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Hints for Graphing Rational Functions

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					                                     Hints for Graphing Rational Functions
                                                          :ÐBÑ
A rational function is any function of the form 0 ÐBÑ œ        where :ÐBÑ and ;ÐBÑ are polynomials. To sketch the graph of such a
                                                          ;ÐBÑ
function, perform the following steps in order.

1). First factor :ÐBÑ and ;ÐBÑ but DO NOT REDUCE the fraction by canceling common factors. Leave the common factors in both
    the numerator and denominator. We assume you can find all rational and all irrational real zeros of both :ÐBÑ and ;ÐBÑ. We
    assume the degree of ;ÐBÑ is at least 1, since otherwise, if ;ÐBÑ is a constant, 0 ÐBÑ is just a polynomial in disguise.

2). Perform a complete sign analysis on 0 ÐBÑÞ When you label the critical numbers, any zeros from factors in the denominator make
    0 ÐBÑ undefined. Only zeros in the numerator :ÐBÑ that do not also appear in the denominator should be labeled as 0's.

3). In this step you will determine which of the undefined points found in step 2Ñ are vertical asymptotes and which are simply
    “holes” in the graph. All undefined points correspond to either vertical asymptotes or holes. For each B œ + value where
    ;Ð+Ñ œ ! we need only consider the multiplicity of the factor ÐB  +Ñ for both :ÐBÑ and ;ÐBÑÞ If :Ð+Ñ Á !ß then the line B œ + is a
    vertical asymptote. In this case, ÐB  +Ñ is not a factor of :ÐBÑÞ If :Ð+Ñ œ !, then ÐB  +Ñ is a factor of :ÐBÑ and if the multiplicity
    of this factor is larger than or equal to the multiplicity of this same factor for ;ÐBÑ then B œ + corresponds to a hole. When
    ÐB  +Ñ is common factor of both :ÐBÑ and ;ÐBÑ and its multiplicty is greater in ;ÐBÑ, then the line B œ + is a vertical asymptote.
    Further label each undefined point as a hole or a vertical asymptote.

%)Þ If 0 ÐBÑ has any horizontal asymptote, it has only one horizontal asymptote. If the degree of :ÐBÑ is strictly larger than the
    degree of ;ÐBÑ then 0 ÐBÑ does not have any horizontal asymptote. If the degrees of :ÐBÑ and ;ÐBÑ are the same, then the
                                                                        :8
    horizontal asymptote is the line C œ 5 where the constant 5 œ            where :8 is the leading coefficient of :ÐBÑ and ;8 is the
                                                                        ;8
    leading coefficient of ;ÐBÑÞ If the degree of :ÐBÑ is less than the degree of ;ÐBÑ then the line C œ !, which is the B-axis, is the
    horizontal asymptote. In this case, and in this case only, the sign analysis in step 2Ñ will tell you whether the extreme ends of the
    graph lie either above or below the asymptote. Otherwise, pick a large positive number and a large negative number and compare
    0 ÐBÑ with 5 to determine whether the graph lies above or below the ends of the line that is the horizontal asymptote. You can
    determine if and where 0 ÐBÑ crosses or touches its horizontal asymptote by solving the equation 0 ÐBÑ œ 5 where C œ 5 is the
    horizontal asymptote. The equation 0 ÐBÑ œ 5 may have no solutions, or it may have one or more solutions.

5). If the degree of the original :ÐBÑ polynomial is larger than the degree of the original ;ÐBÑ polynomial, then from step 4) we know
    the graph of 0 ÐBÑ does not have a horizontal asymptote. If the degree of :ÐBÑ is exactly one more than the degree of ;ÐBÑ then
    the graph of 0 ÐBÑ may have what is called a slant asymptote. Even in this case, a rational function does not have to have a slant
    asymptote, but like a horizontal asymptote, if 0 ÐBÑ has a slant asymptote, it has only one slant asymptote. The slant asymptote is
    the line C œ 7B  ,ß where 7B  , is the quotient polynomial that results when you perform long division of polynomials by
    dividing ;ÐBÑ into :ÐBÑÞ After performing the long division it helps to write 0 ÐBÑ œ 7B  ,  </7+38./<ÐBÑ where  ;ÐBÑ
    </7+38./<ÐBÑ is another polynomial whose degree is strictly smaller than that of ;ÐBÑÞ We implicitly assume the degree of ;ÐBÑ
    is at least 1. If </7+38./<ÐBÑ is the constant 0, then the graph of 0 ÐBÑ does not have a slant asymptote because 0 ÐBÑ is just a
    line with holes in it wherever ;ÐBÑ œ !Þ If </7+38./<ÐBÑ is not the constant !, then the line C œ 7B  , is a true slant
    asymptote for 0 ÐBÑÞ You determine whether the graph is above or below the slant asymptote just like you do for a horizontal
    asymptote by picking a large positive number and a large negative number for B. The sign of the expression </7+38./<ÐBÑ for large
                                                                                                                      ;ÐBÑ
    B determines whether the graph of 0 ÐBÑ is above or below the line C œ 7B  ,Þ The graph of 0 ÐBÑ will touch or cross its slant
    asymptote line wherever </7+38./<ÐBÑ œ ! and ;ÐBÑ Á !Þ In particular, if </7+38./<ÐBÑ is a polynomial of degree # or more,
    then the graph of 0 ÐBÑ could touch or cross its slant asymptote more than once. It is also possible that simultaneously
    </7+38./<ÐBÑ œ ;ÐBÑ œ ! in which case the fraction </7+38./<ÐBÑ can and should be reduced.
                                                                ;ÐBÑ


Try graphing the following rational functions:

             B#  *                        2                         B%  $B$  $B#  (B  '                        #B$  &B#  #!B  %
 1 0 ÐBÑ œ              2 0 ÐBÑ œ     $  #B#  $B
                                                         3 0 ÐBÑ œ                                      4 0 ÐBÑ œ
             B$                    B                                      B#  B  #                                   B#  $B  "!


             B#  B  '                        B$  %B#  $B                           B$  $B#  )B  #!                          B%  "
 5 0 ÐBÑ œ                         6 0 ÐBÑ œ                              7 0 ÐBÑ œ                                    8 0 ÐBÑ œ
              B#  #B                           B#  %B  $                               #B#  #B  #%                            B$  "
Solutions:

             B#  *       ÐB  $ÑÐB  $Ñ
 1 0 ÐBÑ œ             œ                    œ B  $ provided B Á $Þ
              B$            ÐB  $Ñ
This graph is just a line with a hole in it at B œ $Þ




                   2               #                 #
 2 0 ÐBÑ œ                 œ                œ                 . The sign analysis for 0 ÐBÑ appears below.
             B$  #B#  $B   BÐB#  #B  $Ñ   BÐB  $ÑÐB  "Ñ




Note that the B-axis is a horizontal asymptote and the sign analysis can be used to tell whether the extremes of the graph are above or
below this line. The graph of 0 ÐBÑ appears below. The three vertical asymptotes are B œ "à B œ !; and B œ $Þ
              B%  $B$  $B#  (B  '
 3 0 ÐBÑ œ                             . We begin by guessing a couple of zeros for the numerator polynomial and applying synthetic
                     B#  B  #
substitution. This helps us completely factor the numerator. The denominator is also easy to factor.

                                         #                                                   "
           "    $   $    (     '                                          "   "   &    $
               #   #    "!    '                                              "    #   $
           "    "   &     $     !                                          "   #   $    !

          ÐB  #ÑÐB$  B#  &B  $Ñ   ÐB  #ÑÐB  "ÑÐB#  #B  $Ñ   ÐB  #ÑÐB  "ÑÐB  $ÑÐB  "Ñ
0 ÐBÑ œ                             œ                             œ
                ÐB  #ÑÐB  "Ñ               ÐB  #ÑÐB  "Ñ                ÐB  #ÑÐB  "Ñ

œ ÐB  $ÑÐB  "Ñ provided B Á # and B Á "Þ The sign analysis for 0 ÐBÑ appears below.




Note that the graph of 0 ÐBÑ is just a parabola with two holes in it.




               #B$  &B#  #!B  %
 4 0 ÐBÑ œ                         . We can guess one root to the numerator polynomial by peforming synthetic substitution.
                   B#  $B  "!

                                                                        #
                                                #   &   #!     %
                                                    %    ")    %
                                                #   *    #     !

          ÐB  #ÑÐ#B#  *B  #Ñ
0 ÐBÑ œ
              ÐB  &ÑÐB  #Ñ

To completely factor the numerator we must solve the quadratic equation: 2B#  *B  # œ !.
                                                 * „ È*(                                                *  È*(
When we do this, we find the two roots are B œ            Þ We will call the two roots + and , where + œ          ¸ %Þ("##
           *  È*(
                                                     %                                                       %
and , œ             ¸ !Þ#"##. Then + and , are critical numbers and we may write 0 ÐBÑ in its completely factored form as:
               %

          ÐB  #ÑÐB  +ÑÐB  ,Ñ
0 ÐBÑ œ
             ÐB  &ÑÐB  #Ñ



The sign analysis for 0 ÐBÑ is:




Because the points 5 and + œ %Þ("## are so close together it may be difficult to read the space between these points, but the sign
over the short interval between & and %Þ("## is +.

We note that the line B œ & is the only vertical asymptote for 0 ÐBÑ. 0 ÐBÑ is undefined and has a hole at B œ #Þ Since the degree of
the numerator is exactly one more than the degree of the denominator we look for a slant asymptote by performing long division.

                             #B         "
 B#        $B        "!    #B$       &B#         #!B        %
                             #B$       'B#         #!B
                                         B#         !B         %
                                         B#         $B        "!
                                                      $B         '


                         $B  '                    $ÐB  #Ñ                    $
0 ÐBÑ œ Ð#B  "Ñ                  œ Ð#B  "Ñ                 œ Ð#B  "Ñ          provided B Á #
                      B#  $B  "!              ÐB  &ÑÐB  #Ñ              ÐB  &Ñ

This shows the line C œ #B  " is a slant asymptote. The graph of 0 ÐBÑ appears below. The slant asymptote is not drawn.
             B#  B  '   ÐB  $ÑÐB  #Ñ
 5 0 ÐBÑ œ              œ
              B#  #B        BÐB  #Ñ

The sign analysis for 0 ÐBÑ is:




                                                                                     "
Since the degrees of both the numerator and denominator are the same, the line C œ     œ " is a horizontal asymptote.
                                                                                     "

To determine whether the graph is above or below the horizontal asymptote on the far left, we let B œ "!! and compute
    Ð103Ñ † Ð*)Ñ     10094
Cœ                  œ         ¸ !Þ*89607  "Þ On the left, the graph is below the line C œ "Þ
    Ð"!!ÑÐ"!#Ñ       "!#!!

To determine whether the graph is above or below the horizontal asymptote on the far right, we let B œ "!! and compute
    Ð*(Ñ † Ð102Ñ   9894
Cœ               œ       ¸ 1.00959  1Þ On the right, the graph is above the line C œ "Þ
     Ð"!!ÑÐ98Ñ     9800

The graph of 0 ÐBÑ appears below. The lines B œ ! and B œ # are both vertical asymptotes.
             B$  %B#  $B   BÐB#  %B  $Ñ   BÐB  $ÑÐB  "Ñ
 6 0 ÐBÑ œ                 œ                œ                 œ B provided B Á " and B Á $Þ
              B#  %B  $    ÐB  $ÑÐB  "Ñ    ÐB  $ÑÐB  "Ñ

This graph is just a line with two holes in it. Note that even though the degree of the numerator is exactly one more than the degree of
the denominator, there is no slant asymptote. The graph of 0 ÐBÑ appears below.




             B$  $B#  )B  #!
 7 0 ÐBÑ œ
                #B#  #B  #%

We begin by trying to factor the numerator. We guess B œ # is a root and try synthetic substitution.

                                                              #
                                    "   $     )     #!
                                         #     #    #!
                                    "   "    "!      !

                               ÐB  #ÑÐB#  B  "!Ñ   ÐB  #ÑÐB#  B  "!Ñ
So we may now write 0 ÐBÑ œ                         œ
                                  #ÐB#  B  "#Ñ         #ÐB  %ÑÐB  $Ñ


           "  È%"                          "  È%"
The discriminant of the quadratic B#  B  "! œ ! is %" so the remaining two roots of the numerator are irrational.

We let + œ             ¸ 2.70156 and , œ             ¸ 3.70156 denote the two roots to this quadratic.
               #                                #

               ÐB  #ÑÐB  +ÑÐB  ,Ñ
Then 0 ÐBÑ œ                         and the sign analysis for 0 ÐBÑ is:
                  #ÐB  %ÑÐB  $Ñ




The two undefined points correspond to vertical asymptotes, B œ $ and B œ %Þ
Since the degree of the numerator polynomial is exactly one more than that of the denominator, we perform long division to see if
there is a slant asymptote for this graph.

                                           "
                                           #B         "
              #B#          #B       #%   B $
                                                     $B#          )B         #!
                                           B$         B#          "#B
                                                       #B#         %B        #!
                                                       #B#         #B        #%
                                                                     #B         %

Since the remainder is nonzero we have a slant asymptote and we may write

                           #B  %                        #ÐB  #Ñ                        B#
0 ÐBÑ œ Ð " B  "Ñ 
          #                            œ Ð " B  "Ñ 
                                           #                          œ Ð " B  "Ñ 
                                                                          #
                       #ÐB  %ÑÐB  $Ñ                #ÐB  %ÑÐB  $Ñ                ÐB  %ÑÐB  $Ñ

The line C œ " B  " is the slant asymptote.
             #


The graph of 0 ÐBÑ appears below. This particular function has the unusual property that it crosses its slant asymptote at the point
Ð#ß !Ñ and this point is also an B-intercept in the graph. The graph lies below the slant asymptote when B  $ and it lies above the
slant asymptote when B  %Þ
              B%  "     ÐB#  "ÑÐB#  "Ñ     ÐB  "ÑÐB  "ÑÐB#  "Ñ   ÐB  "ÑÐB#  "Ñ
 8 0 ÐBÑ œ      $"
                     œ           #  B  "Ñ
                                            œ            #  B  "Ñ
                                                                     œ                 provided B Á "Þ
              B        ÐB  "ÑÐB               ÐB  "ÑÐB                 B#  B  "



The sign analysis for 0 ÐBÑ is:




We can see that at B œ " we have a hole in the graph.

Since the degree of the numerator is exactly one more than that of the denominator we perform long division to write 0 ÐBÑ in the
form:

              B"             ÐB  "Ñ † "             "
0 ÐBÑ œ B           œB                        œB #     provided B Á "Þ
              B$  "     ÐB  "Ñ † ÐB#  B  "Ñ    B B"

From this last equation we can determine that 0 ÐBÑ has a slant asymptote that is the line C œ BÞ
The graph of 0 ÐBÑ is shown below. It is interesting to note that 0 ÐBÑ lies all on one side (above) its slant asymptote.

				
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