10 inches. If the mass is pulled 5 inches then released , the mass will oscillate back and forth. Over time the amplitude of the oscillations will decrease over time to zero. This describes the right end behavior of the length of the spring L ( as \(t \to \infty, L \to 0\) ) with a horizontal asymptote occurring at \(L=10\) as shown in the graph below.
Focus: The Zeros of a Rational Function
When is a fraction zero? Fractions such as \(\dfrac{0}{3} \), \(\dfrac{0}{8} \), and \(\dfrac{0}{2} \) are all equal to zero. A fraction is zero when the numerator is zero and the denominator is not. If the denominator is also zero, such as \(\dfrac{0}{0} \), the fraction is undefined.
The zeros of a rational function occur when the numerator is zero at a defined value.
Find the zeros of the rational functions.
- \(f(x)=\dfrac{x-1}{x-4}\)
- \(g(x)=\dfrac{x^2-x-6}{x^2-4}\)
Solution
- The zeros occur when the numerator \(x-1=0\). Solving for \(x\) we have \(x=1\). Since the denominator \(x-4 = 1-4 = 3 \ne 0 \) at \(x=1\), \(x=1\) is a zero of \(f(x)\).
- Setting the numerator to zero and solving by factoring \[\begin{array}{l} {x^2-x-6=0} \\ {(x-3)(x+2) = 0} \\ {x-3=0 \text{ and } x=+2} \\ {\text{so }x=3 \text{ and } x = -2} \end{array}\nonumber\]We have possible zeros \( x=3 \) and \( x =-2 \). Since the denominator \(x^2-4 = 3^2-4 = 5 \ne 0 \) at \(x=3\), \(x=3\) is a zero of \(g(x)\). However, since the denominator \(x^2-4 = (-2)^2-4 = 0 \) at \(x=-2\), \(x=-2\) is not a zero of \(g(x)\). \(g\) is undefined at \(x=-2\).
Focus: The Undefined Values of Rational Functions
Whereas all polynomials are defined for all real values of \(x\), some rational functions may be undefined at certain values of \(x\) since division by zero is undefined as we saw in example 4.5.1. The rational function \(f(x)=\dfrac{x-1}{x-3} \)is undefined when the denominator \(x-3=0\). This occurs when \(x=3\).
The undefined values of a rational function occur when the denominator is zero.
Identify where the undefined values occur for each function.
- \(f(x)=\dfrac{x+2}{2x-3} \)
- \(g(x)=\dfrac{x-1}{x^2+5x+6} \)
Solution
The undefined values occur when the denominator of a rational function is zero.
- Setting the denominator to zero and solving for \(x\) \[\begin{array}{l} {\text{den.}=2x-3=0} \\ {2x= 3} \\ {x=\dfrac{3}{2}} \end{array}\nonumber\]So \(f(x)\) is undefined at \( x=\dfrac{3}{2} \).
- Setting the denominator to zero and solving by factoring for \(x\) \[\begin{array}{l} {\text{den.}=x^2+5x+6=0} \\ {(x+2)(x+3)=0} \\ {x+2=0 \text{ and } x + 3 = 0} \\ {x=-2 \text{ and } x=-3} \end{array}\nonumber\]So \(g(x)\) is undefined at \( x=-2 \text{ and } x=-3 \).
This brings up an important question, what happens on the graph near an undefined value? Let's explore two different cases. A fraction can be simplified or not. A simplified fraction, such as \( \dfrac{2}{5}\), has no common factor among the numerator and denominator whereas an unsimplified fraction, such as \( \dfrac{4}{6}\), does have a common factor of two among the numerator and denominator.
Case: Simplified Rational Functions
Let's explore behavior of the graph of a simplified rational function \(y=f(x)=\dfrac{1}{x} \) near the undefined value at \(x=0\). Now \(x\) can near zero from values above (or greater than) zero or from values below (or less than) zero. Notice what happens to the value of \(y=\dfrac{1}{x} \) as \(x\) approaches zero from each side in the table below.
| x | 1 | \(\dfrac{1}{10} =0.1\) | \(\dfrac{1}{100}=0.01 \) | \(\dfrac{1}{1000}=0.001 \) | \( \to 0 \) |
|---|---|---|---|---|---|
| \(y=\dfrac{1}{x} \) | 1 | 10 | 100 | 1000 | \( \to \infty \) |
| x | -1 | \(-\dfrac{1}{10} =-0.1\) | \(-\dfrac{1}{100}=-0.01 \) | \(-\dfrac{1}{1000}=-0.001 \) | \( \to 0 \) |
|---|---|---|---|---|---|
| \(y=\dfrac{1}{x} \) | -1 | -10 | -100 | -1000 | \( \to -\infty \) |
Notice as \(x\) approaches zero from above, the value of \(y\) approaches infinity. Also, as \(x\) approaches zero from below, the value of \(y\) approaches negative infinity.
The graph of \(y=f(x)=\dfrac{1}{x} \) near the undefined value at \(x=0\) is shown below.
Notice as \(x\) approaches zero from above, the value of \(y\) approaches infinity, the graph nears the horizontal line \(x=0\) on the top portion of the graph. Also, as \(x\) approaches zero from below, the value of \(y\) approaches negative infinity, the graph also nears the vertical line \(x=0\) on the bottom portion of the graph. The line \(x=0\) is called a vertical asymptote of the function \(f(x)=\dfrac{1}{x} \). The line \(x=0\) is not a part of the graph, but rather describes what the part of the graph nears as \(x\) nears zero. The notation \(x \to 0^+\) to indicate that x is approaching zero from above. The notation \(x \to 0^-\) to indicate that x is approaching zero from below.
The function \(y=f(x)\) has a vertical asymptote at an undefined value \(x=a\) if as \(x\) approaches \(a\) , \(y\) approaches \( \infty\) or \( -\infty \) from above or below \(x=a\). On the graph of \(y=f(x) \) , the graph nears the vertical line \(x=a\).
Let's explore behavior of the graph of another the simplified rational function \(y=f(x)=\dfrac{1}{x^2} \) near the undefined value at \(x=0\) in the tables and graph below.
| x | 1 | \(\dfrac{1}{10} =0.1\) | \(\dfrac{1}{100}=0.01 \) | \(\dfrac{1}{1000}=0.001 \) | \( \to 0 \) |
|---|---|---|---|---|---|
| \(y=\dfrac{1}{x^2} \) | 1 | 100 | 10,000 | 1,000,000 | \( \to \infty \) |
| x | -1 | \(-\dfrac{1}{10} =-0.1\) | \(-\dfrac{1}{100}=-0.01 \) | \(-\dfrac{1}{1000}=-0.001 \) | \( \to 0 \) |
|---|---|---|---|---|---|
| \(y=\dfrac{1}{x^2} \) | -1 | -100 | -10,000 | -1,000,000 | \( \to -\infty \) |
Notice as \(x\) approaches zero from above and below, the value of \(y\) approaches infinity. The graph of \(y=f(x)=\dfrac{1}{x^2} \) near the undefined value at \(x=0\) is shown below.
Notice as \(x\) approaches zero from above and below, the value of \(y\) approaches infinity, the graph nears the horizontal line \(x=0\) on the top portion of the graph. The graph also approaches the vertical line \(x=0\). The line \(x=0\) is a vertical asymptote of \(y=\dfrac{1}{x^2} \). On the graph of the two functions \(y=\dfrac{1}{x} \) and \(y=\dfrac{1}{x^2} \) we see that the graph can either approach \( \infty\) or \( -\infty\) on either side of the vertical asymptote depending on the function itself.
If \(y=f(x)=\dfrac{1}{x^2} \), then \(y=g(x)= f(x-2)=\dfrac{1}{(x-2)^2} \) is a horizontal shift of \(y=\dfrac{1}{x^2} \) two units to the right. This horizontal shift shifts the vertical asymptote two units to the right to \(x=2\) where \(g(x) = \dfrac{1}{(x-2)^2} \) is undefined as shown in the graph below.
In general, \(y=g(x)= f(x-a)=\dfrac{1}{(x-a)^2} \) is a horizontal shift of \(y=\dfrac{1}{x^2} \) \(a\) units to the right. This horizontal shift shifts the vertical asymptote \(a\) units to the right to \(x=a\) where \(g(x) = \dfrac{1}{(x-a)^2} \) is undefined. In general, if \(y=f(x)\) is a simplified rational function and the value of \(x\) is nearing an undefined value at \(x=a\), then the factor \( (x-a) \) in the denominator is nearing zero and so the whole denominator is approaching zero. Furthermore, since the rational function is simplified, \( (x-a) \) is not a factor of the numerator and so the numerator is approaching a non-zero value. Therefore, the value of the rational function \(y=f(x)\) gets larger and larger in size approaching either positive or negative infinity as the value of \(x\) as approaches the undefined value at \(x=a\).
At an undefined value \(x=a\) of a simplified rational function, a vertical asymptote occurs at \(x=a\) on the graph.
Although we can identify that a vertical asymptote occurs at an undefined value of a simplified rational function, we don't have enough information to determine which direction the graph approaches on either side of the vertical asymptote. The function value could approach positive or negative infinity as \(x\) nears the undefined value. For example, \(y=\dfrac{x-3}{x-1} \) is undefined at \(x=1\) since this value of \(x\) makes the denominator zero. If we knew that \(f\) was positive below the undefined value at \(x=1\) on the interval \( (-\infty,1) \), then we could conclude that the value of \(y\) approaches positive infinity as \(x\) nears \(x=1\) from below. If we knew that \(f\) is negative above \(x=1\) on the interval \( (1,3) \), then we could conclude that the value of \(y\) approaches negative infinity as \(x\) nears \(x=1\) from above. With this information, we can graph the function near the vertical asymptote as shown below.
Now we have almost all the tools needed to graph simplified rational functions.
Let \(y=\dfrac{x+2}{x-3} \).
- Identify the undefined value(s). Determine if a vertical asymptote occurs at the undefined value.
- Find the zero(s).
- Describe the end behavior.
- Use part a-c and the sign chart for \(f\) to graph the function.
Solution
- The undefined value(s) occur when the denominator is zero. Setting the denominator to zero and solving \[x-3=0 \text{ and so } x = 3 \nonumber\]So \(f\) is undefined at \(x=3\). Since \(f\) is a simplified rational function, a vertical asymptote occurs at \(x=3\).
- The zeros of a simplified rational function occur when the numerator is zero. Setting the numerator to zero and solving \[x+2=0 \text{ and so } x = -2 \nonumber\]So \(f\) has a zero of \(x=-2\) with multiplicity one.
- The end behavior of \(f\) resembles the end behavior of the ratio of leading terms \(y=\dfrac{x}{x}=1\). So as \(x \to \pm \infty \), \(y \to 1 \). The graph has a horizontal asymptote at \(y=1\).
Let \(y=\dfrac{x-3}{x^2-3x+2} \).
- Identify the undefined value(s). Determine if a vertical asymptote occurs at the undefined value.
- Find the zero(s).
- Describe the end behavior.
- Use part a-c and the sign chart for \(f\) to graph the function.
Solution
- The undefined value(s) occur when the denominator is zero. Setting the the denominator to zero and solving by factoring \[\begin{array}{l} {\text{den.}=x^2-3x+2=0} \\ {(x-2)(x-1)=0} \\ {x-2=0 \text{ and } x - 1 = 0} \\ {x=2 \text{ and } x=1} \end{array} \nonumber\]So \(f\) is undefined at \(x=2\) and \(x = 1\). Since \(f\) is a simplified rational function, a vertical asymptote occurs at \(x=2\) and \(x=1\).
- The zeros of a simplified rational function occur when numerator is zero. Setting the the denominator to zero and solving \[\begin{array}{l} {\text{den.}=x-3=0} \\ {x=3 } \end{array} \nonumber\]So \(f\) has a zero of \(x=3\) with multiplicity one.
- The end behavior of \(f\) resembles the end behavior of the ratio of leading terms \(y=\dfrac{x}{x^2}=\dfrac{1}{x}\). This is a power in the denominator which approaches \(y=0\) as \(x\) approaches \( \pm \infty \). The graph has a horizontal asymptote at \(y=0\)
d. 
In general, we may not know at first glance whether a rational function is simplified or not. It is critical that we verify that the function is simplified before we determine whether a vertical asymptote occurs at an undefined value. We will explore what happens at undefined values of unsimplified rational functions later in this section.
Let \(y=\dfrac{x^2-16}{x+1} \).
- Identify the undefined value(s). Determine if a vertical asymptote occurs at the undefined value.
- Find the zero(s).
- Describe the end behavior.
- Use part a-c and the sign chart for \(f\) to graph the function.
- Answer
-
- \(f\) is undefined at \(x=-1\). A vertical asymptote occurs at \(x=1\) since \(f(x) =\dfrac{(x-4)(x-4)}{x+1} \) is a simplified.
- The zeros are \(x= \pm 4 \) each with multiplicity one.
- The end behavior of \(f\) resembles the end behavior of \(y=x\). This is a odd power which opens up on the right (as \(x \to \infty , y \to \infty \) ) and down on the left (as \(x \to -\infty, y \to -\infty \) ).
- The graph has a horizontal asymptote at \(y=0\).
These examples highlight an important piece of graphing rational functions. We need the sign of each interval along with the zeros, undefined values, and end behavior to graph a simplified rational function.
Focus: Determining The Sign of a Rational Function and More Graphing Rational Functions
Unlike polynomials where we could use the graph to determine the sign of the function on different intervals, we need the sign of the rational function (or other information) in order to graph it. How can we determine the sign of each interval for a rational function? We can use an extension of the test point method we used to determine the sign of polynomials. Notice on the graph of the function \(f(x)=\dfrac{x+2}{x-3} \) in example 4.5.3 (shown below left), the function changes sign only at the zero \(x=-2\) and the undefined value \(x=3\). Whereas the graph of the function \(f(x)=\dfrac{1}{x^2}\) in the introductory example (show below middle), the function did not change sign at the undefined value at \(x=0\). The function was positive above and below \(x=0\). We learned in section 4.3 that a polynomial may or may not change sign at a zero. Consider, the graph of \(y=x^2\) (below right). The function does not change sign at the zero \(x=0\). Notice between consecutive zeros and/or undefined values the graphs do not change sign.
A rational function will only change sign at a zero or an undefined value. Between consecutive zeros and/or undefined values the function does not change signs. It is either strictly increasing or strictly decreasing.
Since the rational function will only change sign at a zero or an undefined value, we can determine the sign of \(f\) in an interval between consecutive zeros and/or and undefined value using a test point. If the value of \(f\) is positive at the test point, the function is positive on that interval since it can only change sign at the zero or the undefined value. If the value of \(f\) is negative at the test point, the function is negative on that interval for the same reason. With this adaption of the test point method, we have a strategy to build a sign chart for \(f\) and determine the sign of \(f\) on each interval.
Let \(f(x)=\dfrac{2x-1}{x-4} \). Determine where \(f\) is positive and where \(f\) is negative.
Solution
First, we need to find the zeros and undefined values where \(f\) can change sign. Setting the denominator to zero and solving to find the undefined value(s) \[\begin{array}{l} {\text{den.}=x-4=0} \\ {x=4} \end{array} \nonumber\]So \(f\) has an undefined value at \(x=4\). Setting the numerator to zero and solving to find the zero(s) \[\begin{array}{l} {\text{den.}=2x-1=0} \\ {2x=1} \\ {x=\dfrac{1}{2}} \end{array} \nonumber\]So \(f\) has a zero at \(x=\dfrac{1}{2}\) since \(f\) is defined here.
The zero and undefined value break up the x-axis in to three intervals. Now we need to evaluate \(f\) at a test point to determine the sign of each interval.
| Interval | \( (-\infty,\dfrac{1}{2}) \) | \( (\dfrac{1}{2},4) \) | \( (4, \infty) \) |
|---|---|---|---|
| Test Point | 0 | 1 | 5 |
| Value | \(\dfrac{2(0)-1}{(0)-4} = \dfrac{1}{4} \) | \(\dfrac{2(1)-1}{(1)-4} = \dfrac{1}{-3} \) | \(\dfrac{2(5)-1}{(5)-4} = \dfrac{9}{1}=9 \) |
| Sign of The Interval | positive | negative | positive |
So \(f\) is positive on \( (-\infty,\dfrac{1}{2}) \cup (4,\infty) \) and negative on the interval \( (\dfrac{1}{2},4) \). Although not necessary in this problem, it may be helpful to display this result with a sign chart for \(f\) when graphing \(f\).
Let \(f(x)=\dfrac{x+1}{x^2-9} \).
- Identify the undefined value(s). Determine if a vertical asymptote occurs at the undefined value.
- Find the zero(s).
- Identify the end behavior.
- Determine the sign of each interval.
- Use parts (a)-(d) to graph \(f\).
Solution
- Setting the denominator to zero and solving to find the undefined value(s) \[\begin{array}{l} {\text{den.}=x^2-9=0} \\ {(x-3)(x+3)=0} \\{x-3=0 \text{ and } x+3 = 0} \\ {x=3 \text{ and } x=-3} \end{array} \nonumber\]Since the rational function \(f(x)=\dfrac{x+1}{(x-3)(x+3)} \) is simplified, vertical asymptotes occur at \(x=3\) and \(x=3\).
- Setting the numerator to zero and solving to find the zero(s) \[\begin{array}{l} {\text{den.}=x+1=0} \\ {x=-1} \end{array} \nonumber\]So the zero occurs at \(x=-1\).
- The end behavior of \(f\) resembles the end behavior of the ratio of leading terms \(y=\dfrac{x}{x^2}=\dfrac{1}{x}\). This is a power in the denominator which approaches \(y=0\) as \(x\) approaches \( \infty \). Therefore a horizontal asymptote occurs at \(y=0\).
- The zero and undefined value break up the x-axis in to three intervals. Evaluate \(f\) at a test point to determine the sign of each interval.
Building a sign chart for f using the test point method Interval \( (-\infty,-3) \) \( (-3,-1) \) \( (-1, 3) \) \( (3, \infty) \) Test Point -4 -2 0 4 Value \(\dfrac{(-4)+1}{(-4)^2-9}= \dfrac{-3}{7}\) \(\dfrac{(-2)+1}{(-2)^2-9}= \dfrac{1}{5}\) \(\dfrac{(0)+1}{(0)^2-9}= \dfrac{1}{-9}\) \(\dfrac{(4)+1}{(4)^2-9}= \dfrac{5}{16}\) Sign of The Interval negative positive negative positive So \(f\) is positive on \((-3,-1) \cup (3,\infty) \) and negative on the interval \((-\infty,-3) \cup (-1, 3) \).
Since we have a sign chart for \(f\) in the example above, the multiplicity of the zero is redundant information in determining whether the graph crosses the x-axis or not at the zero as illustrated in example 4.5.6 above. If a zero has odd multiplicity and the graph crosses the x-axis at the zero, the sign changes at the zero. If a zero has even multiplicity and the graph does not cross the x-axis at the zero, the sign does not change at the zero. The multiplicity of a zero will help confirm what the sign chart for \(f\) already determines for us, whether the graph crosses the x-axis or not at the zero.
Alternatively if we are given the graph of a rational function or information that would help us graph the function, we can use these same concepts to build a formula for the rational function.
Identify a possible formula for a rational function whose graph is
Solution
Since the rational function has a zero of \(x=1\), \(x=1\) is a zero of the numerator and so \( (x-1) \) is a factor of the numerator. Since there is a vertical asymptote at \(x=3\), \(x=3\) is a zero of the denominator and so\( (x-3) \) is a factor of the denominator. However, the rational function \(f(x)=\dfrac{x-1}{x-3} \) has a horizontal asymptote at \(y=1\) since the end behavior of \(y=\dfrac{(x-1)}{x-3} \) has the same end behavior as \(y=\dfrac{x}{x}=1 \). We can use a stretch factor \(a\) to alter end behavior. The end behavior of \(y=\dfrac{a(x-1)}{x-3} \) has the same end behavior as \(y=\dfrac{ax}{x}=a \). Therefore \(a=2\) and so \[f(x)=\dfrac{-2(x-1)}{x-3} \nonumber \]
Let \(f(x)=\dfrac{3x-1}{2x+5} \).
- Identify the undefined value(s). Determine if a vertical asymptote occurs at the undefined value.
- Find the zeros
- Identify the end behavior
- Determine the sign of each interval
- Use parts (a)-(d) to graph \(f\)
- Answer
-
- The undefined value occurs at \( x = \dfrac{-5}{2} \) Since the rational function is simplified, a vertical asymptote occurs at \(\dfrac{-5}{2}\).
- \(x=\dfrac{1}{3} \) is the zero.
- The end behavior of \(f\) resembles the end behavior of \(y=\dfrac{3x}{2x}=\dfrac{3}{2}\). Therefore \(y \to \dfrac{3}{2}\) as \(x \to \infty \). So a horizontal asymptote occurs at \(\dfrac{3}{2}\).
- Using the test point method, \(f\) is positive on \( (-\infty,\dfrac{-5}{2}) \cup (\dfrac{1}{3}, \infty) \) and negative on the interval \((\dfrac{-5}{2},\dfrac{1}{3}) \)
Building a sign chart for f using the test point method Interval \( (-\infty,\dfrac{-5}{2}) \) \( (\dfrac{-5}{2},\dfrac{1}{3}) \) \( (\dfrac{1}{3}, \infty) \) Test Point -3 0 1 Value 10 \(\dfrac{-1}{5}\) \(\dfrac{2}{9} \) Sign of The Interval positive negative positive 
Identify a possible formula for a rational function whose graph is
- Answer
-
\(f(x)=\dfrac{1.5(x-2)}{(x+3)(x+1)} \). Notice that in this case, we had to use the y-intercept to find the stretch factor \(a\) in \(f(x)=\dfrac{a(x-2)}{(x+3)(x+1)} \) since the end behavior is unchanged by a stretch factor. The end behavior resembles \(f(x)=\dfrac{ax}{x^2}=\dfrac{a}{x} \). This is a power in the denominator where \(y \to 0 \) as \(x \to \infty \) no matter what value we have for \(a\).
Focus: Applications of Vertical Asymptotes
Vertical Asymptotes describe many phenomena in applications in economics, physics, and engineering. Let's explore one.
Suppose the cost of removing \(p\) percent of a pollutant from 1000 gallons of water is modeled by the function \(C=f(p)=\dfrac{1.125p+1250}{100-p} \) where 0<\(p\)<100. Identify the undefined value(s). Interpret the meaning of the vertical asymptote in terms of cost and the percent of pollutants removed.
Solution
The undefined value occurs at \(p=100\) which makes the denominator \(100-p = 0 \). A zero occurs when the numerator \(1.125p+1250 = 0\) Solving for \(p\) we have a zero at \(p =\dfrac{-1250}{1.125} \approx -1111.11 \) which is not in the domain of \(f\). Therefore the cost function does not change sign in the domain 0<\(p\)<100. Since \(f(1)=12.6 \), \(f\) is positive on this interval. Therefore, the value of \(C\) approaches \(+\infty\) as \(p\) approaches 100 and the graph nears the vertical asymptote as shown below. Effectively this means that as the percent removed approaches 100%, the cost of removing it gets infinitely large. In this scenario, it is unrealistic that all of the pollutant can be removed given the cost.
Focus: The Undefined Values of Simplified Rational Functions
A fraction such as \(\dfrac{4}{6}\) is unsimplified since it contains a common factor of two in the numerator and denominator. It can be simplified as \[\dfrac{4}{6}=\dfrac{2 \cdot 2}{2 \cdot 3} = \dfrac{2}{3} \nonumber \]However, zero can't be a common factor in a fraction since the fraction would be undefined\[\dfrac{0 \cdot 2}{0 \cdot 3} = \dfrac{0}{0} \nonumber \]
For example, the function \(f(x)=\dfrac{x^2-4}{x-2}\) is undefined at \(x=2\) since this value makes the denominator zero. Notice, the fraction is unsimplified since there is a common factor of \( (x-2) \). We can simplify \(f(x)=\dfrac{x^2-4}{x-2}=\dfrac{(x+2)(x-2)}{x-2}=x+2 \) as long as the common factor \( (x-2) \ne 0 \) which occur when \(x \ne 2\). Therefore the functions \(y=\dfrac{x^2-4}{x-2}\) and \(y=x+2\) are equivalent when \(x \ne 2 \). The graph of \(y=x+2\) is a line with slope \(m=2\) and y-intercept at \(y=2\). Since the function \(f\) is not defined at \(x=2\), we need to remove the point on the line at \(x=2\) leaving a hole in the graph below represented by an open dot.
In general, to simplify an unsimplified rational function, the zero corresponding the common factor in the numerator and denominator will need to be removed from the domain of the simplified fraction for the two functions to be equivalent. On the graph of the simplified fraction, the corresponding point will need to be removed from the graph leaving a hole.
On the graph of an unsimplified rational function, a hole occurs at the undefined value \(x=a\) corresponding to the zero of a common factor \( (x-a) \) of the numerator and denominator.
Let \(f(x)=\dfrac{x^{2}-1 }{x^2-3x+2 }\).
- Identify the undefined value(s). Determine what happens on the graph near each undefined value.
- Use the zeros, the undefined values, the end behavior and a sign chart for \(f\) to graph the function.
Solution
- Setting the denominator is equal to zero and solving by factoring to find the undefined values\[\begin{array}{l} {x^2-3x+2 =0} \\ {(x-2)(x-1)=0} \\ {x=2,\; 1} \end{array}\nonumber\]To determine what happens at each undefined value, we need to simplify the fraction \[f(x)=\dfrac{x^{2}-1 }{x^2-3x+2 }=\dfrac{(x-1)(x+1)}{(x-1)(x-2) }=\dfrac{x+1 }{x-2 } \nonumber\]provided that \(x \ne 2\) and \(x \ne 1\). Since the undefined value \(x=1\) corresponding to the common factor \( (x-1) \) in the numerator and denominator, a hole occurs on the graph of \(f(x)\) at \(x=1\). Since \(x=2\) is an undefined value of the resulting simplified fraction \(y=\dfrac{x+1 }{x-2 }\), a vertical asymptote occurs at \(x=2\).
- Setting the numerator of the simplified fraction is equal to zero and solving to find the zero(s) \[\begin{array}{l} {x+1 =0} \\ {x=-1} \end{array}\nonumber\]So \(f\) has a zero at \(x=-1\). The end behavior resembles the end behavior of \(y=\dfrac{x}{x}=1\). Therefore as \(x \to \pm \infty \), \(y \to 1\). So \(f\) has a horizontal asymptote at \(y=1\). Building a sign chart for \(f\) using the test point method shown below, we see that \(f\) is positive on \( (-\infty,-1) \cup (2, \infty) \) and negative on \( (-1,1) \cup (1,2) \)
Building a sign chart for f using the test point method Interval \( (-\infty,-1) \) \( (-1,1) \) \( (1, 2) \) \( (2, \infty) \) Test Point -2 0 1.5 3 Value \(\dfrac{(-2)+1}{(-2)-2} = \dfrac{1}{4} \) \(\dfrac{(0)+1}{(0)-2} = \dfrac{1}{-2} \) \(\dfrac{(1.5)+1}{(1.5)-2} = -5 \) \(\dfrac{(-2)+1}{(-2)-2} = \dfrac{1}{4} \) Sign of The Interval positive negative negative positive
It is possible that a rational function could change sign at an undefined value corresponding to hole on the graph. That is why we tested the function \(f\) in example 4.5.9 in the intervals before and after the hole at \(x=1\).
Focus: Solving Rational Inequalities
Solving the rational inequality \( \dfrac{x-3}{x-1}>0 \) is equivalent to finding where \( y=\dfrac{x-3}{x-1} \) is positive. Solving the rational inequality \( \dfrac{x-3}{x-1}<0 \) is equivalent to finding where \( y=\dfrac{x-3}{x-1} \) is negative. On the otherhand, the inequality \( \dfrac{1}{x}>x \) can't be solved in this manner since \(x\) and \( \dfrac{1}{x} \) may be either positive or negative depending on the value of \(x\). However, if we get zero on one side of this inequality \( \dfrac{1}{x}-x>0 \), then we can solve this inequality by determining where \( y=\dfrac{1}{x}-x \) is positive.
A rational inequality of the form \(y=\dfrac{f(x)}{g(x)}>0 \) or \(y=\dfrac{f(x)}{g(x)}<0 \) can be solved by determining the sign of the rational function.
We can use the sign chart for \(f\) using the test point method to determine the sign of \(f\) on each interval instead of graphing the entire function.
Solve the inequality \(\dfrac{x+2}{x^2-3x}>0 \)
Solution
We can solve this inequality by determining where \(y=\dfrac{x+2}{x^2-3x} \) is positive using a sign chart. Setting the denominator to zero and solving to find the undefined value(s) \[\begin{array}{l} {x^2-3x =0} \\ {x(x-3)=0} \\ {x=0 \text{ and } x=3} \end{array}\nonumber\]Setting the numerator to zero and solving to find the zero(s) \[\begin{array}{l} {x+2 =0} \\ {x=-2} \end{array}\nonumber\]Building a sign chart for \(f\) using the test point method shown below
| Interval | \( (-\infty,-2) \) | \( (-2,0) \) | \( (0, 3) \) | \( (3, \infty) \) |
|---|---|---|---|---|
| Test Point | -3 | -1 | 1 | 4 |
| Value | \(\dfrac{(-3)+2}{(-3)^2-3(-3)} = \dfrac{-1}{18} \) | \(\dfrac{(-1)+2}{(-1)^2-3(-1)} = \dfrac{1}{4} \) | \(\dfrac{(1)+2}{(1)^2-3(1)} = \dfrac{3}{-2} \) | \(\dfrac{(4)+2}{(4)^2-3(4)} = \dfrac{6}{4} = \dfrac{3}{2}\) |
| Sign of The Interval | negative | positive | negative | positive |
Therefore, the solution to the inequality is \( (-2,0) \cup (3, \infty) \).
Solve the inequality \(\dfrac{1}{x} \le x \)
Solution
First, getting zero on one side. \[\dfrac{1}{x} - x \le 0 \nonumber \] We can solve this inequality by determining where \(y=\dfrac{1}{x} - x \) is negative or zero using a sign chart. However, this function is not written as a single fraction that we can analyze as a rational function. Getting a common denominator and adding we have \[y=\dfrac{1}{x} - x = \dfrac{1}{x} - \dfrac{x^2}{x}= \dfrac{1-x^2}{x}\nonumber \]The undefined values occur when the denominator is zero. So \(x\) is undefined at \(x=0\). The zeros occur when the numerator is zero (and the denominator is not zero) \[\begin{array}{l} {1-x^2 =0} \\ {(1-x)(1+x) =0} \\ {1-x =0 \text{ and } 1 + x=0 } \\{x=1 \text{ and } x= -1} \end{array}\nonumber\] Building a sign chart for \(f\) using the test point method shown below
| Interval | \( (-\infty,-1) \) | \( (-1,0) \) | \( (0, 1) \) | \( (1, \infty) \) |
|---|---|---|---|---|
| Test Point | -2 | -0.5 | 0.5 | 2 |
| Value | \(\dfrac{1-(-2)^2}{(-2)} = \dfrac{3}{2} \) | \(\dfrac{1-(0.5)^2}{(-0.5)} = \dfrac{-3}{2} \) | \(\dfrac{1-(0.5)^2}{(0.5)} = \dfrac{3}{2} \) | \(\dfrac{1-(2)^2}{(2)} = \dfrac{-3}{2} \) |
| Sign of The Interval | positive | negative | positive | negative |
Therefore, the solution to the inequality is \( [-1,0) \cup [1, \infty) \).
Solve the inequality \(\dfrac{x-4}{x+2} \ge 0 \)
- Answer
-
The solution to the inequality is \( (-\infty,-2) \cup [4, \infty) \).
Important Topics of this Section
- The End Behavior of Rational Functions
- Horizontal Asymptotes
- Zeros of Rational Functions
- Vertical Asymptotes
- Rational Functions
- Holes
- Graphing Rational Functions
- Rational Inequalities