Rational Functions:

Vertical and Horizontal Asymptotes

Lesson Overview

Instructor Presentation

Introduction

Rational functions are ratios (or quotients) of polynomial functions.  Rational functions are of the form

({{({{R(x) = ((N(x))/(D(x)))}})}})

where N(x) and D(x) are polynomial functions and D(x) ≠0.

Two simple examples are below.

Plot[(x+1)/(x-3),{x,-4,10},PlotRange->{-7,7},PlotStyle->{Thick,Red}]

Plot[((x^2)+4x+4)/(x+3),{x,-10,10},PlotRange->{-10,10},PlotStyle->{Thick,Red}]

We will explore the following scenarios in this lesson...

What makes a function have a vertical asymptote and where is it?

What makes a function have a horizontal asymptote and where is it?

parameters={a,c,d,x}

Concept Exploration

Domain

To begin speaking about rational functions, we must first discuss the domain of a function.  The domain of a function is the set of all numbers which produce an function value that is defined.  Any x-value that does not yield a defined value is not in the domain of the function.  What this implies is that something interest happens at this x-value on the graph of the function.  We will explore this a bit later.

To find the domain of the function:

Set the denominator D(x) equal to zero, if possible.

These values are then excluded from the domain of the function.

If D(x) cannot equal zero, then the domain is all real numbers.

Let's look at a couple of examples:

Find the domain of f(x) = ((x + 4)/(4x - 2)).

Set D(x) = 3x - 2 equal to zero.				4x - 2 = 0
								        x = (2/4)
So, the domain is all real numbers such that x ≠(1/2), or, more formally, {x | x ≠ (1/2)}.

Find the domain of f(x) = (x^3) - 2(x^2) + 7.

Since D(x) = 1, essentially, and since 0 ≠ 1, there are no domain restrictions.
The domain is "all real numbers," or symbolically, domain = ℝ.

Example 1

Find the domains of the given rational functions.

1)  y = (((x^2) - 8)/(7x + 3))

2)  f(x) = (1/(2(x^2) - 5x - 12))

3)  f(x) = (x^4) - 256

Concept Exploration

Vertical Asymptotes

What happens if we graph the functions from the domain page?

Since we cannot graph where the domain restricts (i.e. an x-value that leaves the function undefined, or with a denominator equal to 0 cannot occur on the actual graph), this is shown as a vertical asymptote.

Plot[((x+4)/(4x-2)),{x,-2,5},PlotRange->{-5,5},PlotStyle->{Thick,Red},Ticks->{Range[-2,5,(1/2)], Range[-5,5,1]}]

To see just what the function values are doing very close to x = (1/2), let's look at this table of values:

data=Table[{x,((x+4)/(4x-2))},{x,{0.4,.49,.499,.4999,0.5001,.501,.51,.6}}];
headers={TraditionalForm[x],TraditionalForm[HoldForm[((x+4)/(4x-2))]]};
Options[DisplayTable]=Join[{TableLabel->"Table label"},Options[Grid]];fulldata=Prepend[data,headers];
DisplayTable[data_List,opts:OptionsPattern[]]:=CellPrint[TextCell[Labeled[Grid[data,FilterRules[{opts},Options[Grid]],Frame->All,ItemSize->10,Alignment->Center,BaseStyle->{FontFamily->"Helvetica",FontSize->18},Background->LightYellow,Frame->All,Spacings->{2,0.62}],Style[OptionValue[TableLabel],FontFamily->"Times",16]],"Output",ShowStringCharacters->False,Background->None]]

DisplayTable[fulldata,TableLabel->""]

As you'll see, the closer we get to x = (1/2)from the left side, the function values tend toward -∞.  As we approach x = (1/2)from the right side, the function values tend toward ∞.  So, the asymptotes not only indicate where there is an undefined value for the function, but also indicate the graphs behavior at infinity.

Concept Exploration

Horizontal Asymptotes

Click on the buttons on the following demonstration and answer the questions below.

Manipulate[Plot[(a*x+1)/(x-2),{x,-14,18},PlotRange->{-4,8},PlotLabel->(a*x+1)/(x-2)],{a,{-1,0,1,2,3}}]

What do you see happening as the coefficient in front of the x in the numerator increases?

The "line" about which the graph tends to level out toward positive and negative infinity increases, or moves up in y-values

A general expansion of the Rational Function denoted earlier is as follows...

({{({{R(x) = ((N(x))/(D(x))) = (((Subscript[a, n])(x^n) + (Subscript[a, n-1])(x^(n-1)) + ... + (Subscript[a, 2])(x^2) + (Subscript[a, 1])x + (Subscript[a, 0]))/((Subscript[b, m])(x^m) + (Subscript[b, m-1])(x^(m-1)) + ... + (Subscript[b, 2])(x^2) + (Subscript[b, 1])x + (Subscript[b, 0])))}})}})

The most important ideas we will extract from this complicated looking rational function are below...

If n < m, the rational function has a horizontal asymptote at y = 0 (the x-axis).

If n = m, the rational function has a horizontal asymptote at y = ((Subscript[a, n])/(Subscript[b, m])).

If n > m, the rational function has no horizontal asymptote.

Example 2

For each of the rational functions below, predict the existence and location of the vertical and horizontal asymptotes.  Then plot to see if you were correct.

1)  f(x) = (1/((x^2) - 9))

Prediction :

2)  g(x) = ((2(x^2) - 3x - 2)/((x^2) + 4))

Prediction :

3)  h(x) = (((x^2) + 4x - 7)/(x + 2))

Prediction :

4)  Is it possible to create a rational function with more than one horizontal asymptote?  If so, create one in the Input box below.  If not, explain why in the Answer box below.

Example 3

For each of the scenarios, create a function that satisfies the given conditions and plot it.

A rational function with...

1)  a vertical asymptote at x = -3, and the x-axis serving as the horizontal asymptote

2)  Vertical asymptotes at x = -2 and x = 1 and a horizontal asymptote at x = 3

3)  No vertical asymptotes, no horizontal asymptotes, y-intercept at (0,2).

4)  Vertical asymptotes at x = -3 and x = 3.  No horizontal asymptotes.

Answer Key

Example 1

Find the domains of the given rational functions.

1)  y = (((x^2) - 8)/(7x + 3))

{x | x ≠ -(3/7)} or all real numbers except  - (3/7).

2)  f(x) = (1/(2(x^2) - 5x - 12))

{x | x ≠ - (5/2)and x ≠ 4} or all real numbers except - (5/2) and 4.

3)  f(x) = (x^4) - 256

All real numbers.

Example 2

1)  f(x) = (1/((x^2) - 9))

Prediction : vertical asymptotes at x = -3, 3 and horizontal asymptote at the x-axis.

Plot[(1/((x^2)-9)),{x,-10,10},PlotRange->{-10,10},PlotStyle->{Thick,Purple}]

2)  g(x) = ((2(x^2) - 3x - 2)/((x^2) + 4))

Prediction : No vertical asymptotes, horizontal asymptote at y = 2.

Plot[((2(x^2)-3x-2)/((x^2)+4)),{x,-20,30},PlotRange->{-5,5},PlotStyle->{Thick, Red}]

3)  h(x) = (((x^3) + 4x - 7)/(x + 2))

Prediction : Vertical asymptote at x = -2, no horizontal asymptotes.

Plot[(((x^2)+4x-7)/(x+2)),{x,-10,10},PlotRange->{-40,40},PlotStyle->{Thick,Green}]

4)  Is it possible to create a rational function with more than one horizontal asymptote?  If so, create one in the Input box below.  If not, explain why in the Answer box below.

not algebraically possible.

While it is possible to create a function with two horizontal asymptotes, it cannot be done as a rational function.  Logistic models have this characteristic (see below.)

Plot[(1200/(1+39*Exp[-.9x])),{x,-5,10},PlotStyle->{Thick,Orange}]

Example 3

For each of the scenarios, create a function that satisfies the given conditions and plot it.

A rational function with...

1)  a vertical asymptote at x = -3, and the x-axis serving as the horizontal asymptote

Plot[(1/(x+3)),{x,-10,10},PlotRange->{-15,15},PlotStyle->{Thick,Pink}]

2)  Vertical asymptotes at x = -2 and x = 1 and a horizontal asymptote at x = 3

Plot[((3(x^2)+2)/((x+2)*(x-1))),{x,-10,10},PlotRange->{-5,30},PlotStyle->{Thick,Blue}]

3)  No vertical asymptotes, no horizontal asymptotes, y-intercept at (0,2).

Plot[(((x^3)+2)/((x^2)+1)),{x,-10,10},PlotRange->{-5,5},PlotStyle->{Thick,Yellow}]

4)  Vertical asymptotes at x = -3 and x = 3.  No horizontal asymptotes.

Plot[((x^3)/((x^2)-9)),{x,-10,10},PlotStyle->{Thick,Brown}]