Rational Functions

Slant asymptotes

Lesson Overview

Instructor Presentation

Introduction

As a quick review from the last Rational Function lesson plan, we learned that...

Rational functions R(x)=((N(x))/(D(x))) contain vertical asymptotes when the denominator of a rational function (in simplest or reduced form) can be set equal to zero.  The vertical asymptotes are precisely D(x)=0.

Plot[((x+1)/(x-3)),{x,-4,10},PlotRange->{-7,7},PlotStyle->{Thick,Red},PlotLabel ->Style[Row[{Style["y",Italic]," = ",ToString[((x+1)/(x-3)),TraditionalForm]}],20]]

Rational functions R(x)=((N(x))/(D(x))) contain horizontal asymptotes when the degree of N(x) is less than or equal to that of D(x).  Here are two examples for the generic rational function...

({{({{((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])))}})}})

When n = m, there exists a horizontal asymptote at y =(Subscript[a, n])/(Subscript[b, m]).

Plot[{((6x-1)/(3x-2)),2},{x,-4,10},PlotRange->{-7,7},PlotStyle->{{Thick,Blue},{Thick,Red}},PlotLabel ->Style[Row[{Style["y",Italic]," = ",ToString[((6x-1)/(3x-2)),TraditionalForm]}],20]]

When n < m, there exists a horizontal asymptote at y = 0.

Plot[{(1/((x^2)-9)),0},{x,-5,5},PlotRange->{-4,4},PlotStyle->{{Thick,Blue},Red},PlotLabel ->Style[Row[{Style["y",Italic]," = ",ToString[(1/((x^2)-9)),TraditionalForm]}],20]]

Of course, this begs the question, "What happens when the degree of N(x) is greater than the degree of D(x)?"

parameters={a,c,d,x}

Concept Exploration

Divide (x^3)  by ((x^2) - 9)

💡 To divide polynomials, use the Function PolynomialQuotientRemainder.

PolynomialQuotientRemainder[(x^3),(x^2)-9,x]

This means that ((x^3)/((x^2)-9))=((9 x)/((x^2)-9))+x

Now, plot ((x^3)/((x^2)-9)).

Plot[((x^3)/((x^2)-9)),{x,-10,10},PlotRange->{-14,14},PlotStyle->{Thick,Green}]

What happens if we add in the "quotient" portion of the answer (sans remainder).

Plot[{((x^3)/((x^2)-9)),x},{x,-10,10},PlotRange->{-14,14},PlotStyle->{{Thick,Green},{Thick,Blue}}]

The non-remainder portion of the polynomial division problem serves as the "slant asymptote" or "oblique asymptote."

Example 1

Determine which (if any) asymptotes the following rational functions have.  Then, plot the rational function.  Give the equation that represents each asymptote.  Make sure to include in your plot the line that represents the horizontal or slant asymptote in a different color.

1)  y = (((x^2) - 2x + 3)/(x - 5))

Vertical Asymptote:

Horizontal Asymptote:

Slant Asymptote:

Plot:

2)  y = ((3(x^2) - 2x + 4)/((x^2) - 4x + 5))

Vertical Asymptote:

Horizontal Asymptote:

Slant Asymptote:

Plot:

Example 2

Create a rational function with the following characteristics.  Then plot your function.  Make sure to graph any horizontal or slant asymptote in a different color.

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

Function :

Plot:

2)  Vertical asymptote at x = -2; slant asymptote at y = 3x - 1

Function :

Plot:

3)  Is it possible to create a rational function with both a horizontal asymptote and a slant asymptote?  If so, create one in the Input box below.  If not, explain why in the Answer box below.

Concept exploration — nonlinear slant asymptotes

We've shown slant asymptotes as lines.  What happens when the difference in the degrees of N(x) and D(x) is greater than 1?

Run the following model to make a conclusion.

Manipulate[Plot[{(((x^n)+2)/(x-1)),(((x^n)+2)/(x-1))-(3/(x-1))},{x,-5,5},PlotRange->{-20,20},PlotStyle->{{Thick,Blue},Red},PlotLabel ->Style[Row[{Style["y",Italic]," = ",ToString[(((x^n)+2)/(x-1)),TraditionalForm]}],20]],{n,{0,1,2,3,4}}]

What happens?  (Expand this cell.)

The asymptote is a polynomial of degree n - m.

Example 3

For each of the following rational functions, predict the appearance of the asymptote.  Then, divide the polynomials to prove this is true.  Finally, plot the rational function along with the asymptote on the same set of axes.  Make sure the asymptote is colored differently than the rational function.

1)  f(x) = (((x^3) + (x^2) - 2x + 5)/(x + 2))

Prediction:

Quotient and remainder:

PolynomialQuotientRemainder[

Plot:

2)  f(x) = ((2(x^2) - 4x +  7)/(x - 6))

Prediction:

Quotient and remainder:

PolynomialQuotientRemainder[

Plot:

3)  f(x) = ((0.001(x^4) + 0.2x + 1)/(x - 3))

Prediction:

Quotient and remainder:

PolynomialQuotientRemainder[

Plot:

Answer Key

Example 1

Determine which (if any) asymptotes the following rational functions have.  Then, plot the rational function.  Give the equation that represents each asymptote.  Make sure to include in your plot the line that represents the horizontal or slant asymptote in a different color.

1)  y = (((x^2) - 2x + 3)/(x - 5))

Vertical Asymptote:

x = 5

Horizontal Asymptote:

none

Slant Asymptote:

PolynomialQuotientRemainder[(x^2)-2x+3,x-5,x]

y = x + 3

Plot:

Plot[{(((x^2)-2x+3)/(x-5)),x+3},{x,-10,25},PlotStyle->{{Thick,Blue},Red}]

2)  y = ((3(x^2) - 2x + 4)/((x^2) - 4x + 5))

Vertical Asymptote:

no vertical asymptotes

💡 We need to solve D(x) = 0.  Use Mathematica's MouseOver command.

Solve[(x^2)-4x+5==0,x]

💡 Since both "vertical asymptotes" involve complex numbers, this means that there are no vertical asymptotes over the real numbers.

Horizontal Asymptote:

y = 3

Slant Asymptote:

none

Plot:

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

Example 2

Create a rational function with the following characteristics.  Then plot your function.  Make sure to graph any horizontal or slant asymptote in a different color.

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

Function :

f(x) = ((2(x^2))/((x + 2)(x - 3) )) =  ((2(x^2))/((x^2) - x - 6))

Plot:

Plot[{((2(x^2))/((x^2)-x-6)),2},{x,-80,80},PlotStyle->{{Thick,Blue},Red}]

2)  Vertical asymptote at x = -2; slant asymptote at y = 3x - 1

Function :

f(x) = ((3(x^2) + 5x - 3)/(x + 2))

💡 The vertical asymptote is created from D(x) = x + 2.  If D(x) is multiplied to (3x - 1), we get the result of 3(x^2) + 5x - 2. However, this only gives a line with a discontinuity point at x = -2.  A simple way to create a function that fits the given criteria is to alter the constant term slightly.

Plot:

Plot[{((3(x^2)+5x-3)/(x+2)),3x-1},{x,-10,5},PlotStyle->{{Thick,Blue},Red},PlotRange->{-25,15}]

3)  Is it possible to create a rational function with both a horizontal asymptote and a slant asymptote?  If so, create one in the Input box below.  If not, explain why in the Answer box below.

not possible

The criteria for a rational function having a horizontal asymptote and a slant asymptote are mutually exclusive.  The degree of N(x) cannot both be equal to and greater than the degree of D(x).

Example 3

For each of the following rational functions, predict the appearance of the asymptote.  Then, divide the polynomials to prove this is true.  Finally, plot the rational function along with the asymptote on the same set of axes.  Make sure the asymptote is colored differently than the rational function.

1)  f(x) = (((x^3) + (x^2) - 2x + 5)/(x + 2))

Prediction:

The slant asymptote will be a parabola

Quotient and remainder:

PolynomialQuotientRemainder[(x^3)+(x^2)-2x+5,x+2,x]

Quotient: y = (x^2) - x; Remainder: (5/(x + 2))

Plot:

Plot[{(((x^3)+(x^2)-2x+5)/(x+2)),(x^2)-x},{x,-10,10},PlotStyle->{{Thick,Blue},Red},PlotRange->{-10,70}]

2)  f(x) = ((2(x^2) - 4x +  7)/(x - 6))

Prediction:

Slant asymptote will be a line.

Quotient and remainder:

PolynomialQuotientRemainder[2(x^2)-4x+7,x-6,x]

Quotient: y = 2x - 8 ; Remainder: (55/(x - 6))

Plot:

Plot[{((2(x^2)-4x+7)/(x-6)),2x+8},{x,-10,25},PlotStyle->{{Thick,Blue},Red}]

3)  f(x) = ((0.001(x^4) + 0.2x + 1)/(x - 3))

Prediction:

The slant asymptote will take the form of a cubic polynomial.

Quotient and remainder:

PolynomialQuotientRemainder[.001(x^4)+.2x+1,x-3,x]

Quotient: 0.001(x^3) + 0.003(x^2) + 0.009x + 0.227 ; Remainder: (1.681/(x - 3))

Plot[{((.001(x^4)+.2x+1)/(x-3)),.001(x^3)+.003(x^2)+.009x+.227},{x,-20,15},PlotStyle->{{Thick,Blue},Red},PlotRange->{-5,5}]