Curving Fitting with 6-9 Polynomial Functions Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 2 Holt Algebra 2

Warm Up Find a line of best fit for the data. 1. x 2 8 15 21 24 y 70 62 80 190 160 y = 5.45x + 36.12 2. x 38 42 44 35 49 y 92 80 75 81 68 y = –1.28x + 132.66

Objectives Use finite differences to determine the degree of a polynomial that will fit a given set of data. Use technology to find polynomial models for a given set of data.

The table shows the closing value of a stock index on the first day of trading for various years. To create a mathematical model for the data, you will need to determine what type of function is most appropriate. In Lesson 5-8, you learned that a set of data that has constant second differences can be modeled by a quadratic function. Finite difference can be used to identify the degree of any polynomial data.

Example 1A: Using Finite Differences to Determine Degree Use finite differences to determine the degree of the polynomial that best describes the data. x 4 6 8 10 12 14 y –2 4.3 8.3 10.5 11.4 11.5 The x-values increase by a constant 2. Find the differences of the y-values. y –2 4.3 8.3 10.5 11.4 11.5 First differences: 6.3 4 2.2 0.9 0.1 Not constant Second differences: –2.3 –1.8 –1.3 –0.8 Not constant Third differences: 0.5 0.5 0.5 Constant The third differences are constant. A cubic polynomial best describes the data.

Example 1B: Using Finite Differences to Determine Degree Use finite differences to determine the degree of the polynomial that best describes the data. x –6 –3 3 6 9 y –9 16 26 41 78 151 The x-values increase by a constant 3. Find the differences of the y-values. y –9 16 26 41 78 151 First differences: 25 10 15 37 73 Not constant Second differences: –15 5 22 36 Not constant Third differences: 20 17 14 Not constant Fourth differences: –3 –3 Constant The fourth differences are constant. A quartic polynomial best describes the data.

Check It Out! Example 1 Use finite differences to determine the degree of the polynomial that best describes the data. x 12 15 18 21 24 27 y 3 23 29 31 43 The x-values increase by a constant 3. Find the differences of the y-values. y 3 23 29 31 43 First differences: 20 6 0 2 12 Not constant Second differences: –14 –6 2 10 Not constant Third differences: 8 8 8 Constant The third differences are constant. A cubic polynomial best describes the data.

Once you have determined the degree of the polynomial that best describes the data, you can use your calculator to create the function.

Example 2: Using Finite Differences to Write a Function The table below shows the population of a city from 1960 to 2000. Write a polynomial function for the data. Year 1960 1970 1980 1990 2000 Population (thousands) 4,267 5,185 6,166 7,830 10,812 Step 1 Find the finite differences of the y-values. First differences: 918 981 1664 2982 Second differences: 63 683 1318 Third differences: 620 635 Close

Example 2 Continued Step 2 Determine the degree of the polynomial. Because the third differences are relatively close, a cubic function should be a good model. Step 3 Use the cubic regression feature on your calculator. f(x) ≈ 0.10x3 – 2.84x2 + 109.84x + 4266.79

Check It Out! Example 2 The table below shows the gas consumption of a compact car driven a constant distance at various speed. Write a polynomial function for the data. Speed 25 30 35 40 45 50 55 60 Gas (gal) 23.8 25.2 25.4 27 30.6 37 Step 1 Find the finite differences of the y-values. First differences: 1.2 0.2 –0.2 0.4 1.6 3.6 6.4 Second differences: –1 –0.4 0.6 1.2 2 2.8 Third differences: 0.6 1 0.6 0.8 0.8Close

Check It Out! Example 2 Continued Step 2 Determine the degree of the polynomial. Because the third differences are relatively close, a cubic function should be a good model. Step 3 Use the cubic regression feature on your calculator. f(x) ≈ 0.001x3 – 0.113x2 + 4.134x + 24.867

Often, real-world data can be too irregular for you to use finite differences or find a polynomial function that fits perfectly. In these situations, you can use the regression feature of your graphing calculator. Remember that the closer the R2-value is to 1, the better the function fits the data.

Example 3: Curve Fitting Polynomial Models The table below shows the opening value of a stock index on the first day of trading in various years. Use a polynomial model to estimate the value on the first day of trading in 2000. Year 1994 1995 1996 1997 1998 1999 Price ($) 683 652 948 1306 863 901 Step 1 Choose the degree of the polynomial model. Let x represent the number of years since 1994. Make a scatter plot of the data. The function appears to be cubic or quartic. Use the regression feature to check the R2-values. cubic: R2 ≈ 0.5833 quartic: R2 ≈ 0.8921 The quartic function is more appropriate choice.

Example 3 Continued Step 2 Write the polynomial model. The data can be modeled by f(x) = 32.23x4 – 339.13x3 + 1069.59x2 – 858.99x + 693.88 Step 3 Find the value of the model corresponding to 2000. 2000 is 6 years after 1994. Substitute 6 for x in the quartic model. f(6) = 32.23(6)4 – 339.13(6)3 + 1069.59(6)2 – 858.99(6) + 693.88 Based on the model, the opening value was about $2563.18 in 2000.

Check It Out! Example 3 The table below shows the opening value of a stock index on the first day of trading in various years. Use a polynomial model to estimate the value on the first day of trading in 1999. Year 1994 1995 1996 2000 2003 2004 Price ($) 3754 3835 5117 11,497 8342 10,454 Step 1 Choose the degree of the polynomial model. Let x represent the number of years since 1994. Make a scatter plot of the data. The function appears to be cubic or quartic. Use the regression feature to check the R2-values. cubic: R2 ≈ 0.8624 quartic: R2 ≈ 0.9959 The quartic function is more appropriate choice.

Check It Out! Example 3 Continued Step 2 Write the polynomial model. The data can be modeled by f(x) = 19.09x4 – 377.90x3 + 2153.24x2 – 2183.29x + 3871.46 Step 3 Find the value of the model corresponding to 1999. 1999 is 5 years after 1994. Substitute 5 for x in the quartic model. f(5) = 19.09(5)4 – 377.90(5)3 + 2153.24(5)2 – 2183.29(5) + 3871.46 Based on the model, the opening value was about $11,479.76 in 1999.

x 8 10 12 14 16 18 y 7.2 1.2 –8.3 –19.1 –29 –35.8 Lesson Quiz: Part I 1. Use finite differences to determine the degree of the polynomial that best describes the data. x 8 10 12 14 16 18 y 7.2 1.2 –8.3 –19.1 –29 –35.8 cubic

Lesson Quiz: Part II 2. The table shows the opening value of a stock index on the first day of trading in various years. Write a polynomial model for the data and use the model to estimate the value on the first day of trading in 2002. Year 1994 1996 1998 2000 2001 2004 Price ($) 2814 3603 5429 3962 4117 3840 f(x) = 7.08x4 – 126.92x3 + 595.95x2 – 241.81x + 2780.54; about $3003.50