splash screen. lesson menu five-minute check (over lesson 2-1) then/now new vocabulary example...

Five-Minute Check (over Lesson 2-1)

Then/Now

New Vocabulary

Example 1: Graph Transformations of Monomial Functions

Key Concept: Leading Term Test for Polynomial End Behavior

Example 2: Apply the Leading Term Test

Key Concept: Zeros and Turning Points of Polynomial Functions

Example 3: Zeros of a Polynomial Function

Key Concept: Quadratic Form

Example 4: Zeros of a Polynomial Function in Quadratic Form

Example 5: Polynomial Function with Repeated Zeros

Key Concept: Repeated Zeros of Polynomial Functions

Example 6:Graph a Polynomial Function

Example 7: Real-World Example: Model Data Using Polynomial Functions

Over Lesson 2-1

A. Graph f (x) = 3x 3.

A.

B.

C.

D.

Over Lesson 2-1

B. Analyze f (x) = 3x 3.

A. D = (–∞, ∞) R = [0, ∞), intercept: 0, ,

continuous for all real numbers, decreasing: (–∞, 0) ,

increasing: (0, ∞)

B. D = (–∞, ∞) R = (–∞, ∞), intercept: 0, ,

continuous for all real numbers, decreasing: (–∞, 0),

increasing: (0, ∞)

C. D = (–∞, ∞) R = (–∞, ∞), intercept: 0, ,

continuous for all real numbers, decreasing: (–∞, ∞)

D. D = (–∞, ∞) R = (–∞, ∞), intercept: 0, ,

continuous for all real numbers, increasing: (–∞, ∞)

Over Lesson 2-1

Solve .

A. 7

B. 3, 7

C.

D. no solution

Over Lesson 2-1

A. –14

B. –2

C. 88

D. no solution

Solve .

Over Lesson 2-1

A. –3, 4

B. 3

C. 6, 1

D. 6

You analyzed graphs of functions. (Lesson 1-2)

• Graph polynomial functions.

• Model real-world data with polynomial functions.

• polynomial function

• polynomial function of degree n

• leading coefficient

• leading-term test

• quartic function

• turning point

• quadratic form

• repeated zero

• multiplicity

Graph Transformations of Monomial Functions

A. Graph f (x) = (x – 3)5.

This is an odd-degree function, so its graph is similar to the graph of y = x

3. The graph of f(x) = (x – 3)5 is the graph of y = x

5 translated 3 units to the right.

Answer:

Graph Transformations of Monomial Functions

B. Graph f (x) = x 6 – 1.

This is an even-degree function, so its graph is similar to the graph of y = x

2. The graph of f(x) = x 6 – 1 is the

graph of y = x 6 translated 1 unit down.

Answer:

Graph f (x) = 2 + x 3.

A.

B.

C.

D.

Apply the Leading Term Test

A. Describe the end behavior of the graph of f (x) = 3x

4 – x 3 + x

2 + x – 1 using limits. Explain your reasoning using the leading term test.

The degree is 4, and the leading coefficient is 3.

Because the degree is even and the leading

coefficient is positive, .

Answer:

Apply the Leading Term Test

B. Describe the end behavior of the graph of f (x) = –3x

2 + 2x 5 – x

3 using limits. Explain your reasoning using the leading term test.

Write in standard form as f (x) = 2x 5 – x

3 – 3x 2. The

degree is 5, and the leading coefficient is 2. Because

the degree is odd and the leading coefficient is

positive, .

Answer:

Apply the Leading Term Test

C. Describe the end behavior of the graph of f (x) = –2x

5 – 1 using limits. Explain your reasoning using the leading term test.

The degree is 5 and the leading coefficient is –2.

Because the degree is odd and the leading coefficient

is negative, .

Answer:

Describe the end behavior of the graph of g (x) = –3x

5 + 6x 3 – 2 using limits. Explain your

reasoning using the leading term test.

A. Because the degree is odd and the leading coefficient negative, .

B. Because the degree is odd and the leading coefficient negative, .

C. Because the degree is odd and the leading coefficient negative, .

D. Because the degree is odd and the leading coefficient negative, .

Zeros of a Polynomial Function

State the number of possible real zeros and turning points of f (x) = x

3 + 5x 2 + 4x. Then

determine all of the real zeros by factoring.

The degree of the function is 3, so f has at most 3 distinct real zeros and at most 3 – 1 or 2 turning points. To find the real zeros, solve the related equation f (x) = 0 by factoring.

x 3 + 5x

2 + 4x= 0Set f (x) equal to 0.

x(x 2 + 5x + 4)= 0

Factor the greatest common factor, x.

x(x + 4)(x + 1) = 0Factor completely.

Zeros of a Polynomial Function

So, f has three distinct real zeros, x = 0, x = –4, and x = –1. This is consistent with a cubic function having at most 3 distinct real zeros.

CHECK You can use a graphing calculator to graph f (x) = x

3 + 5x 2 + 4x

and confirm these zeros. Additionally, you can see that the graph has 2 turning points, which is consistent with cubic functions having at most 2 turning points.

Zeros of a Polynomial Function

Answer: The degree is 3, so f has at most 3 distinct real zeros and at most 2 turning points. f (x) = x

3 + 5x 2 + 4x = x (x + 1)(x + 4), so

f has three zeros, x = 0, x = –1, and x = –4.

State the number of possible real zeros and turning points of f (x) = x

4 – 13x 2 + 36. Then

determine all of the real zeros by factoring.

A. 4 possible real zeros, 3 turning points; zeros 2, –2, 3, –3

B. 4 possible real zeros, 2 turning points; zeros 4, 9

C. 3 possible real zeros, 2 turning points; zeros 2, 3

D. 4 possible real zeros, 4 turning points; zeros –2, –3

Zeros of a Polynomial Function in Quadratic Form

State the number of possible real zeros and turning points for h (x) = x

4 – 4x 2 + 3. Then

determine all of the real zeros by factoring.

The degree of the function is 4, so h has at most 4 distinct real zeros and at most 4 – 1 or 3 turning points. This function is in quadratic form because x

4 – 4x 2 + 3 = (x

2)2 – 4(x 2) + 3. Let u = x

2.

(x2)2 – 4(x2) + 3

= 0

Set h(x) equal to 0.

u2 – 4u + 3

= 0

Substitute u for x2.

(u – 3)(u – 1)

= 0

Factor the quadratic expression.

Zeros of a Polynomial Function in Quadratic Form

Factor completely.

(x 2 – 3)(x

2 – 1)

= 0

Substitute x 2 for u.

, x = –1, x = 1 Solve for x.

h has four distinct real zeros, , –1, and 1. This is consistent with a quartic function. The graph of h (x) = x

4 – 4x 2 + 3 confirms this. Notice that there are

3 turning points, which is also consistent with a quartic function.

Zero Product Propertyor x + 1 = 0 or x – 1 = 0

Zeros of a Polynomial Function in Quadratic Form

Answer: The degree is 4, so h has at most 4 distinct real zeros and at most 3 turning points.

h (x) = x 4 – 4x

2 + 3 = (x 2 – 3)(x – 1)(x + 1), so h has four distinct real zeros, x = ,

x = –1, and x = 1.

State the number of possible real zeros and turning points of g (x) = x

5 – 5x 3 – 6x. Then

determine all of the real zeros by factoring.

A. 3 possible real zeros, 2 turning points;real zeros 0, –1, 6

B. 5 possible real zeros, 4 turning points; real zeros 0,

C. 3 possible real zeros, 3 turning points; real zeros 0, –1, .

D. 5 possible real zeros, 4 turning points; real zeros 0, 1, –1,

Polynomial Function with Repeated Zeros

State the number of possible real zeros and turning points of h (x) = x

4 + 5x 3 + 6x

2. Then determine all of the real zeros by factoring.

The degree of the function is 4, so it has at most 4 distinct real zeros and at most 4 – 1 or 3 turning points. Find the real zeros.

x 4 + 5x

3 + 6x 2= 0

Set h(x) equal to 0.

x 2(x

2 + 5x + 6)= 0Factor the greatest common factor, x

2.

x 2(x + 3)(x + 2) = 0

Factor completely.

Polynomial Function with Repeated Zeros

The expression above has 4 factors, but solving for x yields only 3 distinct real zeros, x = 0, x = –3, and x = –2. Of the zeros, x = 0 is repeated.

The graph of h (x) = x 4 + 5x

3 + 6x 2 shown here

confirms these zeros and shows that h has three turning points. Notice that at x = –3 and x = –2, the graph crosses the x-axis, but at x = 0, the graph is tangent to the x-axis.

Answer: The degree is 4, so h has at most 4 distinct real zeros and at most 3 turning points. h (x) = x

4 + 5x 3 + 6x

2 = x 2(x + 2)(x + 3), so

h has three zeros, x = 0, x = –2, and x = –3. Of the zeros, x = 0 is repeated.

Polynomial Function with Repeated Zeros

State the number of possible real zeros and turning points of g (x) = x

4 – 4x 3 + 4x

2. Then determine all of the real zeros by factoring.

A. 4 possible real zeros, 3 turning points; real zeros 0, 2

B. 4 possible real zeros, 3 turning points; real zeros 0, 2, –2

C. 2 possible real zeros, 1 turning point; real zeros 2, –2

D. 4 possible real zeros, 3 turning points; real zeros 0, –2

A. For f (x) = x(3x + 1)(x – 2) 2, apply the leading-

term test.

Graph a Polynomial Function

The product x(3x + 1)(x – 2)2 has a leading term of x(3x)(x)2 or 3x4, so f has degree 4 and leading coefficient 3. Because the degree is even and the leading coefficient is positive, .

Answer:

B. For f (x) = x(3x + 1)(x – 2) 2, determine the zeros

and state the multiplicity of any repeated zeros.

Graph a Polynomial Function

The distinct real zeros are x = 0, x = , and x = 2.

The zero at x = 2 has multiplicity 2.

Answer: 0, , 2 (multiplicity: 2)

C. For f (x) = x(3x + 1)(x – 2) 2, find a few additional

points.

Choose x-values that fall in the intervals determined by the zeros of the function.

Graph a Polynomial Function

Answer: (–1, 18), (–0.1, –0.3087), (1, 4), (3, 30)

D. For f (x) = x(3x + 1)(x – 2) 2, graph the function.

Graph a Polynomial Function

Plot the points you found. The end behavior of the

function tells you that the graph eventually rises to the

left and to the right. You also know that the graph

crosses the x-axis at nonrepeated zeros x = and

x = 0, but does not cross the x-axis at repeated zero

x = 2, because its multiplicity is even. Draw a

continuous curve through the points as shown in the

figure.

Answer:

Graph a Polynomial Function

Determine the zeros and state the multiplicity of any repeated zeros for f (x) = 3x(x + 2)2(2x – 1)3.

A. 0, –2 (multiplicity 2), (multiplicity 3)

B. 2 (multiplicity 2), – (multiplicity 3)

C. 4 (multiplicity 2), (multiplicity 3)

D. –2 (multiplicity 2), (multiplicity 3)

Model Data Using Polynomial Functions

A. POPULATION The table to the right shows a town’s population over an 8-year period. Year 1 refers to the year 2001, year 2 refers to the year 2002, and so on. Create a scatter plot of the data, and determine the type of polynomial function that could be used to represent the data.

Model Data Using Polynomial Functions

Answer: cubic function;

Enter the data using the list feature of a graphing calculator. Let L1 be the number of the year. Then create a scatter plot of the data. The curve of the scatter plot resembles the graph of a cubic equation, so we will use a cubic regression.

Model Data Using Polynomial Functions

B. POPULATION The table below shows a town’s population over an 8-year period. Year 1 refers to the year 2001, year 2 refers to the year 2002, and so on. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth, and state the correlation coefficient.

Model Data Using Polynomial Functions

Using the CubicReg tool on a graphing calculator and rounding each coefficient to the nearest thousandth yields f (x) = 10.020x

3 – 176.320x 2 + 807.469x + 4454.786.

The value of r 2 for the data is 0.89, which is close to 1, so

the model is a good fit. We can graph the complete (unrounded) regression by sending it to the Y= menu. In the Y= menu, pick up this regression equation by entering VARS , Statistics, EQ. Graph this function and the scatter plot in the same viewing window. The function appears to fit the data reasonably well.

Model Data Using Polynomial Functions

Answer: f (x) = 10.020x 3 – 176.320x

2 + 807.469x + 4454.786; r

2 = 0.89

Model Data Using Polynomial Functions

C. POPULATION The table below shows a town’s population over an 8-year period. Year 1 refers to the year 2001, year 2 refers to the year 2002, and so on. Use the model to estimate the population of the town in the year 2012.

Model Data Using Polynomial Functions

Answer: 6069

Use 12 for the year 2012 and use the CALC feature on a calculator to find f (12). The value of f (12) is 6069.1926, so the population in 2012 will be about 6069.

Model Data Using Polynomial Functions

D. POPULATION The table below shows a town’s population over an 8-year period. Year 1 refers to the year 2001, year 2 refers to the year 2002, and so on. Use the model to determine the approximate year in which the population reaches 10,712.

Model Data Using Polynomial Functions

Answer: 2015

Graph the line y = 10,712 for Y2. Then use 5: intersect on the CALC menu to find the point of intersection of y = 10,712 with f (x). The intersection occurs when x ≈ 15, so the approximate year in which the population will be 10,712 is 2015.

BIOLOGY The number of fruit flies that hatched after day x is given in the table. Write a polynomial function to model the data set. Round each coefficient to the nearest thousandth, and state the correlation coefficient. Use the model to estimate the number of fruit flies hatched after 8 days.

A. y = 12.014x 2 – 72.940x + 5.3; r = 0.84; 190

B. y = 20.833x 3 + 125.786x

2 + 251.238x + 195.714; r 2 = 0.99;

40,922

C. y = 10x4 + 60.833x 3 + 141.5x

2 + 202.667x + 182; r 2 = 1;

82,966

D. y = 70.893x – 20.672; r = 0.829; 346