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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

Lesson MenuA. 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: (–∞, ∞)

B. Analyze f(x) = 3x3.

5–Minute Check 1You analyzed graphs of functions. (Lesson 1-2)

- Graph polynomial functions.
- Model real-world data with polynomial functions.

- 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 = x3. The graph of f(x) = (x – 3)5 is the graph of y = x5 translated 3 units to the right.

Answer:

Example 1Graph Transformations of Monomial Functions

B. Graph f(x) = x6– 1.

This is an even-degree function, so its graph is similar to the graph of y = x2. The graph of f(x) = x6 – 1 is the graph of y = x6 translated 1 unit down.

Answer:

Example 1The 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

A. Describe the end behavior of the graph of f(x) = 3x4 – x3 + x2 + x – 1 using limits. Explain your reasoning using the leading term test.

Example 2Write in standard form as f(x) = 2x5 – x3 – 3x2. 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

B. Describe the end behavior of the graph of f(x) = –3x2 + 2x5–x3 using limits. Explain your reasoning using the leading term test.

Example 2The degree is 5 and the leading coefficient is –2. Because the degree is odd and the leading coefficient is negative, .

Answer:

Apply the Leading Term Test

C. Describe the end behavior of the graph of f(x) = –2x5– 1 using limits. Explain your reasoning using the leading term test.

Example 2A. 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, .

Describe the end behavior of the graph of g(x) = –3x5 + 6x3 – 2 using limits. Explain your reasoning using the leading term test.

Example 2Zeros of a Polynomial Function

State the number of possible real zeros and turning points of f(x) = x3 + 5x2 + 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.

x3 + 5x2 + 4x = 0 Set f(x) equal to 0.

x(x2 + 5x + 4) = 0 Factor the greatest common factor, x.

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

Example 3Zeros 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) = x3 + 5x2 + 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.

Example 3Zeros 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) = x3 + 5x2 + 4x = x(x + 1)(x + 4), so f has three zeros, x = 0, x = –1, and x = –4.

Example 3State the number of possible real zeros and turning points of f(x) = x4 – 13x2 + 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

Example 3Zeros of a Polynomial Function in Quadratic Form

State the number of possible real zeros and turning points for h(x) = x4 – 4x2 + 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 x4 – 4x2 + 3 = (x2)2 – 4(x2) + 3. Let u = x2.

(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.

Example 4Zero Product Property

or x + 1 = 0 or x – 1 = 0

,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) = x4 – 4x2 + 3 confirms this. Notice that there are 3 turning points, which is also consistent with a quartic function.

Zeros of a Polynomial Function in Quadratic Form

(x2 – 3)(x2 – 1) = 0 Substitute x2 for u.

Example 4Answer:The degree is 4, so h has at most 4 distinct real zeros and at most 3 turning points. h(x) = x4 – 4x2 + 3 = (x2 – 3)(x – 1)(x + 1), so h has four distinct real zeros, x = , x = –1, and x = 1.

Zeros of a Polynomial Function in Quadratic Form

Example 4A. 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,

State the number of possible real zeros and turning points of g(x) = x5 – 5x3 – 6x. Then determine all of the real zeros by factoring.

Example 4Polynomial Function with Repeated Zeros

State the number of possible real zeros and turning points of h(x) = x4 + 5x3 + 6x2. 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.

x4 + 5x3 + 6x2 = 0 Set h(x) equal to 0.

x2(x2 + 5x + 6) = 0 Factor the greatest common factor, x2.

x2(x + 3)(x + 2) = 0 Factor completely.

Example 5Polynomial 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) = x4 + 5x3 + 6x2 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.

Example 5Polynomial Function with Repeated Zeros

Answer:The degree is 4, so h has at most 4 distinct real zeros and at most 3 turning points. h(x) = x4 + 5x3 + 6x2 = x2(x + 2)(x + 3), so h has three zeros, x = 0, x = –2, and x = –3. Of the zeros, x = 0 is repeated.

Example 5State the number of possible real zeros and turning points of g(x) = x4 – 4x3+ 4x2. 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

Example 5The 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:

Graph a Polynomial Function

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

Example 6The distinct real zeros are x = 0, x = , and x = 2. The zero at x = 2 has multiplicity 2.

Answer:0, , 2 (multiplicity: 2)

Graph a Polynomial Function

B. For f(x) = x(3x + 1) (x – 2)2, determine the zeros and state the multiplicity of any repeated zeros.

Example 6C. 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.

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

Example 6Plot 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.

Graph a Polynomial Function

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

Example 6A. 0, –2 (multiplicity 2), (multiplicity 3)

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

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

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

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

Example 6Model 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.

Example 7Model Data Using Polynomial Functions

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.

Answer:cubic function;

Example 7Model 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.

Example 7Using the CubicReg tool on a graphing calculator and rounding each coefficient to the nearest thousandth yields f(x) = 10.020x3 – 176.320x2 + 807.469x + 4454.786. The value of r2 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

Example 7Model Data Using Polynomial Functions

Answer:f(x) = 10.020x3 – 176.320x2 + 807.469x + 4454.786; r2 = 0.89

Example 7Model 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.

Example 7Model Data Using Polynomial Functions

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.

Answer:6069

Example 7Model 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.

Example 7Model Data Using Polynomial Functions

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.

Answer:2015

Example 7BIOLOGY 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 = 12.014x2 – 72.940x + 5.3; r = 0.84; 190

B. y = 0.2x4 – 1.683x3 + 13.3x2 – 32.6; r2 = 0.75; 476

C. y = 0.907x3 – 4.331x2 + 12.535x – 2.262; r2 = 0.99; 285

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

Example 7
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