warm-up: simplify

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Warm-up: Simplify. 2 2 2 5 8 99 3 7 98 7 1 3 0 6 2 4 5 3 1.3 3 2. 3. 5 7 6.810 4. (2.5x10)(1.8x10) 5. 1.510 45 6. 74 7. 9 x x x y x x y

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Warm-up: Simplify. Warm-up: Answers. Check Homework. p. 533 # 15-43 odd Quiz 5.1 tomorrow Objective(5.2A) : Classify polynomials and evaluate polynomials using synthetic substitution. A Polynomial Function is a function of the form. f(x) = a n x n + a n-1 x n-1 + …+a 1 x + a 0 - PowerPoint PPT Presentation

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Page 1: Warm-up:  Simplify

Warm-up: Simplify.

222 5

8

993 7

98

7 130 62 4

5 31. 3 3 2. 3. 5 7

6.8 104. (2.5x10 )(1.8x10 ) 5.1.5 10

456. 74 7.9

xx

x yxx y

Page 2: Warm-up:  Simplify

Warm-up: Answers

27 6

2

222 5

8

993 7

98

7 130 6

10

3

18 2 54

5 31. 3 3 2. 3. 5 7

6.8 104. (2.5x10 )(1.8x10 ) 5.1.5 10

456

3 93 57 49

4.5x10 4.533x10

1. 7 54 7.9

xx

x y yy xxx

x

Page 3: Warm-up:  Simplify

Check Homework

p. 533 # 15-43 odd

Quiz 5.1 tomorrow

Objective(5.2A): Classify polynomials and evaluate polynomials using synthetic substitution.

Page 4: Warm-up:  Simplify

A Polynomial Function is a function of the form

• f(x) = anxn + an-1xn-1 + …+a1x + a0

The exponents are all whole numbers, and the coefficients are all real numbers.

an is the leading coefficientn is the degree

Page 5: Warm-up:  Simplify

Degree Type Standard Form Example

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

4 Quartic f (x) = 5x 4 + 4 x

3 + 7 x 2 + 9 x + 1

0 Constant f (x) = 4

3 Cubic f (x) = 5 x 3 + 3 x

2 - x + 6

2 Quadratic f (x) = 5 x 2 + 2 x + 7

1 Linear f (x) = 2x -10

Page 6: Warm-up:  Simplify

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 – 71

2

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

Page 7: Warm-up:  Simplify

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

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

Page 8: Warm-up:  Simplify

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

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.

Page 9: Warm-up:  Simplify

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

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

Page 10: Warm-up:  Simplify

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

Page 11: Warm-up:  Simplify

What is the degree of the monomial? 245 bx

The degree of a monomial is the sum of the exponents of the variables in the monomial. The exponents of each variable are 4 and 2. 4+2 = 6.

The degree of the monomial is 6. The monomial can be referred to as a sixth degree monomial.

Page 12: Warm-up:  Simplify

Classify the polynomials by degree and number of terms.

Polynomiala.

b.

c.

d.

5

42 x

xx 23

14 23 xx

DegreeClassify by

degree

Classify by number of

termsZero Constant Monomial

First Linear Binomial

Second Quadratic Binomial

Third Cubic Trinomial

Page 13: Warm-up:  Simplify

Write the polynomials in standard form and identify the polynomial by degree and number of terms. 23 237 xx a)

b) xx 231 2

3 23 8 11x x x c)

Page 14: Warm-up:  Simplify

• Page 338 (1-5)

Page 15: Warm-up:  Simplify

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.

Just Watch--

Page 16: Warm-up:  Simplify

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.

2 0 –8 5 –7 CoefficientsCoefficients

3

x-value

3 •

SOLUTION

Polynomial instandard form

Page 17: Warm-up:  Simplify

EXAMPLE 3 Evaluate by synthetic substitution

a) Use synthetic substitution to evaluate f (x) from Example 2 when x = 3. f (x) = 2x4 – 5x3 – 4x + 8

SOLUTION

STEP 1 Write the coefficients of f (x) in order of descending exponents. Write the value at which f (x) is being evaluated to the left.

Page 18: Warm-up:  Simplify

EXAMPLE 3 Evaluate by synthetic substitution

STEP 2 Bring down the leading coefficient. Multiply the leading coefficient by the x-value. Write the product under the second coefficient. Add.

STEP 3 Multiply the previous sum by the x-value. Write the product under the third coefficient. Add. Repeat for all of the remaining coefficients. The final sum is the value of f(x) at the given x-value.

Page 19: Warm-up:  Simplify

EXAMPLE 3 Evaluate by synthetic substitution

Synthetic substitution gives f(3) = 23, which matches the result in Example 2.

ANSWER

Page 20: Warm-up:  Simplify

GUIDED PRACTICE for Examples 3 and 4

Use synthetic substitution to evaluate the polynomial function for the given value of x.

b) f (x) = 5x3 + 3x2 – x + 7; x = 2

Write the coefficients of f (x) in order of descending exponents. Write the value at which f (x) is being evaluated to the left.

Page 21: Warm-up:  Simplify

GUIDED PRACTICE for Examples 3 and 4

Synthetic substitution gives f(2) = 57ANSWER

Page 22: Warm-up:  Simplify

GUIDED PRACTICE for Examples 3 and 4

c) g (x) = – 2x4 – x3 + 4x – 5; x = – 1

Write the coefficients of g(x) in order of descending exponents. Write the value at which g (x) is being evaluated to the left.

– 1 – 2 – 1 0 4 – 5

Page 23: Warm-up:  Simplify

GUIDED PRACTICE for Examples 3 and 4

2 –1 1 –5– 1 – 2 – 1 0 4 – 5

– 2 1 –1 5 – 10

Synthetic substitution gives f(– 1) = – 10ANSWER

Page 24: Warm-up:  Simplify

Assignments

Classwork: Practice 5.2 # 1-7

Homework (5.2A): p. 341 # 3-8, 15-23 (15 pts)

Closure: Review exponent rules for quiz tomorrow