\chapter{Polynomial and rational functions}
\section{Fractional Linear functions}
Just like a rational number is a fraction made with whole numbers, a rational function is a
fraction made with polynomials. So for example
\[\frac{1}{x}, \hspace{1ex}\frac{4x+3}{2x-1},\hspace{1ex} \frac{2x^2-6x+4}{x^2-1}\]
are rational functions. But
\[\frac{x-1}{4+\sqrt{x} }\]
is not a rational function, because the denominator is not a polynomial.
\subsection{The reciprocal function}
The simplest rational function is the reciprocal function:
\[f(x)=\frac{1}{x},\]
that we studied as one of the basic parent functions in Unit 1.10.
We will draw its graph again here, and discuss it in more detail.
The first thing we notice is that the domain cannot contain \(0\), because the denominator
cannot be \(0\). Then we also notice that when we take the reciprocal \(1/x\) of numbers,
small numbers become large and large numbers become big, as shown in the table
below:
\[
\begin{array}{c|ccccccccc}
x & 1 & 2 & 0.5 & 4 & 0.25 & 10 & 0.1 & 20 & 0.05 \\
\hline
\displaystyle \frac{1}{x} & 1 & 0.5 & 2 & 0.25 & 4 & 0.1 & 10& 0.05 & 20
\end{array}
\]
This explains the shape of the graph on the right of the \(y\)-axis, where we have
plotted the first \(5\) points in the table:
\[\img{U2_8F1.png}{}{12em}{} \]
If we use negative numbers, we find a similar shape, but below the \(x\)-axis:
\[
\begin{array}{c|ccccccccc}
x & -1 & -2 & -0.5 & -4 & -0.25 & -10 & -0.1 & -20 & -0.05 \\
\hline
\displaystyle \frac{1}{x} & -1 & -0.5 & -2 & -0.25 & -4 & -0.1 & -10& -0.05 & -20
\end{array}
\]
\[\img{U2_8F2.png}{}{12em}{} \]
As \(x\) becomes very large (either positive or negative), the graph gets closer
and closer to the \(x\)-axis. We say that the \(x\)-axis is a
\textit{Horizontal Asymptote} for \(f(x)\), and write for short HA (for Horizontal Asymptote).
So the equation of the HA is \(y=0\) (the equation of the \(x\)-axis).
As \(x\) gets closer and closer to \(0\) (and so it gets close to the
\(y\)-axis, either from the left or from the right), the value of the function becomes either
very large and positive (if we are on the right of \(0\)) or very large and negative
(if we are on the left of \(0\)). We say that the \(y\)-axis is a \textit{Vertical Asymptote}
for \(f(x)\), and write for short VA (for Vertical Asymptote). So the equation of the VA is
\(x=0\) (the equation of the \(y\)-axis).
\[\img{U2_8F3.png}{}{12em}{} \]
The Domain and Range of this function are both \((-\infty,0)\cup(0,\infty)\), as we saw in
Unit 1.10.
There are no \(x\) or \(y\)-intercepts.
We summarize the essential points below:
\[\begin{array}{l|l}
\mbox{Function} & \displaystyle y=\frac{1}{x}\\[1ex]
\hline
\mbox{Domain} & (-\infty,0)\cup (0,\infty)\\[0.5ex]
\mbox{Range} & (-\infty,0)\cup (0,\infty)\\[0.5ex]
\mbox{VA} & x=0\\[0.5ex]
\mbox{HA} & y=0\\[0.5ex]
x-\mbox{intercepts} & \mbox{None}\\[0.5ex]
y-\mbox{intercept} & \mbox{None}
\end{array}
\]
The next type of rational functions we discuss is obtained by applying some simple transformations
to the basic parent \(1/x\).
\subsection{Fractional linear functions}
We will take as base point \(P\) for the parent function the point \((1,1)\), as shown in the picture:
\[\img{U2_8F4.png}{}{12em}{} \]
As for the quadratic functions of Unit 2.1, the most general function obtained
by transforming \(f(x)\) is then
\[ af(x-h)+k=\frac{a}{x-h}+k.
\]
When the parent function is \(1/x\), these transformed functions are called \textit{fractional linear
functions}. The word \textit{linear} refers to the fact that if we re-write the
formula as a single fraction, the polynomials we find in the
numerator and denominator are of degree at most \(1\).
For example,
\[3f(x-1)+2=\frac{3}{x-1}+2=\frac{3}{x-1}+\frac{2(x-1)}{x-1}=\frac{3+2x-2}{x-1}=\frac{2x+1}{x-1}.\]
\begin{example}
Let \(f(x)=1/x\), and \(g(x)=2f(x-1)+1\). So \(g(x)\) is obtained from \(f(x)\) by stretching
vertically by \(2\), then shifting right by \(1\) and up by \(1\).
The formula for \(g(x)\) is
\[g(x)=\frac{2}{x-1}+1.\]
We can draw the graph in two steps. First stretch vertically by 2, so \((1,1)\) becomes
\((1,2)\), and \((-1,-1)\) becomes \((-1,-2)\), then shift right by \(1\) and up by \(1\):
\[\img{U2_8F5.png}{}{12em}{}
\begin{array}{c}
\mbox{stretch}\\
\longrightarrow
\end{array}
\img{U2_8F6.png}{}{12em}{}
\begin{array}{c}
\mbox{shift}\\
\longrightarrow
\end{array} \img{U2_8F7.png}{}{12em}{}\]
When shifting right, the VA \(x=0\) becomes the
vertical line \(x=1\), and when shifting up the HA \(y=0\) becomes the horizontal line \(y=1\).
Note also that the transformed function has \(x\)-and \(y\)-intercepts, at
\((-1,0)\) and \((0,-1)\). Of course we could also have found these from the formula:
\[g(0)=\frac{2}{0-1}+1=-2+1=-1,\]
and solving \(g(x)=0\) we find
\begin{eqnarray*}
\frac{2}{x-1}+1&=& 0\\[1ex]
\frac{2}{x-1}&=& -1\\[1ex]
\cancel{(x-1)}\frac{2}{\cancel{x-1}}&=& -1(x-1)\\[1ex]
2 & = & -(x-1)\\[1ex]
2 & = & -x+1\\[1ex]
x&=& -1
\end{eqnarray*}
From the graph we also find the domain and range:
\[D=(-\infty,1)\cup(1,\infty), \hspace{5ex} R=(-\infty,1)\cup(1,\infty).\]
\end{example}
If we are given the equation of a fractional linear function as a ratio of two polynomials and we want to draw the graph,
we can perform synthetic division and find what the transformation from the basic parent \(1/x\) is,
as shown in the next example.
\begin{example}
Let
\[f(x)=\frac{2x+1}{x+1}.\]
Do the synthetic division:
| \(-1\) |
\(2\) | \(1\) |
| | | \(-2\) |
| | \(2\) | \(-1\) |
So the quotient is \(2\) and the remainder is \(-1\), and we can write \(f(x)\) as
\[f(x)=2-\frac{1}{x+1}=-\frac{1}{x+1}+2.\]
We see from this answer that \(f(x)\) is obtained from \(1/x\) by first reflecting across the
\(x\) axis, (so the base point \((1,1)\) becomes \((1,-1)\) ), then shifting left \(1\) and up \(2\),
so that the HA is \(y=2\) and the VA is \(x=-1\):
\[\img{U2_8F4.png}{}{12em}{} \begin{array}{c}
\mbox{reflect}\\
\longrightarrow
\end{array} \img{U2_8F8.png}{}{12em}{} \begin{array}{c}
\mbox{shift}\\
\longrightarrow
\end{array} \img{U2_8F9.png}{}{12em}{}\]
We see from the graph that the transformed function has \(y\)-intercept \((0,1)\) (that is
just the shifted base point \(P\) of the original graph), and an \(x\)-intercept that we can find
from the formula:
\begin{eqnarray*}
\frac{2x+1}{x+1}&=& 0\\[1ex]
\cancel{(x+1)}\frac{2x+1}{\cancel{x+1}}&=& 0(x+1)\\[1ex]
2x+1 &=& 0\\[1ex]
x=-\frac{1}{2}
\end{eqnarray*}
So the \(x\)-intercept is \((-1/2,0)\).
From the graph we also get domain and range:
\[D=(-\infty,-1)\cup(-1,\infty), \hspace{5ex} R=(-\infty,2)\cup(2,\infty).\]
\end{example}
| Fractional Linear Functions |
|---|
| \begin{itemize}
\item A fractional linear function is a rational function obtained as a simple transformation
of the reciprocal function \(y=1/x\).
\item
Every rational function of form \(\displaystyle y=\frac{ax+b}{cx+d}\) is a fractional linear function.
\item
To re-write \(\displaystyle y=\frac{ax+b}{cx+d}\) as a transformation of \(y=1/x\), divide using synthetic or long division.
\end{itemize} |
Problems
\problem
Let \(f(x)=1/x\), and \(g(x)=2f(x-2)-1\). Draw the graph of \(g(x)\), showing all intercepts and
asymptotes, and give domain and range.
\begin{sol}
We need to stretch the parent function \(1/x\) vertically by \(2\), then shift right by \(2\) and
down by \(1\). So the base point \(P=(1,1)\) will get transformed as: \((1,1)\rightarrow
(1,2)\rightarrow (3,2) \rightarrow (3,1)\), the VA will be shifted to \(x=2\), and the
HA will be shifted to \(y=-1\):
\[\img{U2_8F5.png}{}{12em}{} \begin{array}{c}
\mbox{stretch}\\
\longrightarrow
\end{array} \img{U2_8F6.png}{}{12em}{} \begin{array}{c}
\mbox{shift}\\
\longrightarrow
\end{array} \img{U2_8F10.png}{}{12em}{}\]
We see from the graph that the \(x\)-intercept is \((4,0)\) and the \(y\)-intercept is
\((0,-2)\). We can check from the formula that this is correct:
\[g(0)=2f(-2)-1=\frac{2}{-2}-1=-1-1=-2,\]
and solving \(g(x)=0\) we find
\begin{eqnarray*} \frac{2}{x-2}-1&=& 0\\[1ex]
\frac{2}{x-2}&=& 1\\[1ex]
\cancel{(x-2)}\frac{2}{\cancel{x-2}}&=& 1(x-2)\\[1ex]
2 &=& x-2\\[1ex]
x&=& 4\end{eqnarray*}
Finally we can write the Domain and Range:
\[D=(-\infty,2)\cup(2,\infty), \hspace{5ex} R=(-\infty,-1)\cup(-1,\infty).\]
\end{sol}
\mproblem
Let \(f(x)=1/x\), and \(g(x)=2f(x+2)+1\). Draw the graph of \(g(x)\), showing all intercepts and
asymptotes, and give domain and range.
\problem
Let \(\displaystyle f(x)=\frac{4x-3}{2x-4}\). Draw the graph of \(f(x)\), showing
all intercepts and asymptotes, and find domain and range.
\begin{sol}
Since the denominator is not of form \(x-c\), we use long division.
| |
|
2 |
|
| \(2x\) |
\(-4\) |
\(4x\) |
\(-3\) |
| |
|
\(-4x\) |
\(+8\) |
| | |
|
\(5\) |
The quotient is \(2\) and the remainder is \(5\), and we can write \(f(x)\) as
\[f(x)=\frac{4x-3}{2x-4}=\frac{5}{2x-4}+2.\]
To find the horizontal shift, divide numerator and denominator of the fraction by \(2\) (the coefficient of \(x\)):
\[f(x)=\frac{5/2}{x-2}+2.\]
So \(f(x)\) is obtained from the parent \(1/x\) by stretching vertically by \(5/2=2.5\), then
shifting right \(2\) and up \(2\). This means that the VA will be \(x=2\), and the HA will be \(y=2\).
\[\img{U2_8F5.png}{}{12em}{} \begin{array}{c}
\mbox{stretch}\\
\longrightarrow
\end{array} \img{U2_8F11.png}{}{12em}{} \begin{array}{c}
\mbox{shift}\\
\longrightarrow
\end{array} \img{U2_8F12.png}{}{12em}{}\]
To find the exact value of the intercepts, we use the formula:
\[f(0)=\frac{4(0)-3}{2(0)-4}=\frac{-3}{-4}=0.75,\]
so the \(y\)-intercept is \((0,0.75)\), and solving \(f(x)=0\) we get
\begin{eqnarray*}
\frac{4x-3}{2x-4}&=& 0\\[1ex]
\cancel{(2x-4)} \frac{4x-3}{\cancel{2x-4}}&=& 0(2x-4)\\[1ex]
4x-3&=& 0\\[1ex]
x&=& \frac{3}{4}=0.75.
\end{eqnarray*}
So the \(x\)-intercept is \((0.75,0)\).
From the graph, domain and range are
\[D=(-\infty,2)\cup(2,\infty), \hspace{5ex} R=(-\infty,2)\cup(2,\infty).\]
\end{sol} \mproblem
Let \(\displaystyle f(x)=\frac{2x-1}{2x+2}\). Draw the graph of \(f(x)\), showing
all intercepts and asymptotes, and find domain and range.
\problem
Let \(\displaystyle f(x)=\frac{2x-3}{3-x}\). Draw the graph of \(f(x)\), showing
all intercepts and asymptotes, and find domain and range.
\begin{sol}
To do the synthetic division, we need a binomial of form \(x-c\) in the denominator, but
in this problem we have \(3-x\). To fix this we multiply both numerator and denominator
by \(-1\):
\begin{eqnarray*}
f(x)&=& \frac{(-1)(2x-3)}{(-1)(3-x)}\\[1ex]
&=& \frac{-2x+3}{-3+x}\\[1ex]
&=& \frac{-2x+3}{x-3}
\end{eqnarray*}
Now we can divide \(-2x+3\) by \(x-3\) using synthetic division:
| \(3\) |
\(-2\) | \(3\) |
| | | \(-6\) |
| | \(-2\) | \(-3\) |
The quotient is \(-2\) and the remainder \(-3\). So we can write
\[\frac{2x-3}{3-x}=-2-\frac{3}{x-3}=-\frac{3}{x-3} -2,\]
and we see that \(f(x)\) is the parent \(1/x\) reflected across the \(x\)-axis, stretched
vertically by \(3\), then shifted right \(3\) and down \(2\). The VA will be \(x=3\) and
the HA will be \(y=-2\).
\[\img{U2_8F5.png}{}{12em}{} \begin{array}{c}
\mbox{reflect}\\
\longrightarrow
\end{array} \img{U2_8F8.png}{}{12em}{}\]
\[\begin{array}{c}
\mbox{stretch}\\
\longrightarrow
\end{array} \img{U2_8F13.png}{}{12em}{}
\begin{array}{c}
\mbox{shift}\\
\longrightarrow
\end{array} \img{U2_8F14.png}{}{12em}{}\]
To find the intercepts use the formula:
\[f(0)=\frac{2(0)-3)}{3-0}=\frac{-3}{3}=-1,\]
so the \(y\)-intercept is \((0,-1)\), and solving \(f(x)=0\) we find
\begin{eqnarray*}\frac{2x-3}{3-x}&=& 0\\[1ex]
\cancel{(3-x)}\frac{2x-3}{\cancel{3-x}} & = & 0(3-x)\\[1ex]
2x-3 &=& 0 \\[1ex]
x&=& \frac{3}{2}=1.5.\end{eqnarray*}
So the \(x\)-intercept is \((1.5,0)\).
The domain and range are
\[D=(-\infty,3)\cup(3,\infty) \hspace{5ex} R=(-\infty,-2)\cup (-2,\infty).\]
\end{sol} \mproblem
Let \(\displaystyle f(x)=\frac{3x-6}{1-x}\). Draw the graph of \(f(x)\), showing
all intercepts and asymptotes, and find domain and range.