P.o.D. – Write the standard
{𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘} form or vertex
[𝑦 − 𝑘 = 𝑎(𝑥 − ℎ)2 ] form equation of
the parabola with the indicated vertex
and point.
1.) Vertex(-2,5); point(0,9)
2.) Vertex(4,-1); point(2,3)
3.) Vertex(2,3); point(0,2)
1.) 𝑦 = (𝑥 + 2)2 + 5
2.) 𝑓 (𝑥 ) = (𝑥 − 4)2 − 1
3.) 𝑓 (𝑥 ) =
−1
4
(𝑥 − 2)2 + 3
2.2 – Polynomial Functions of Higher
Degree
Learning Target: be able to find zeros
of a polynomial.
- Polynomials will have continuous
graphs.
- In other words, we could sketch the
graph without lifting our pencil
from the paper.
- We call these types of graphs
“smooth.”
EX: Sketch the graph of each function.
a.) 𝑓 (𝑥 ) = (𝑥 + 5)4
1
b.) 𝑓 (𝑥 ) = (𝑥 − 3)4
4
a.)
b.)
*Study and Memorize “The Leading
Coefficient Test” on page 141 in your
text.
EX: Describe the left hand and right
hand behavior of the graph of each
function.
1
a.) 𝑓 (𝑥 ) = 𝑥 3 − 2𝑥
4
b.) 𝑓 (𝑥 ) = −3.6𝑥 5 + 5𝑥 3 − 1
a.)
The degree is odd
and the leading coefficient is
positive. The graph falls to the
left and rises to the right.
b.)
The degree is odd
and the leading coefficient is
negative. The graph rises to the
left and falls to the right.
*Odd degree functions will go
opposite directions.
*Even degree functions will go the
same direction.
http://www.youtube.com/watch?v=k6JppISqhj4
Zeros (Root) of a Polynomial
Function:
- Solutions to the equation
- Places where the graph crosses the
x-axis.
- We can solve these both
algebraically and graphically.
EX: Find all the real zeros of
𝑓 (𝑥 ) = 𝑥 3 − 12𝑥 2 + 36𝑥. Then
determine the number of turning
points of the graph of the function.
We will begin by factoring.
𝑓 (𝑥 ) = 𝑥(𝑥 2 − 12𝑥 + 36)
= 𝑥(𝑥 − 6)(𝑥 − 6)
Now set each factor equal to 0 and
solve.
X=0
x-6=0
x-6=0
X=6
x=6
This is a 3rd degree equation, so we
should expect 3 solutions.
We can confirm our zeros graphically.
*When a graph is tangent to the x-axis
(touches at one point), the Solution
occurs twice. This is called a Double
Root.
Because the function is a 3rd degree
polynomial, the graph can have at
most 2 turning points.
*We can think of turning points as
relative maximums or minimums.
EX: Sketch the graph of 𝑓(𝑥 ) =
2𝑥 3 − 6𝑥 2 and locate any zeros
graphically on the TI-84.
X=0,0,3
EX: Sketch the graph of 𝑓(𝑥 ) =
−1
4
4
3
9
𝑥 + 𝑥 − 𝑥 2 and locate any
2
3
4
zeros.
x=0,0,3,3
The Intermediate Value Theorem:
If the value of the function changes
from positive to negative or vice
versa, then a zero must exist
somewhere in that interval.
EX: Use the intermediate value
theorem to approximate the real zeros
of 𝑓 (𝑥 ) = −𝑥 5 + 3𝑥 3 − 2𝑥 + 2.
Examine the TABLE.
At x=1 the function is
positive (y=2). At x=2 the function is
negative (y= -10). This means that a
zero must exist somewhere between
x=1 and x=2.
Use the graph to locate the zero.
The approximate zero is
1.65.
*Your calculator operates on the
Intermediate Value Theorem.
http://www.youtube.com/watch?v=eTIPf1KfAjQ
EX: Find a polynomial of 2nd degree
with zeros x= -8 and x= -4.
If -8 and -4 are our zeros, then the
polynomial has factors of (x+8) and
(x+4).
FOIL these factors to create the
polynomial.
(𝑥 + 8)(𝑥 + 4) = 𝑥 2 + 4𝑥 + 8𝑥 + 32
= 𝑥 2 + 12𝑥 + 32
EX: Find a 3rd degree polynomial with
zeros x=0,1,-3
(𝑥 )(𝑥 − 1)(𝑥 + 3)
= (𝑥 2 − 𝑥 )(𝑥 + 3)
= 𝑥 3 + 3𝑥 2 − 𝑥 2 − 3𝑥
= 𝑥 3 + 2𝑥 2 − 3𝑥
EX: Find a 3rd degree equation with
zeros x=2,4
(𝑥 − 2)(𝑥 − 2)(𝑥 − 4)
= (𝑥 2 − 4𝑥 + 4)(𝑥 − 4)
= 𝑥 3 − 4𝑥 2 + 4𝑥 − 4𝑥 2 + 16𝑥 − 16
= 𝑥 3 − 8𝑥 2 + 20𝑥 − 16
Do the Vocabulary Check on page
148.
1.) Continuous
2.) Leading Coefficient Test
3.) N, n-1
4.) Solution, (x-a), x-intercept
5.) Touches, crosses
6.) Standard
7.) Intermediate Value
Upon completion of this lesson, you
should be able to:
1. Find the zeros of continuous
polynomial functions.
2. Describe the end behavior of a
graph.
3. Explain the Intermediate Value
Theorem.
4. Write a polynomial equation
given its zeros.
For more information on the
Intermediate Value Theorem, visit
http://www.mathsisfun.com/algebra/intermediate-valuetheorem.html
HW Pg. 148 6-72 6ths, 90,
105-114 3rds