Graphing Polynomial Functions Goal: Evaluate and graph polynomial functions.
Unit 3 Polynomial Functions
Transcript of Unit 3 Polynomial Functions
Unit 3Polynomial Functions
Friday, January 31, 2020
Polynomial Function
A polynomial function is define by a polynomial expression in x given in the standard form;
y = anxn + an-1x
n-1 + … +a1x + a0
The value of n must be a positive integer
an-1, ..., a1, a0 are called coefficients and these
are real numbers
Kinds of Polynomial FunctionPolynomial can be classified by the degree of the polynomial as shown below
Polynomial Function in Standard Form Degree
Name of Function
y = mx + b 1 Linear
y = ax2 + bx + c 2 Quadratic
Y = ax3 + bx2 + cx + d 3 Cubic
Y = ax4 + bx3 + cx2 + dx + e 4 Quartic
Y = ax5 + bx4 + cx3 + dx2 + ex 5 Quintic
The largest exponent within the polynomial determines the degree of the polynomial.
Leading CoefficientThe leading coefficient is the coefficient of the first term in a polynomial when the terms are written in descending order by degrees.
For example, the quartic function f(x) = -2x4 + x3 – 5x2 – 10 has a leading coefficient of -2.
Graphs of Polynomial Function
Exploring the graphs of polynomial function
Linear
Function
Quadratic
Function
Cubic
Function
Quartic
Function
-10 -8 -6 -4 -2 2 4 6 8 10
-10
-8
-6
-4
-2
2
4
6
8
10
-10 -8 -6 -4 -2 2 4 6 8 10
-10
-8
-6
-4
-2
2
4
6
8
10
-10 -8 -6 -4 -2 2 4 6 8 10
-10
-8
-6
-4
-2
2
4
6
8
10
-5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10
-60
-55
-50
-45
-40
-35
-30
-25
-20
-15
-10
-5
5
10
Cubic PolynomialsLook at the two graphs and discuss the questions given
below.
1. How can you check to see if both graphs are functions?
3. What is the end behaviour for each graph?
4. Which graph do you think has a positive leading coeffient? Why?
5. Which graph do you think has a negative leading coefficient? Why?
2. How many x-intercepts do graphs A & B have?
Graph BGraph A
-10 -8 -6 -4 -2 2 4 6 8 10
-10
-8
-6
-4
-2
2
4
6
8
10
-10 -8 -6 -4 -2 2 4 6 8 10
-10
-8
-6
-4
-2
2
4
6
8
10
Cubic Polynomials
Equation
Factored form & Standard form
X-Intercepts
Sign of Leading
Coefficient
End Behaviour
Domain and Range
Factoredy=(x+1)(x+4)(x-2)
Standardy=x3+3x2-6x-8
-4, -1, 2 Positive
As x, y and x-,
y-
Domain
{x| x Є R}
Range
{y| y Є R}
Factoredy=-(x+1)(x+4)(x-2)
Standardy=-x3-3x2+6x+8
-4, -1, 2Negativ
e
As x, y-
and
x-, y
Domain
{x| x Є R}
Range
{y| y Є R}
The following chart shows the properties of the graphs on the left.
-5 -4 -3 -2 -1 1 2 3 4 5
-12
-10
-8
-6
-4
-2
2
4
6
8
10
12
-5 -4 -3 -2 -1 1 2 3 4 5
-12
-10
-8
-6
-4
-2
2
4
6
8
10
12
Quartic PolynomialsLook at the two graphs and discuss the questions given below.
1. How can you check to see if both graphs are functions?
3. What is the end behaviour for each graph?
4. Which graph do you think has a positive leading coeffient? Why?
5. Which graph do you think has a negative leading coefficient? Why?
2. How many x-intercepts do graphs A & B have?
Graph BGraph A
-5 -4 -3 -2 -1 1 2 3 4 5
-10
-8
-6
-4
-2
2
4
6
8
10
12
14
-5 -4 -3 -2 -1 1 2 3 4 5
-14
-12
-10
-8
-6
-4
-2
2
4
6
8
10
Quartic Polynomials
Equation
Factored form & Standard form
X-Intercepts
Sign of Leading
Coefficient
End Behaviour
Domain and Range
Factoredy=(x-4)2(x-1)(x+1)
Standardy=x4-8x3+15x2+8x-16
-1,1,4 Positive
As x, y and x-, y
Domain
{x| x Є R}
Range
{y| y Є R,
y ≥ -16.95}
Factoredy=-(x-4)2(x-1)(x+1)
Standardy=-x4+8x3-15x2-8x+16
-1,1,4Negativ
e
As x, y-
and
x-, y-
Domain
{x| x Є R}
Range
{y| y Є R,
y ≤ 16.95}
The following chart shows the properties of the graphs on the left.
-5 -4 -3 -2 -1 1 2 3 4 5
-15
-12
-9
-6
-3
3
6
9
12
15
18
-5 -4 -3 -2 -1 1 2 3 4 5
-18
-15
-12
-9
-6
-3
3
6
9
12
15
Polynomial Equation
If y = f(x) is a polynomial function then f(x) = 0 is a polynomial equation. Example
1.) F(x) = 2x4 ‒ 3x3 + x ‒ 4
2.) F(x) = 4x4 + 3x3 + 2x + 1
2x4 ‒ 3x3 + x ‒ 4 = 0
4x4 + 3x3 + 2x + 1 = 0
polynomial function polynomial equation
Note: The solution to a polynomial equation is called the zero of the polynomial function.
Remainder TheoremIf a polynomial function P(x) is divided by x –a, then the remainder is number P(a) where the function is evaluated at x = a
1.) F(x) = 2x4 ‒ 3x3 + x ‒ 4 divided by (x – 2)
3.) F(x) = 2x³ + 1 ‒ x + 3x² divided by (x + 2)
4.) F(x) = 2x³ ‒ 3x² ‒ x + 2 divided by (x – 1)
2.) F(x) = 4x4 + 3x3 + 2x + 1 divided by (x + 1)
Find the remainder of each
R = 6
R = 0
R = -1
R = 0
Factor TheoremIf a polynomial function P(x) is divided by x – a, and the remainder P(a) = 0 if and only if x – a is a factor of P(x).
Or in other words,• – If f(a) = 0, then x – a is a factor of f(x).• – If x – a is a factor of f(x), then f(a) = 0
• If we know a factor, we know a zero!• If we know a zero, we know a factor!
Fundamental Theorem of Algebra
Every polynomial function of degree n≥1 has at least one complex zero.Every polynomial function of degree n≥1 has exactly n complex zeros counting multiplicities.Example:1.) F(x) = 2x4 ‒ 3x3 + x ‒ 4 Degree 4
has exactly 4 complex zeros or solution
2.) F(x) = 2x³ + 1 ‒ x + 3x² Degree 3
has exactly 3 complex zeros or solution
Solving Polynomial Equation
• Solving Polynomial Equation in Factored Form
• Example:
1.) (x – 2)(2x – 1)(x + 1) = 0
x – 2 = 0
x = 2
2x – 1 = 0
2x = 1
2 2
x =1
2
x + 1 = 0
x = -1
Solving Polynomial Equation
• Solving Polynomial Equation in Factored Form
• Example:
2.) (x + 3)(3x + 2)(x - 5) = 0
x + 3 = 0
x = -3
3x + 2 = 0
3x = -2
3 3
x =-2
3
x - 5 = 0
x = 5
Finding Polynomial Function or Equation
Finding a Polynomial Function or Equation given its Roots, Zeros or Solution.
Example:
1. Find a Polynomial function whose zeros are x = 1 and x = -3
Solution:
f(x) =
(x – 1) (x + 3)
f(x) = x2 + 3x - 1x - 3
f(x) = x2 + 2x- 3 or x2 + 2x – 3 = 0
Finding Polynomial Function or Equation
Example:
2. Find a Polynomial function whose zeros are x = -2, x = -1 and x = 2
Solution:
f(x) =
(x + 2) (x + 1)
f(x) = (x2 + 1x + 2x + 2)
f(x) = (x2 + 3x + 2) (x – 2)
(x - 2)
(x - 2)
= x3 - 2x2+ 3x2- 6x + 2x - 4
= x3 + 1x2 - 4x - 4
f(x)
f(x)
Unit 3Solving Polynomial
Functions / Equations
Friday, January 31, 2020
Solving Polynomial Equation
Solving for the solutions or zeroes of a polynomial equation or function can be done by:
1. Graphing calculator, Synthetic Division and Factoring.
2. Rational Root Theorem, Synthetic Division and Factoring.
Solving Polynomial Equation
Solving by Graphing Calculator, Synthetic Division and Factoring
Example: Find the zeroes or solution
1. f(x) = 3x3 + x2 – 3x – 1
Step 1: Use graphing calculator to get as many exact zero as possible by doing y = 3x3
+ x2 – 3x - 1 and do 2nd TRACE ZERO.
Step 2: Use synthetic Division to reduced the function to a lower degree.
Rational Root/ Zero Theorem
The Rational Root/ Zero Theorem gives a list of possible rational zeros of a polynomial function. Equivalently, the theorem gives all possible rational roots of a polynomial equation.
The Rational Zero Theorem
If f(x) = anxn + an-1x
n-1 +…+ a1x + a0 has integer
coefficients and (where is reduced) is a
rational zero, then p is a factor of the constant term
a0 and q is a factor of the leading coefficient an.
p
q
p
q
Using the Rational Root/ Zero Theorem
List all possible rational zeros of f (x) = 15x3 + 14x2 - 3x – 2.
Solution The constant term is –2 and the leading coefficient is 15.
1 2 1 2 1 25 53 3 15 15
Factors of the constant term, 2Possible rational zeros
Factors of the leading coefficient, 15
1, 2
1, 3, 5, 15
1, 2, , , , , ,
-=
=
=
Divide 1
and 2
by 1.
Divide 1
and 2
by 3.
Divide 1
and 2
by 5.
Divide 1
and 2
by 15.
There are 16 possible rational zeros. The actual solution set to f(x) = 15x3 +
14x2 - 3x – 2 = 0 is {-1, -1/3, 2/5}, which contains 3 of the 16 possible solutions.
Solve: x4 - 6x2 - 8x + 24 = 0.
Solution Because we are given an equation, we will use the
word "roots," rather than "zeros," in the solution process. We
begin by listing all possible rational roots.
Factors of the constant term, 24Possible rational zeros
Factors of the leading coefficient, 1
1, 2 3, 4, 6, 8, 12, 24
1
1, 2 3, 4, 6, 8, 12, 24
=
=
=
EXAMPLE: Solving a Polynomial Equation
Solve: x4 - 6x2 - 8x + 24 = 0.
Solution The graph of f(x) = x4 - 6x2 - 8x + 24 is shown the
figure below. Because the x-intercept is 2 it is a zero or root, we
will test 2 by synthetic division and show that it is a root of the
given equation.
x-intercept: 2
The zero remainder
indicates that 2 is a root
of x4 - 6x2 - 8x + 24 = 0.
2 1 0 -6 -8 24
2 4 -4 -24
1 2 -2 -12 0
EXAMPLE: Solving a Polynomial Equation
EXAMPLE: Solving a Polynomial Equation
Solve: x4 - 6x2 - 8x + 24 = 0.
Solution Now we can rewrite the given equation in
factored form.
(x – 2)(x3 + 2x2 - 2x - 12) = 0 This is the result obtained from
the synthetic division.
x – 2 = 0 or x3 + 2x2 - 2x - 12 = 0 by zero property.
x4 - 6x2 + 8x + 24 = 0 This is the given equation.
we must continue by factoring
x3 + 2x2 - 2x - 12 = 0
EXAMPLE: Solving a Polynomial Equation
Solve: x4 - 6x2 - 8x + 24 = 0.
Solution Because the graph turns around at 2, this
means that 2 is a root of even multiplicity. Thus, 2 must
also be a root of x3 + 2x2 - 2x - 12 = 0. We use 2 again
for synthetic division
x-intercept: 2
2 1 2 -2 -12
2 8 12
1 4 6 0
These are the coefficients
of x3 + 2x2 - 2x - 12 = 0.
The zero remainder indicates that 2 is a root ofx3 + 2x2 - 2x - 12 = 0.
EXAMPLE: Solving a Polynomial Equation
Solve: x4 - 6x2 - 8x + 24 = 0.
Solution Now we can solve the original equation as
follows.
(x – 2)(x3 + 2x2 - 2x - 12) = 0 This was obtained from the first
synthetic division.
x4 - 6x2 + 8x + 24 = 0 This is the given equation.
(x – 2)(x – 2)(x2 + 4x + 6) = 0 This was obtained from the
second synthetic division.
x – 2 = 0 or x – 2 = 0 or x2 + 4x + 6 = 0 by zero property.
x = 2 x = 2 x2 + 4x + 6 = 0 Solve.
EXAMPLE: Solving a Polynomial Equation
Solve: x4 - 6x2 - 8x + 24 = 0.
Solution We can use the quadratic formula to solve x2 + 4x + 6 = 0.
Let a = 1, b = 4, and c = 6.
24 4 4 1 6
2 1
- -=
We use the quadratic formula because x2 + 4x + 6 = 0
cannot be factored.
2 4
2
b b acx
a
- -=
Simplify.2 2i= -
Multiply and subtract under the radical. 4 8
2
- -=
4 2 2
2
i- = - = - =8 4(2)( 1) 2 2i
The solution set of the original equation is {2, -2 - i -2 + i }.2,i 2i
If f (x) = anxn + an-1xn-1 + … + a2x
2 + a1x + a0 be a
polynomial with real coefficients.
1. The number of positive real zeros of f is either equal to
the number of sign changes of f (x) or is less than that
number by an even integer. If there is only one variation in
sign, there is exactly one positive real zero.
2. The number of negative real zeros of f is either equal to
the number of sign changes of f (-x) or is less than that
number by an even integer. If f (-x) has only one variation in
sign, then f has exactly one negative real zero.
Descartes' Rule of Signs
Descartes' Rule of Signs
Example: State the possible positive and negative zeros of the polynomial function
1.) f(x) = 3x4 – 5x3 – 12x2 + 20x
Positive: 2 or 0 positive zeros
+ - - +
f(-x) = 3(-x)4 – 5(-x)3 – 12(-x)2 + 20(-x)
f(-x) = 3x4 + 5x3 – 12x2 – 20x
+ + - -Negative: 1 negative zeros
Unit 3Theorem on Roots of Polynomial
Equations
Friday, January 31, 2020
•The Conjugate Root Theorem
•The Irrational Root Theorem
Conjugate Root Theorem
If the complex number a + bi is a root of a
polynomial equation, then its conjugate a - bi is also a root
Note conjugate root always come in pairs
Example
i
i
23
23
-
+
i
i
- i
i
3
3
-
Conjugate Root Theorem
Example Problem
Write a polynomial function with the given zeros: 2, i,
1.)
f(x) =
(x - 2) (x - i)
f(x) = (x2 + 1)
(x + i)
(x - 2)
= x3 - 2x2 + 1x - 2f(x)
Solution
Conjugate Root Theorem
Example Problem
Write a polynomial function with the given zeros: 3, 4i
2.)
= x3 - 3x2 + 16x - 48f(x)
Solution
Conjugate Root Theorem
Example Problem
Write a polynomial function with the given zeros: 4, 2 - 3i
3.)
f(x) =
(x - 4) (x – 2 + 3i)
f(x) = (x2 -4x + 13)
(x - 2 – 3i)
(x - 4)
= x3 - 4x2 + 13xf(x)
Solution
- 4x2 + 16x- 52
= x3 - 8x2 + 29xf(x) - 52
Conjugate Root Theorem
Example Problem
Write a polynomial function with the given zeros: 3, 1 + 2i
4.)
= x3 - 5x2 + 11xf(x) - 15
Irrational Root Theorem
a b+If the irrational number is a root
of a polynomial equation, then a b- is
also a root
Note
Irrationals root always come in pairs
Unit 3Polynomial Inequalities
Friday, January 31, 2020
Polynomial Inequalities
They can be written in the forms:
0f x 0f x 0f x 0f x
To solve the inequality f (x) > 0 is to find the values of x that make f (x) positive
To solve the inequality f (x) < 0 is to find the values of x that make f (x) negative
If the expression f (x) is a product, we can find its sign by looking at the sign of each of its factors.
Polynomial Inequalities
Determine the real number values of x that cause the given function to be (a) zero, (b) positive, (c) negative.
223 1 4f x x x x= + + -
Real zeros: x = –3, 4
We use a sign chart to find the sings of f(x) at any values of x:
–3 4
2
- + - 2
+ + - 2
+ + +
Negative Positive Positive
The function is positive on 3,4 4,-
The function is negative on , 3- -
Polynomial InequalitiesFor the same function, give the solution of each of the following:
223 1 4f x x x x= + + -
3,4 4, - 223 1 4 0x x x+ + -
3, - 223 1 4 0x x x+ + -
, 3 - - 223 1 4 0x x x+ + -
, 3 4 - - 223 1 4 0x x x+ + -
Polynomial InequalitiesSolve the given inequality analytically.
3 22 7 10 24 0x x x- - +
Let 3 22 7 10 24f x x x x= - - +
Use the RZT to list the possible rational zeros!!!
1 31, 2, 3, 4, 6, 8, 12, 24, ,
2 2
Look at the graph x = 4 seems like a zero of f(x)
Polynomial Inequalities
Solve the given inequality analytically.
3 22 7 10 24 0x x x- - +
4 2 –7 –10 24
8 4 –24
2 1 –6 0
24 2 6f x x x x= - + -
4 2 3 2f x x x x= - - +
Zeros: 4, 3
2, - 2
Polynomial InequalitiesSolve the given inequality analytically.
3 22 7 10 24 0x x x- - +
4 2 3 2f x x x x= - - +
Now, for the sign chart:
–2 Positive Negative x3/2 4 PositiveNegative
Solution: 2,3 2 4,-
Polynomial Inequalities
Solve the given inequality analytically.
1.) (x – 2)(2x – 1)(x + 1) > 0
2.) (x + 3)(3x + 2)(x - 5) ≤ 0
3.) 3x3 – 13x2 + 13x - 3 ≥ 0
x = 1, x = 3 and x = 1/3
Polynomial Inequalities with Unusual Answers…
•Give the solution to each of the following:
2 27 2 1 0x x+ + ,-
2 27 2 1 0x x+ + ,-
•The graph???
2 27 2 1 0x x+ +
2 27 2 1 0x x+ +
•No Solution
•No Solution
Polynomial Inequalities with Unusual Answers…
•Give the solution to each of the following:
22 3 3 2 5 0x x x- + + , 5 2 5 2,- - -
•The graph???
22 3 3 2 5 0x x x- + + ,-
22 3 3 2 5 0x x x- + + •No Solution
22 3 3 2 5 0x x x- + + 5 2x = -