Rational functions
A rational function is a fraction of two polynomials:
${r(x)=\dfrac{P(x)}{Q(x)}}$
The denominator ${Q(x)}$ cannot be ${0}$. So the domain of ${r}$ is all real numbers except the zeros of ${Q}$. Near those numbers, and far out to the left and right, the graph of ${r}$ approaches lines called asymptotes. This page shows how to find them and how to sketch the graph.
The graph of ${y=1/x}$
The simplest rational function is ${y=\dfrac{1}{x}}$. Its domain is all ${x\ne 0}$. Look at what happens for small positive ${x}$:
| ${x}$ | ${0.1}$ | ${0.01}$ | ${0.001}$ |
|---|---|---|---|
| ${\dfrac{1}{x}}$ | ${10}$ | ${100}$ | ${1000}$ |
As ${x}$ approaches ${0}$ from the right, ${\dfrac{1}{x}}$ increases without bound. From the left, it decreases without bound. And as ${x}$ gets very large, ${\dfrac{1}{x}}$ approaches ${0}$. For example, ${\dfrac{1}{1000}=0.001}$.
We use arrows to write this. ${x\to a^+}$ means "${x}$ approaches ${a}$ from the right" (with ${x>a}$). ${x\to a^-}$ means "${x}$ approaches ${a}$ from the left" (with ${x<a}$). So for ${y=\dfrac{1}{x}}$:
${\dfrac{1}{x}\to\infty}$ as ${x\to 0^+}$, and ${\dfrac{1}{x}\to -\infty}$ as ${x\to 0^-}$.
${\dfrac{1}{x}\to 0}$ as ${x\to\infty}$ and as ${x\to-\infty}$.
Asymptotes
An asymptote is a line that a graph approaches more and more closely. There are two main kinds:
The line ${x=a}$ is a vertical asymptote if ${r(x)\to\infty}$ or ${r(x)\to-\infty}$ as ${x\to a^+}$ or ${x\to a^-}$.
The line ${y=b}$ is a horizontal asymptote if ${r(x)\to b}$ as ${x\to\infty}$ or ${x\to-\infty}$.
For ${y=\dfrac{1}{x}}$, the vertical asymptote is ${x=0}$ (the ${y}$-axis), and the horizontal asymptote is ${y=0}$ (the ${x}$-axis).
Example 1: Sketch the graph of ${r(x)=\dfrac{x-1}{x-3}}$.
Solution:
Divide ${x-1}$ by ${x-3}$. The quotient is ${1}$, and the remainder is ${2}$:
${r(x)=1+\dfrac{2}{x-3}}$
So the graph is the graph of ${y=\dfrac{1}{x}}$, after three transformations (see graphs of functions):
- a vertical stretch by ${2}$,
- a shift ${3}$ units to the right,
- a shift ${1}$ unit up.
The asymptotes move with the graph. The vertical asymptote is ${x=3}$, and the horizontal asymptote is ${y=1}$.
The ${x}$-intercept is where the numerator is ${0}$: ${x=1}$. The ${y}$-intercept is ${r(0)=\dfrac{-1}{-3}=\dfrac{1}{3}}$.
Vertical asymptotes and holes
A vertical asymptote comes from a zero of the denominator. But first, cancel any factor that the numerator and denominator share:
Write ${r(x)=\dfrac{P(x)}{Q(x)}}$ in lowest terms. Then the vertical asymptotes are the lines ${x=a}$, where ${a}$ is a zero of the denominator.
A canceled factor does not give an asymptote. Instead, it gives a hole: a single missing point in the graph.
Example 2: Find the vertical asymptotes and the holes of
${r(x)=\dfrac{x^2-1}{x^2-x}}$.
Solution:
Factor the numerator and the denominator:
${r(x)=\dfrac{(x-1)(x+1)}{x(x-1)}}$
The domain leaves out ${x=0}$ and ${x=1}$. The factor ${x-1}$ cancels:
${r(x)=\dfrac{x+1}{x}}$, for ${x\ne 1}$.
In lowest terms, the denominator is ${x}$. So ${x=0}$ is a vertical asymptote.
The canceled factor ${x-1}$ gives a hole at ${x=1}$. To find its height, put ${x=1}$ into the simplified form: ${\dfrac{1+1}{1}=2}$. The hole is at ${(1,2)}$.
Horizontal asymptotes
Far out, only the leading terms matter (see polynomial functions and their graphs). So compare the degrees. Let ${n}$ be the degree of the numerator, and ${m}$ the degree of the denominator:
If ${n<m}$, the horizontal asymptote is ${y=0}$.
If ${n=m}$, the horizontal asymptote is ${y=\dfrac{a}{b}}$, where ${a}$ and ${b}$ are the leading coefficients of the numerator and the denominator.
If ${n>m}$, there is no horizontal asymptote.
Why? Take ${r(x)=\dfrac{3x^2-3}{x^2-4}}$, with ${n=m=2}$. Divide the numerator and the denominator by ${x^2}$:
$r(x)=\dfrac{3-\dfrac{3}{x^2}}{1-\dfrac{4}{x^2}}$
As ${x\to\pm\infty}$, the small fractions approach ${0}$. So ${r(x)\to\dfrac{3}{1}=3}$.
Example 3: Find the asymptotes of each function.
(a) ${f(x)=\dfrac{5x+1}{x^2+2}}$
(b) ${g(x)=\dfrac{3x^2-3}{x^2-4}}$
Solution:
(a) The denominator ${x^2+2}$ is never ${0}$. So there is no vertical asymptote. The degree of the numerator is ${1}$, less than ${2}$. So the horizontal asymptote is ${y=0}$.
(b) Factor:
${g(x)=\dfrac{3(x-1)(x+1)}{(x-2)(x+2)}}$
No factor cancels. The denominator is ${0}$ at ${x=2}$ and ${x=-2}$. These are the vertical asymptotes. Both degrees are ${2}$, so the horizontal asymptote is ${y=\dfrac{3}{1}=3}$.
Slant asymptotes
If the degree of the numerator is exactly one more than the degree of the denominator, divide. The quotient is a linear function ${y=mx+b}$, and the remainder part approaches ${0}$ far out. So the graph approaches the line ${y=mx+b}$. This line is called a slant asymptote.
Example 4: Find the asymptotes of ${r(x)=\dfrac{x^2+1}{x-1}}$.
Solution:
The denominator is ${0}$ at ${x=1}$, and the numerator is not. So ${x=1}$ is a vertical asymptote.
The numerator has degree ${2}$, one more than the denominator. Divide with synthetic division, using ${c=1}$ and the coefficients ${1}$, ${0}$, ${1}$. The quotient is ${x+1}$, and the remainder is ${2}$:
${r(x)=x+1+\dfrac{2}{x-1}}$
As ${x\to\pm\infty}$, the fraction ${\dfrac{2}{x-1}}$ approaches ${0}$. So the slant asymptote is ${y=x+1}$.
Graphing rational functions
To sketch the graph of a rational function:
- Factor the numerator and the denominator. Cancel common factors, and note any holes.
- Intercepts: the ${x}$-intercepts are the zeros of the numerator (in lowest terms). The ${y}$-intercept is ${r(0)}$, if ${0}$ is in the domain.
- Asymptotes: find the vertical asymptotes, and the horizontal or slant asymptote.
- Test values: the zeros of the numerator and the denominator split the ${x}$-axis into intervals. A test value in each interval shows whether the graph is above or below the axis there.
- Sketch the graph. Near each asymptote, follow the side the test values give.
Example 5: Sketch the graph of ${r(x)=\dfrac{3x^2-3}{x^2-4}}$ from Example 3(b).
Solution:
From Example 3(b), we have
${r(x)=\dfrac{3(x-1)(x+1)}{(x-2)(x+2)}}$.
There are no holes. The vertical asymptotes are ${x=\pm 2}$, and the horizontal asymptote is ${y=3}$.
Intercepts: the ${x}$-intercepts are ${1}$ and ${-1}$. The ${y}$-intercept is ${r(0)=\dfrac{-3}{-4}=0.75}$.
Test values: the numbers ${-2}$, ${-1}$, ${1}$, and ${2}$ split the axis into five intervals:
| Interval | Test ${x}$ | ${r(x)}$ |
|---|---|---|
| ${(-\infty,-2)}$ | ${-3}$ | ${4.8}$, above |
| ${(-2,-1)}$ | ${-1.5}$ | ${\approx -2.1}$, below |
| ${(-1,1)}$ | ${0}$ | ${0.75}$, above |
| ${(1,2)}$ | ${1.5}$ | ${\approx -2.1}$, below |
| ${(2,\infty)}$ | ${3}$ | ${4.8}$, above |
Also, ${r}$ is even, because only even powers of ${x}$ appear. So the graph is symmetric about the ${y}$-axis.
Watch out: A graph never crosses a vertical asymptote, because the function is not defined there. But a graph can cross a horizontal asymptote. The asymptote only describes the far left and the far right.
An application
Example 6: A small company makes phone cases. It pays ${\$500}$ for the machine setup, plus ${\$2}$ for each case. What is the average cost per case when it makes ${x}$ cases? What happens when ${x}$ is very large?
Solution:
The total cost of ${x}$ cases is ${500+2x}$ dollars. The average cost per case is the total cost divided by the number of cases:
${A(x)=\dfrac{500+2x}{x}}$, for ${x>0}$.
For example, for ${100}$ cases, the average cost is
${A(100)=\dfrac{700}{100}=7}$ dollars.
For ${1000}$ cases, it is
${A(1000)=\dfrac{2500}{1000}=2.50}$ dollars.
The numerator and the denominator both have degree ${1}$. So the horizontal asymptote is ${y=\dfrac{2}{1}=2}$. As the company makes more and more cases, the average cost approaches ${\$2}$ per case. The setup cost is shared by more and more cases, so its part becomes small.
Summary
- A rational function is ${\dfrac{P(x)}{Q(x)}}$. Its domain leaves out the zeros of ${Q}$.
- In lowest terms, each zero ${a}$ of the denominator gives a vertical asymptote ${x=a}$. A canceled factor gives a hole instead.
- Horizontal asymptote: ${y=0}$ if ${n<m}$; ${y=\dfrac{a}{b}}$ (leading coefficients) if ${n=m}$; none if ${n>m}$.
- If ${n=m+1}$, divide: the quotient line is a slant asymptote.
- To graph: factor, find the holes, intercepts, and asymptotes, then use test values between them.