Download Chapter 3- Polynomial and Rational Functions

16 Days
One Day

How does a mountain range relate to
polynomials and their graphs?

Definition: If f is a polynomial function and
f(a) ≠f(b) for a<b, then f takes on every value
between f(a) and f(b) in the interval [a, b].
Facts about IVT:
 f is continuous from f(a) to f(b)
 If f(a) and f(b) have opposite signs (1 positive
and 1 negative), there is at least one number
c between a and b such that f(c)=0.

Step 1
◦ Identify a and b values

Step 2
◦ Plug a and b into f(x) and solve

Step 3
◦ If f(a) and f(b) have opposite signs, you know that
f(c)=0 for at least one real number between a & b


Using the Intermediate Value Theorem
Show that f(x)= x⁵+2x⁴-6xᶟ+2x-3 has a zero
between 1 and 2.
◦ Step 1: identify a and b
 a=1, b=2
◦ Step 2: solve f(a) and f(b)
 f(a)= f(1)= 1+2-6+2-3=-4
 f(b)=f(2)=32+32-48+4-3=17
◦ Step 3: do f(a) and f(b) have opposite signs?
 f(a)=-4 and f(b)=17, yes!
◦ Conclusion:
 there is a c where f(c)=0 between 1 and 2

Sketching graphs of degree greater than 2
◦ Step 1
 Find the zeros
◦ Step 2
 Create a table showing intervals of positive or negative
signs for f(x).
◦ Step 3
 Find where f(x)>0 and where f(x)<0

Construct a table showing intervals of
positive or negative values of f(x)
Interval
(4, )
 ,1
(1,4)
Sign of (x-1)
-
+
+
Sign of (x-4)
-
-
+
Sign of f(x)
+
-
+
Position of
graph
Above xaxis
Below xaxis
Above xaxis
•This means that f(x)>0 if x is in  ,1and (4, )
•This means that f(x)<0 if x is in (1,4) .

Sketching the graph of a polynomial function
of degree 3
◦ Ex

f ( x)  x 3  x 2  4 x  4
Step 1: Find Zeros
◦ Group terms
f ( x)  ( x3  x 2 )  (4 x  4)
◦ Factor out x² and -4
f ( x)  x 2 ( x  1)  4( x  1)
◦ Factor out (x+1)
f ( x)  ( x 2  4)( x  1)
◦ Difference of squares
f ( x)  ( x  2)( x  2)( x  1)
◦ Therefore, the zeros are -2, 2, and -1

Step 2: Create a Table
Interval
Sign of
x+2
Sign of
x+1
Sign of
x-2
Sign of
f(x)
Position
of graph
(,2)
(2,1)
(1,2)
(2, )



f(x)>0 if x is in(2,1)  (2, )
f(x)<0 if x is in(,2)  (1,2)
My version of graph (there are multiple ways
to do this graph)
y
x

Ex
Find the intervals

Sketch graph

◦ f(x) is below x-axis when x is in (,3)and (0,2)
◦ f(x) is above x-axis when x is in (3,1), (1,0), and (2, )
y
x

Steps
◦ Step 1: Assign f(x) to Y1 on a graphing calculator
◦ Step 2: Set x and y bounds large enough to see
from [-15,-15] and [-15,15]
 This allows us to gauge where the zeros lie from a
broad perspective
◦ Step 3: Readjust bounds once you know where
zeros are more likely to be found
◦ Step 4: Use zero or root feature on calculator to
estimate the real zero





Ex/ Estimate the real zeros of
f ( x)  x 3  4.6 x 2  5.72 x  0.656
Graph this function as Y1 in
calculator
Set bounds as [-15,15] by
[-15,15].
Readjust to where zeros
might exist, you may use
[-1,3] by [-1,3]
Find actual root by using zero or
foot feature on your calculator
Actual root is 0.127


P. 227
#1,5,7,15,16,18,23,25,34,41,50
If you need help, check with the
person beside you.
Two Days




If g(x) is a factor of f(x), then f(x) is divisible
by g(x).
4
x
For example,  81 is divisible by each of the
2
2
x

9
,
x
 9, x  3, and x  3 .
following
4
2
x

81
x
Notice that
is not divisible by  3x  1 .
However, we can use Polynomial Long
Division to find a quotient and a remainder.

We will first divide 178 by 8 to find the
quotient and remainder.

Step 1: Make sure the polynomial is written in descending order. If any
terms are missing, use a zero to fill in the missing term (this will help
with the spacing).

Step 2: Divide the term with the highest power inside the division
symbol by the term with the highest power outside the division symbol.

Step 3: Multiply (or distribute) the answer obtained in the previous step
by the polynomial in front of the division symbol.

Step 4: Subtract and bring down the next term.

Step 5: Repeat Steps 2, 3, and 4 until there are no more terms to bring
down.


Step 6: Write the final answer. The term remaining after the last subtract
step is the remainder and must be written as a fraction in the final
answer.
http://www.youtube.com/watch?v=smsKMWf8ZCs

Find the quotient of x 4  81 and x 2  3x  1 .
x 2  3x  1 x 4  0 x 3  0 x 2  0 x  81
x 2  3x
x 2  3 x  1 x 4  0 x 3  0 x 2  0 x  81
x 4  3x 3  x 2
x2
x 2  3x  1 x 4  0 x 3  0 x 2  0 x  81
x 4  3x 3  x 2
x2
x 2  3 x  1 x 4  0 x 3  0 x 2  0 x  81
x 4  3x 3  x 2
 3x 3  x 2  0 x
 3x 3  x 2  0 x
 3x 3  9 x 2  3x
8 x 2  3 x  81

Once we have a quotient and remainder, we
must write our final answer..
x 2  3x  8
x 2  3x  1 x 4  0 x 3  0 x 2  0 x  81
x 4  3x 3  x 2
 3x  x  0 x
3
2
 3x 3  9 x 2  3x
8 x 2  3 x  81
8 x 2  24 x  8
 21x  89
Final Answer:
x 4  81
 21x  89
2
 x  3x  8  2
2
x  3x  1
x  3x  1

3
2
Divide 2 x  7 x  2 x  9 by 2 x  3.
2 x  3 2 x3  7 x 2  2 x  9
x2  2x  2
2 x  3 2 x3  7 x 2  2 x  9
2 x 3  3x 2
4x2  2x
4x2  6x
 4x  9
 4x  6
15
2 x3  7 x 2  2 x  9
15
Solution :
 x2  2x  2 
2x  3
2x  3

3
2
Divide x  3x  4 x  12 by x 2  x  6.
x 2  x  6 x 3  3x 2  4 x  12
x2
x 2  x  6 x 3  3 x 2  4 x  12
x3  x 2  6x
2 x 2  2 x  12
2 x 2  2 x  12
0
x 3  3x 2  4 x  12
Solution :
 x2
2
x  x6


If a polynomial f(x) is divided by x-c, the the
remainder is f(c).
Example:
If f ( x)  x3  3x 2  x  5, use the remainder theorem to find f(-2).
x  2 x 3  3x 2  x  5
x2  x 1
x  2 x 3  3x 2  x  5
x3  2 x 2
 x2  x
Therefore, f (-2)  3.
 x2  2x
 x5
x2
3


A polynomial f(x) has a factor x-c if and only
if f(c)=0.
Example:
Show that x - 2 is a factor of f ( x)  x3  4 x 2  3x  2.

Find a polynomial f(x) of degree 3 that has
zeros 3, -1, and 1.

pg 236 (# 1,2,8,17-20)

http://www.mesacc.edu/~scotz47781/mat12
0/notes/divide_poly/long_division/long_divis
ion.html

Synthetic Division is
the “shortcut”
method for dividing
polynomials.
Let' s compare dividing x 4  x 3  2x  2 by x  1 using both long
division and synthetic division.
Long Division
Synthetic Division
x  2x  2x  4
x  1 x 4  x 3  0 x 2  2x  2
3
2
1
1 1 0  2 2
1 2  2 4
x4  x3
 2x 3  0 x 2

  2x 3  2x 2

1 2 2 4 6


x3  2 x 2  2 x  4
2x 2  2x
 2x 2  2x

 4x  2
  4 x  4 
6

Use synthetic division to find the quotient q(x)
and remainder r if f ( x)  4 x 4  x3  3x 2  5, is divided by x  3.

Using synthetic division to find zeros.
◦ What must we show for a value to be a zero of f(x)?
Think Factor Theorem…
Show that -11 is a zero of f ( x)  x3  8x 2  29 x  44.

Use synthetic division to find f(3) if
f ( x)  3x 4  x3  3x 2  x  15

You should now recognize the following
equivalent statements. If f(a)=b, then:
◦ 1. The point (a,b) is on the graph of f.
◦ 2. The value of f at x=a equals b; ie f(a)=b.
◦ 3. If f(x) is divided by x-a, then the remainder is b.

Additionally, if b=0 then the following are also
equivalent.
◦
◦
◦
◦
1.
2.
3.
4.
The
The
The
The
number a is a zero of the function f.
point (a,0) is on the graph of f; a is an x-int.
number a is a solution of the equation f(x)=0.
binomial x-a is a factor of the polynomial f(x).

pg 237 (# 21-29 odd,32,35,38)
Two Days

If a polynomial f(x) has positive degree and
complex coefficients, then f(x) has at least
one complex zero.

If f(x) is a polynomial of degree n>0, then
there exist n complex numbers c1 , c2 ,..., cn such
that
f ( x)  a( x  c1 )( x  c2 )...( x  cn ),

Where a is the leading coefficient of f(x).
Each number ck is a zero of f(x).

A polynomial of degree n>0 has at most n
different complex zeros.
f
f
f

Find the zeros of the polynomial, state the
multiplicity of each zero, find the y-int, and
sketch the graph.
f ( x) 
1
( x  2)( x  3)3 ( x  1) 2
25

Find a polynomial f(x) in factored for that has
degree 3; has zeros 3, 1, and -1; and
satisfies f(-2)=3.


If f(x) is a polynomial of degree n>0 and if a
zero of multiplicity m is counted m times,
then f(x) has precisely n zeros.
Exactly how many zeros exist for the
following polynomial?
f ( x)  3( x  2)( x  3)3 ( x  1) 2 ( x  2)5

pg 249 (# 1-7 odd, 11-15 odd, 19-23 odd)

If f(x) is a polynomial with real coefficients
and a nonzero constant term, then:
◦ 1. The number of positive real zeros of f(x) either
is equal to the number of variations of sign in f(x)
or is less than that number by an even integer.
◦ 2. The number of negative real zeros of f(x) either
is equal to the number of variations of sign in f(-x)
or is less than that number by an even integer.



We can use the N-Zeros Thm to determine
the total number of zeros possible.
Use Descartes’ Rule of Signs to determine the
total possible number of positive and
negative zeros (and all lesser combinations).
Any unaccounted for zeros must then be
imaginary zeros.

Find the total number of zeros possible for
f(x)=0 where
f ( x)  3 x 5  2 x 4  3 x 3  6 x  5
Number of positive solutions
Number of negative solutions
Number of imaginary solutions
Total number of solutions

Suppose that f(x) is a polynomial with real
coefficients and a positive leading coefficient
and that f(x) is divided synthetically by x-c.
Then:
◦ 1. If c>0 and if all numbers in the third row of the
division process are either positive or zero, the c is a
upper bound for the real zeros of f(x).
◦ 2. If c<0 and if the numbers in the third row of the
division process are alternately positive and negative
(a 0 can be counted as positive OR negative), then c
is a lower bound for the real zeros of f(x).

Determine the smallest and largest integers
that are upper and lower bounds for the real
solutions of f(x)=0 if
f ( x)  x 4  x 3  2 x 2  3 x  6

Suppose f ( x)  an x  an1 x  ...  a1 x  a0 is a
polynomial with real coefficients. All of the
real zeros of f(x) are in the interval
n
( M , M ),
wher e M 
n 1
max  an , an 1 ,..., a1 , a0 
an
1

Determine the interval in which all of the real
solutions of f(x)=0 exist if
f ( x)  3 x 5  2 x 4  3 x 3  6 x  5


Finding Zeros Practice WS (# 1-9)
Quiz Tomorrow
Two Days

If a polynomial f(x) of degree n>1 has real
coefficients and if z  a  bi with b≠0 is a
complex zero of f(x), then the conjugate
.z  a  bi is also a zero of f(x).

Find a polynomial of degree 3 with real
coefficients where 2 and (3-i) are zeros of the
polynomial.

Every polynomial with real coefficients and
positive degree n can be expressed as a
product of linear and quadratic factors with
real coefficients such that the quadratic
factors are irreducible over ℝ.

If the polynomial
f ( x)  an x n  an 1 x n 1  ...  a1 x  a0
has integer coefficients and if c/d is a rational
zero of f(x) such that c and d have no
common prime factor, then:
1. the numerator, c, of the zero is a factor of
the constant term
2. the denominator d of the zero is a factor of
the leading coefficient

All possible rational zeros of a polynomial are
of the form:
Possible Rational Zeros 
factors of the constant t erm a0
factors of the leading coefficien t an

Find all possible rational zeros using rational
zeros theorem. Then factor and find any
remaining zeros.
f ( x)  3x 4  2 x3  x 2  12 x  4


pg 259 (# 7,18,19,23,24,29)
Top Shelf pg 14 (#1,2,5,6,8; Due 11/11)

Prentice Hall p 46 Fundamental Thm Algebra
WS (# 1-4,8,9,10,13,17)
Four Days

A function f is a rational function if
g ( x)
f ( x) 
h( x )
where g(x) and h(x) are polynomials.

The domain of f consists of all real numbers
except the zeros of the denominator h(x).

Find the domain of the following rational
functions:
1
f ( x) 
2x  3
D :  , 32    32 , 
4x 1
g ( x)  2
x  3x  2
x 3  27
g ( x)  2
x 9
D :  ,2   2,1  1, 
D:

Simplify and graph the following rational
expression:
x 2  5x  6
f ( x) 
x2

Definition: The line x=a is a vertical
asymptote for the graph of the function f if
f ( x)   or f ( x)   as x  a  or x  a 
as x approaches a from either the left of the
right.

Create an xy-table and look at the values of
f(x) as x approaches the vertical asymptote.
1
f ( x) 
x 1

Definition: The line y=c is a horizontal
asymptote for the graph of the function f if
f ( x)  c as x   or x  
Let f be a rational function given by
n x  an x n  an 1 x n 1    a1 x  a0
f x  

d x  bm x m  bm 1 x m 1    b1 x  b0
where nx  and d  x  are polynomial s with no common factors.
1. When n  m, the x - axis  y  0  is the horizontal asymptote.
an
2. When n  m, the line y 
(ratio of leading coeffients ) is
bm
the horizontal asymptote.
3. When n  m, there is no horizontal asymptote.
Let f be a rational function given by
n x  an x n  an 1 x n 1    a1 x  a0
f x  

d  x  bm x m  bm 1 x m 1    b1 x  b0
where n x  and d  x  are polynomial s with no common factors.
1. When the degree of the numerator is less than the
degree of the denominato r,  y  0  is the horizontal asymptote.
2. When the degree of the denominato r is equal to
an
the degree of the numerator, the line y 
bm
(ratio of leading coeffients ) is the horizontal asymptote.
3. When the degree of the numerator is greater
than the degree of the denominato r, there is no horizontal
asymptote.

If a factor can be reduced that would cause a
vertical asymptote there will be a hole instead
of a vertical asymptote!
x 1
f ( x)  2
x  5x  4
2
f ( x) 
x  14
5 x  10
3x 2  1
f ( x)  2
x 9
f ( x) 
x 1
x 2  5x  6
f ( x) 
x 1
x 2  5x  6

pg 276 (# 1,2,3,5,6,8,11,15)
◦ Only find the Domain, VA, HA, and reduced form of
the rational function.
◦ We will graph them tomorrow!
g  x  an x n  an 1 x n 1    a1 x  a0
Assume that f  x  

h x  bm x m  bm 1 x m 1    b1 x  b0
where g  x  and h x  are polynomial s with no common factors.
1. Find the x - intercepts , i.e. all real zeros of g(x).
Plot the correspond ing points on the x - axis.
2. Find the real zeros of the denominato r h(x). For each real
zero a, sketch the vertical asymptote x  a with a dashed line.
3. Find the y - intercept f( 0 ), if it exists, and plot the point ( 0,f( 0 )).
4. Apply the theorem on horizontal asymptotes . If there is a horizontal
asymptote y  c, sketch it with a dashed line.
5. If there is a HA y  c, determine whether it intersects the graph.
The x - coordinate s of the points of intersecti on are the solutions
to the equation f(x)  c. Plot these points, if they exist.
6. Sketch the graph of f in each of the regions in the xy - plane determined
by the VAs in guideline 2. If necessary, use the sign of specific function
value s to tell whether t he graph is above or below the x - axis or the HA.
Use guideline 5 to decide whether t he graph approaches the HA from above or below.
4x 1
f ( x) 
2x  3
x 1
f ( x)  2
x  2x  3
x 2  25
f ( x)  2
x  3x  4

pg 46 Dennison PreCalc (find VA, HA, SA, yint, graph # 5, 8) pg 276(# 21, 25)

pg 48 Dennison PreCalc (find VA, HA, SA, yint) pg 277 (# 29,31)


pg 276 (# 7,12,32,35,41,42)
Quiz on Graphs Tomorrow
◦ (No Graphing Calculators!)


Tomorrow nights homework:
pg 279 (# 1,4,9,10,11,13,16-19,24,25,27,
29,31,35)