Title: Lesson 1–1
Author: Teri Willard
Last saved by: Glencoe Mathematics

Lesson 6-4

Example 1  Find Degree and Leading Coefficients
State the degree and leading coefficient of each polynomial in one variable. If it is not a polynomial in one variable, explain why.
a. –2a^3 + a – 

This is not a polynomial. The term  cannot be written in the form x^n, where n is a nonnegative integer.

b. b^5 + b^2 – 8

This is a polynomial in one variable. The degree is 5, and the leading coefficient is 1.

c. mn + 3m – 7n

This is not a polynomial in one variable. It contains two variables, m and n.

d. 11 – 5c + 2c^6 – 4c^3

Rewrite the expression so the powers of c are in decreasing order.
2c^6 – 4c^3 – 5c + 11
This is a polynomial in one variable. The degree is 6, and the leading coefficient is 2.

Example 2  Evaluate a Polynomial Function
GEOMETRY In a polygon with n sides, for n > 3, the number of diagonals can be found by using the function d(n) = , where d(n) is the number of diagonals and n is the number of sides.
a. Draw polygons for n = 4, 5, 6, and 7. Show that the polynomial function d(n) =  gives the total number of diagonals for each figure.

First, draw the polygons and count the number of diagonals.

n = 4 | n = 5 | n = 6 | n = 7
diagonals = 2 | diagonals = 5 | diagonals = 9 | diagonals = 14

Next, find the values of d(4), d(5), d(6), and d(7).

d(n) = 
d(4) = (4)^2 – (4)
= 8 – 6
= 2

d(n) = 
d(5) = (5)^2 – (5)
= 12.5 – 7.5
= 5

d(n) = 
d(6) = (6)^2 – (6)
= 18 – 9
= 9

d(n) = 
d(7) = (7)^2 – (7)
= 24.5 – 10.5
= 14

You can see that the number of diagonals for each figure in the diagrams is the same as the number of diagonals d(n) found by evaluating the function.

b. Find the number of diagonals for a decagon (a ten–sided polygon).

d(n) = 	Given function
d(10) = (10)^2 – (10)	Replace n with 10.
= 50 – 15	Evaluate.
= 35	Simplify.

Example 3  Functional Values of Variables
Find –3[q(4b – 1)] if q(x) = x^2 – x + 1.

First, evaluate q(4b – 1) by replacing x in q(x) with 4b – 1.
q(x) = x^2 – x + 1	Original function
q(4b – 1) = (4b – 1)^2 – (4b – 1) + 1	Replace x with 4b – 1.
= 16b^2 – 8b + 1 – 4b + 1 + 1	Evaluate (4b – 1)^2 and –1(4b – 1).
= 16b^2 – 12b + 3	Simplify.

To evaluate –3[q(4b – 1)], multiply q(4b – 1) by –3.
–3[q(4b – 1)] = –3(16b^2 – 12b + 3)	Substitute 16b^2 – 12b + 3 for q(4b – 1).
 = –48b^2 + 36b – 9	Distributive Property

Example 4  Graphs of Polynomial Functions
For each graph,
•	describe the end behavior.
•	determine whether it represents an odd–degree or an even–degree polynomial function, and
•	state the number of real zeros.
a.	b.	

a.  • 	As x  + ∞, f(x)  + ∞. As x  – ∞, f(x)  + ∞.
	•	It is an even–degree polynomial function.
	•	The graph does not intersect the x–axis, so it has no real zeros.

b.  •	As x  + ∞, f(x)  + ∞. As x  – ∞, f(x)  – ∞.
	•	It is an odd–degree polynomial function. 
	•	The graph intersects the x–axis at three points, so the function has three real zeros.