Polynomials and Graphs through Real-World Analogies
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16 Days
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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)
•
This means that f(x)>0 if x is in
•
This means that f(x)<0 if x is in .
Sketching the graph of a polynomial function
of degree 3
◦
Ex
Step 1: Find Zeros
◦
Group terms
◦
Factor out x² and -4
◦
Factor out (x+1)
◦
Difference of squares
◦
Therefore, the zeros are -2, 2, and -1
Step 2: Create a Table
f(x)>0 if x is in
f(x)<0 if x is in
My version of graph (there are multiple ways
to do this graph)
x
y
Ex
Find the intervals
◦
f(x) is below x-axis when x is in
◦
f(x) is above x-axis when x is in
Sketch graph
x
y
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
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
We will first divide 2172 by 3 to find the
quotient and remainder.
If g(x) is a factor of f(x), then f(x) is divisible
by g(x).
For example, is divisible by each of the
following .
Notice that is not divisible by .
However, we can use Polynomial Long
Division to find a quotient and a 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 and .
Once we have a quotient and remainder, we
must write our final answer..
Final Answer:
Divide by .
Divide by .
If a polynomial f(x) is divided by x-c, the the
remainder is f(c).
Example:
A polynomial f(x) has a factor x-c if and only
if f(c)=0.
Example:
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.
Use synthetic division to find the quotient q(x)
and remainder r
Using synthetic division to find zeros.
◦
What must we show for a value to be a zero of f(x)?
Think Factor Theorem…
Use synthetic division to find f(3) if
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. The number a is a zero of the function f.
◦
2. The point (a,0) is on the graph of f; a is an x-int.
◦
3. The number a is a solution of the equation f(x)=0.
◦
4. The 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 such
that
Where a is the leading coefficient of f(x).
Each number is a zero of f(x).
A polynomial of degree n>0 has at most n
different complex zeros.
Find the zeros of the polynomial, state the
multiplicity of each zero, find the y-int, and
sketch the graph.
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?
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
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
Suppose is a
polynomial with real coefficients. All of the
real zeros of f(x) are in the interval
Determine the interval in which all of the real
solutions of f(x)=0 exist if
Finding Zeros Practice WS (# 1-9)
Quiz Tomorrow
Two Days
If a polynomial f(x) of degree n>1 has real
coefficients and if with b≠0 is a
complex zero of f(x), then the conjugate
. 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
has integer coefficients and if 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:
Find all possible rational zeros using rational
zeros theorem. Then factor and find any
remaining zeros.
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
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)
. A
zero of
h(x)
that reduces results in a hole in
the graph instead of a vertical asymptote.
Find the domain of the following rational
functions:
Simplify and graph the following rational
expression:
Definition: The line x=a is a vertical
asymptote for the graph of the function
f
if
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.
Definition: The line y=c is a horizontal
asymptote for the graph of the function
f
if
If a factor can be reduced that would cause a
vertical asymptote there will be a hole instead
of a vertical asymptote!
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!
pg 46 Dennison PreCalc (find VA, HA, SA, y-
int, graph # 5, 8) pg 276(# 21, 25)
pg 48 Dennison PreCalc (find VA, HA, SA, y-
int) 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)
New Chapter that is all about graphing polynomials greater than degree 2
Standards Met:
Explore the relationship between mountain ranges and polynomials, and learn how to apply the Intermediate Value Theorem to find zeros of polynomial functions. This guide covers concepts like the Interval Value Theorem, sketching graphs of higher-degree polynomials, and constructing tables to analyze a function's behavior. Discover practical insights using visuals to deepen your understanding of polynomial functions and their graphs.
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Presentation Transcript
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)= x +2x -6x +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 ( ) 1 , Interval , 4 ( ) ) 4 , 1 ( Sign of (x-1) - - + + Sign of (x-4) - - - - + Sign of f(x) + - - + Position of graph Above x- axis Below x- axis Above x- axis ) 1 , and ) 4 , 1 ( ( This means that f(x)>0 if x is in This means that f(x)<0 if x is in . , 4 ( )
Sketching the graph of a polynomial function of degree 3 Ex Step 1: Find Zeros Group terms Factor out x and -4 Factor out (x+1) Difference of squares Therefore, the zeros are -2, 2, and -1 = + 3 2 ( ) 4 4 f x x x x = + + 3 2 ( ) ( ) ( 4 ) 4 f x x x x = ) ) 1 + )( ) 1 + ) 1 + 2 ( ) x ( x ( 4 x f x ( x x x = 2 ( 4 f = + ) 1 + ( ) ( 2 )( 2 )( f x x x x
Step 2: Create a Table Interval ) 2 , 2 ) 1 , 2 ( ) 2 , 1 ( ( , ( ) Sign of x+2 Sign of x+1 Sign of x-2 Sign of f(x) Position of graph
f(x)>0 if x is in f(x)<0 if x is in My version of graph (there are multiple ways to do this graph) y , 2 ) 1 , ) 2 , 1 ( ( , 2 ( ) 2 ) ( x
Ex Find the intervals f(x) is below x-axis when x is in f(x) is above x-axis when x is in Sketch graph , 3 ) 3 ), 1 ( , ) 2 , 0 ( ), 0 , 1 and ( ( , 2 ( ) and 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 6 . 4 ) ( = x x f + 3 2 . 5 72 . 0 656 x x 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.
We will first divide 2172 by 3 to find the quotient and remainder.
If g(x) is a factor of f(x), then f(x) is divisible by g(x). For example, is divisible by each of the following . 4 2 x 81 , 9 + x , 9 , 3 + 2 and 3 x x x Notice that is not divisible by . 4 + x + 2 81 3 1 x x However, we can use Polynomial Long Division to find a quotient and a remainder.
Step 1: terms are missing, use a zero to fill in the missing term (this will help with the spacing). Step 1: Make sure the polynomial is written in descending order. If any Step 2: symbol by the term with the highest power outside the division symbol. Step 2: Divide the term with the highest power inside the division Step 3: by the polynomial in front of the division symbol. Step 3: Multiply (or distribute) the answer obtained in the previous step Step 4: Step 4: Subtract and bring down the next term. Step 5: down. Step 5: Repeat Steps 2, 3, and 4 until there are no more terms to bring Step 6: step is the remainder and must be written as a fraction in the final answer. Step 6: Write the final answer. The term remaining after the last subtract http://www.youtube.com/watch?v=smsKMWf8ZCs
Divide by . + + + 3 2 2 + x 2 7 2 9 x x x 3 + + + 2 2 2 2 9 x x x x + + 3 2 + + + + 2 3 2 7 3 2 x x 2 3 2 7 2 9 x x x x + 3 2 2 3 x x + 2 4 2 x x + 2 4 6 x x + 4 9 x 4 6 x 15 + + + 3 2 2 7 2 9 15 x x x x = + + 2 Solution : 2 2 x x + + 2 3 2 3 x
Divide by . + 3 2 3 4 12 x x x + x 2 6 x + 2 x + + 2 3 2 + + 2 3 2 6 3 4 12 x x x x x 6 3 4 12 x x x x x + 3 2 6 + x x x 2 2 2 12 x x + 2 2 2 12 x x 0 + x 3 2 3 x 4 12 x x x = + Solution : 2 x + 2 6
If a polynomial f(x) is divided by x-c, the the remainder is f(c). Example: the use 5, 3 ) ( If + + = x x x x f 3 2 remainder theorem to find f(-2). + + + 3 2 2 3 5 x x x x = Therefore, (-2) -17. f
A polynomial f(x) has a factor x-c if and only if f(c)=0. Example: = + + 3 2 Show that x - a is 2 factor of ( ) 4 3 2. f x x x x
Find a polynomial f(x) of degree 3 that has zeros 3, -1, and 1.
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.
4 3 + + Let' compare s dividing x x 2 x 2 by x using 1 both long division Long Division synthetic and Long Division division. Synthetic Division Synthetic Division + x 3 2 x 2 + x 2 x 4 + 1 1 1 0 2 2 + 4 3 2 x 1 x x 0 x 2 2 1 2 2 4 4 3 x x + 3 2 1 2 2 4 6 2 x 0 x ( ) 3 2 2 x 2 x 3 x 2 2 + x 2 2 x 2 x 2 4 x ( ) + 2 2 x 2 x + 4 x 2 ( ) 4 x 4 6
Use synthetic division to find the quotient q(x) and remainder r ) ( if = x f + + 4 3 2 4 3 divided is 5, by . 3 x x x x
Using synthetic division to find zeros. What must we show for a value to be a zero of f(x)? Think Factor Theorem = + + 3 2 Show that - a is 11 zero of ( ) 8 29 44. f x x x x
Use synthetic division to find f(3) if 3 ) ( = x x f + + 4 3 2 3 15 x x x
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. The number a is a zero of the function f. 2. The point (a,0) is on the graph of f; a is an x-int. 3. The number a is a solution of the equation f(x)=0. 4. The binomial x-a is a factor of the polynomial f(x).
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 such that )( ( ) ( 1 x c x a x f = , ,..., c c nc 1 2 )...( ), c x nc 2 Where a is the leading coefficient of f(x). Each number is a zero of f(x). kc
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. 1 ) ( = x x f ) 3 ) 1 + 3 2 ( 2 )( ( x x 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? = ) 3 ) 1 + + 3 2 5 ( ) ( 3 2 )( ( ( ) 2 f x x x x x
If f(x) is a polynomial with real coefficients and a nonzero constant term, then: 1. The number of positive is equal to the number of variations of sign in f(x) or is less than that number by an even integer. positive real zeros of f(x) either 2. The number of negative is equal to the number of variations of sign in f(-x) or is less than that number by an even integer. negative real zeros of f(x) either
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 2 3 ) ( + = x x x f + 5 4 3 3 6 5 x x 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 = + + 4 3 2 ( ) 2 3 6 f x x x x x
