6.3 evaluating-and-graphing-polynomila-functions

6.3

Evaluating and Graphing Polynomial Functions

A polynomial function is a function of the form

f (x) = an x n + an – 1 x

n – 1 +· · ·+ a 1 x + a 0

Where an 0 and the exponents are all whole numbers.

A polynomial function is in standard form if its terms are written in descending order of exponents from left to right.

For this polynomial function, an is the leading coefficient,

a 0 is the constant term, and n is the degree.

an 0

an

an leading coefficient

a 0

a0 constant term n

n

degree

descending order of exponents from left to right.

n n – 1

Degree Type Standard Form

You are already familiar with some types of polynomialfunctions. Here is a summary of common types ofpolynomial functions.

4 Quartic f (x) = a4 x 4 + a 3 x

3 + a 2 x 2 + a 1 x + a

0 Constant f (x) = a

3 Cubic f (x) = a 3 x 3 + a 2 x

2 + a 1 x + a

2 Quadratic f (x) = a 2 x 2 + a 1 x + a

1 Linear f (x) = a1x + a

Identifying Polynomial Functions

Decide whether the function is a polynomial function. If it is,write the function in standard form and state its degree, typeand leading coefficient.

f (x) = x 2

– 3x4 – 712

SOLUTION

The function is a polynomial function.

It has degree 4, so it is a quartic function.

The leading coefficient is – 3.

Its standard form is f (x) = – 3x 4

+ x 2 – 7. 1

2



The function is not a polynomial function because the

term 3

x does not have a variable base and an exponentthat is a whole number.

SOLUTION

f (x) = x 3 + 3

x



SOLUTION

f (x) = 6x 2 + 2 x

–1 + x

The function is not a polynomial function because the term2x

–1 has an exponent that is not a whole number.



SOLUTION

The function is a polynomial function.

It has degree 2, so it is a quadratic function.

The leading coefficient is .

Its standard form is f (x) = x2 – 0.5x – 2.

f (x) = – 0.5 x + x 2 – 2

f (x) = x 2 – 3 x

4 – 712


f (x) = x 3 + 3x

f (x) = 6x2 + 2 x– 1 + x

Polynomial function?

f (x) = – 0.5x + x2 – 2

Using Synthetic Substitution

One way to evaluate polynomial functions is to usedirect substitution.

Use substitution to evaluateUse substitution to evaluate

f (x) = 2 x 4 + 8 x

2 + 5 x 7 when x = 3.

Now use direct substitution:

f (x) = 2 x 4 + 8 x

2 + 5 x 7 when x = 3.

7)3(5)3(8)3(2)( 24 xf

98


One way to evaluate polynomial functions is to usedirect substitution. Another way to evaluate a polynomialis to use synthetic substitution.

Use synthetic division to evaluate

f (x) = 2 x 4 + 8 x

2 + 5 x 7 when x = 3.

Polynomial in standard form


2 x 4 + 0 x

3 + (–8 x 2) + 5 x + (–7)

2 6

6

10

18

35

30 105

98

The value of f (3) is the last number you write,In the bottom right-hand corner.

The value of f (3) is the last number you write,In the bottom right-hand corner.

2 0 –8 5 –7 CoefficientsCoefficients

3

x-value

3 •

SOLUTION

Polynomial instandard form

Start by writing it in standard form



f (x) = x 2 - x

5 + 1 x = -1.



f (x) = x 2 - x

5 + 1 x = -1.

Did you get 3?

HOMEWORK (DAY 1)

Pg # 333 #16 – 26, 30- 34, 38-46

(Evens only)

If “n” is even, the graph of the polynomial is “U-shaped” meaning it is parabolic (the higher the degree, the more curves the graph will have in it).

If “n” is odd, the graph of the polynomial is “snake-like” meaning looks like a snake (the higher the degree, the more curves the graph will have in it).

Let’s talk about the Leading Coefficient Test:

Leading Coefficient Test

Degree is evenDegree is odd

Leading coefficient is positive

Start high,End high

Leading coefficient is negative

Start low,End low

Leading coefficient is positive

Start low,End high

Leading coefficient is negative

Start high,End low

with a positiveleading coefficient

with a negativeleading coefficient

with a positiveleading coefficient

with a negativeleading coefficient

Determine the left and right behavior of the graph of each polynomial function.

f(x) = -x5 +3x4 – x

f(x) = x4 + 2x2 – 3x

f(x) = 2x3 – 3x2 + 5

Determine the left and right behavior of the graph of each polynomial function.

f(x) = -x5 +3x4 – x

f(x) = x4 + 2x2 – 3x

f(x) = 2x3 – 3x2 + 5

Even, Leading coefficient 1 (positive) , starts high ends high

Odd, Leading coefficient 1 (negative) , starts high ends low

ODD , Leading coefficient 2 (positive) , starts LOW ends HIGH

Tell me what you know about the equation…

Odd / Even ?

Leading coefficient Positive or Negative?


Odd / Even ?



Odd / Even ?



Fundamental Thm of AlgebraZeros of Polynomial Functions:1. The graph of f has at most n

real zeros

The “n” deals with the highest exponent!

How many zeros do these graphs have????

GRAPHING POLYNOMIAL FUNCTIONS

The end behavior of a polynomial function’s graphis the behavior of the graph as x approaches infinity(+ ) or negative infinity (– ). The expressionx + is read as “x approaches positive infinity.”

GRAPHING POLYNOMIAL FUNCTIONS

END BEHAVIOR

x

f(x)–3 –2 –1 0 1 2 3

Graphing Polynomial Functions

Graph f (x) = x 3 + x

2 – 4 x – 1.

SOLUTION

To graph the function, make a table of values and plot the corresponding points. Connect the points with a smooth curve and check the end behavior.

x

f(x)–3

–7

–2

3

–1

3

0

–1

1

–3

2

3

3

23


Graph f (x) = x 3 + x

2 – 4 x – 1.

SOLUTION


x

f (x)

–3 –2 –1 0 1 2 3


Graph f (x) = –x 4 – 2x

3 + 2x 2 + 4x.

SOLUTION


x

f (x)

–3

–21

–2

0

–1

–1

0

0

1

3

2

–16

3

–105


Graph f (x) = –x 4 – 2x

3 + 2x 2 + 4x.

SOLUTION


Assignment

Pg 334 # 49 – 52, 54- 64, 66-72 evens

Education

6.3 evaluating-and-graphing-polynomila-functions