9.6: Graphs of Rational Functions
- Page ID
- 45132
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Previously, in the chapters where we discussed functions, we had a function from the library \(f(x) = \dfrac{1}{x}\). Recall, the graph of this function is

We plotted some points we obtained from the table and determined that the domain is all real numbers except for \(x = 0: \{x|x\neq 0\}\) or \((−∞, 0) ∪ (0, ∞)\). We called this function a rational function.
A rational function, \(R(x)\), is a ratio of two polynomials, \(P(x)\) and \(Q(x)\), of the form \[R(x)=\dfrac{P(x)}{Q(x)},\nonumber\] where \(Q(x)\neq 0\).
In this textbook, we only discuss when \(P(x) = 1\) and when \(Q(x)\) is of the form \(x^n\), where \(n\) is a positive integer and \[R(x)=\dfrac{1}{x^n}\nonumber\] For cases when \(P(x)\) is a polynomial other than the constant function \(1\) and \(Q(x)\) is a polynomial other than the power function \(x^n\) is left for future Algebra classes.
Let’s investigate these functions a little further. We know the domain is all real numbers except for \(x = 0\), but let’s look at the graph more closely. Notice, in the graph of \(f(x)\) above, the graph doesn’t intersect the \(y\) axis. Why? Well, let’s set \(y = f(x) = 0\) and solve:
\[f(x)=0=\dfrac{1}{x}\nonumber\]
When is this fraction zero? We know from previous sections that a fraction is zero when the numerator is zero. Will the numerator ever be zero, i.e.,
\[1\stackrel{?}{=}0\nonumber\]
No, never! This means there are no values of \(x\) such that \(y = 0\), and that \(y = 0\) is not in the range of the function.
This is not a coincidence. The fact that \(x\neq 0\) and \(y\neq 0\) for the function \(f(x) =\dfrac{1}{x}\) means that \(f(x)\) has vertical and horizontal asymptotes at \(x = 0\) and \(y = 0\), respectively.
A function, \(R(x)\), has a horizontal asymptote at \(y = 0\) and a vertical asymptote at \(x = 0\) when \(R(x)\) is of the form \[\dfrac{1}{x^n}\nonumber\] We denote these asymptotes by drawing dashed lines for lines \(x = 0\) and \(y = 0\).
Let’s regraph \(f(x) = \dfrac{1}{x}\) showing the horizontal and vertical asymptotes at \(y = 0\) and \(x = 0\), respectively.

So, we see the asymptotes, in red, are dashed lines on the \(x\) and \(y\)-axis and are the lines \(y = 0\) and \(x = 0\). The only case in which the horizontal and vertical asymptotes move left or right, and up or down, respectively, is if there are shifts to the parent function \(f(x)\).
Graph \(R(x)=\dfrac{1}{x^2}\).
Solution
Let’s pick \(x\)-coordinates, and find corresponding \(y\)-values.
\(x\) | \(R(x)=\dfrac{1}{x^2}\) | \((x,R(x))\) |
---|---|---|
\(-3\) | \(f(\color{blue}{-3}\color{black}{})=\dfrac{1}{(-3)^2}\) | \((-3,\dfrac{1}{9})\) |
\(-2\) | \(f(\color{blue}{-2}\color{black}{})=\dfrac{1}{(-2)^2}\) | \((-2,\dfrac{1}{4})\) |
\(-1\) | \(f(\color{blue}{-1}\color{black}{})=1\) | \((-1,1)\) |
\(0\) | \(f(\color{blue}{0}\color{black}{})=\text{undefined}\) | no point |
\(1\) | \(f(\color{blue}{1}\color{black}{})=1\) | \((1,1)\) |
\(2\) | \(f(\color{blue}{2}\color{black}{})=\dfrac{1}{2^2}\) | \((2,\dfrac{1}{4})\) |
\(3\) | \(f(\color{blue}{3}\color{black}{})=\dfrac{1}{3^2}\) | \((3,\dfrac{1}{9})\) |

Plot the ordered-pairs from the table. To connect the points, be sure to connect them from smallest \(x\)-value to largest \(x\)-value, i.e., left to right. The domain of \(R(x)\) is all real numbers except for \(x = 0: \{x|x\neq 0\}\) or \((−∞, 0)∪(0, ∞)\). Since \(R(x)\) has horizontal and vertical asymptotes at \(y = 0\) and \(x = 0\), respectively, let’s draw the lines that represent these asymptotes.
Graphing Rational Functions Using Shifts
Let’s take a look when rational functions’ graphs contain horizontal and vertical shifts. It will be interesting to see the horizontal and vertical asymptotes change due to these shifts.
Given \(R(x)\) is a rational function, a horizontal shift and vertical shift of \(R(x)\) are described below:
\(R(x-h)\) | \(R(x+h)\) | \(R(x)-k\) | \(R(x)+k\) | |
---|---|---|---|---|
Shift | Horizontal shift | Horizontal shift | Vertical shift | Vertical shift |
Units | Shift \(h\) units to the right | Shift \(h\) units to the left | Shift \(k\) units downward | Shift \(k\) units upward |
Asymptotes | Vertical asymptote is \(x=h\) Horizontal asymptote is \(y=0\) |
Vertical asymptote is \(x=h\) Horizontal asymptote is \(y=0\) |
Vertical asymptote is \(x=0\) Horizontal asymptote is \(y=k\) |
Vertical asymptote is \(x=0\) Horizontal asymptote is \(y=k\) |
Graph \(R(x)=\dfrac{1}{x-2}\).
Solution
Let’s start by taking the parent function \(f(x) = \dfrac{1}{x}\). We see that \(R(x) = f(x − 2)\) because we replaced \(x\) with the factor \((x − 2)\). Looking at the table above, we see this is a horizontal shift with \(h = 2\), moving \(2\) units to the right, and the vertical asymptote changes to \(x = 2\).

We can see the gray graph, \(f(x)\), moved two units to the right, in addition to the vertical asymptote. Recall, from the table above, the horizontal asymptote stays \(y = 0\). Hence, the blue graph, \(R(x)\), is the final graph after applying the shifts.
Graph \(K(x)=\dfrac{1}{x}+1\).
Solution
Let’s start by taking the parent function \(f(x) = \dfrac{1}{x}\). We see that \(K(x) = f(x) + 1\) because we added \(1\) to \(f(x)\). Looking at the table above, we see this is a vertical shift with \(k = 2\), moving \(1\) unit upward, and the horizontal asymptote changes to \(y = 1\).

We can see the gray graph, \(f(x)\), moved one unit upward, in addition to the horizontal asymptote. Recall, from the table above, the vertical asymptote stays \(x = 0\). Hence, the blue graph, \(K(x)\), is the final graph after applying the shifts.
Graph \(Q(x)=\dfrac{1}{x+1}-2\).
Solution
Let’s start by taking the parent function \(f(x) = \dfrac{1}{x}\). We see that \(Q(x) = f(x + 1) − 2\) because we replaced \(x\) with the factor \((x + 1)\) and we subtracted \(2\) from \(f(x)\). Looking at the table above, we see \(Q(x)\) has a few shifts: a horizontal shift with \(h = −1\), moving \(1\) unit to the left, a vertical shift with \(k = −2\), moving \(2\) units downward, and the vertical and horizontal asymptotes change to \(x = −1\) and \(y = −2\), respectively.

We can see the gray graph, \(f(x)\), moved one unit to the left and \(2\) units downward in addition to the horizontal and vertical asymptotes. Notice, we had a vertical and horizontal shift. We moved \(f(x)\) one unit left, then \(2\) units down for all points. These shifts cause the asymptotes to move too. In fact, the vertical asymptote moved one unit to the left and the horizontal asymptote moved \(2\) units downward. Hence, the blue graph, \(Q(x)\), is the final graph after applying the shifts.
Graphs Rational Functions Homework
Graph each rational functions using the parent function \(f(x) = \dfrac{1}{x}\). Include the vertical and horizontal asymptotes.
\(R(x)=\dfrac{1}{x-1}\)
\(Q(x)=\dfrac{1}{x+3}\)
\(S(x)=\dfrac{1}{x}+2\)
\(T(x)=\dfrac{1}{x}-4\)
\(U(x)=\dfrac{1}{x+2}-1\)
\(U(x)=\dfrac{1}{x-3}-2\)
\(P(x)=\dfrac{1}{x}+3\)
\(N(x)=\dfrac{1}{x-1}-4\)