1 / 18

6-9

Curving Fitting with Polynomial Functions. 6-9. Warm Up. Lesson Presentation. Lesson Quiz. Holt Algebra 2. Warm Up. 1. For f ( x ) = x 3 + 5, write the rule for g ( x ) = f ( x – 1) – 2 and sketch its graph. g ( x ) = ( x – 1) 3 + 3.

colorado-alexander
Download Presentation

6-9

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

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

  2. Warm Up 1.For f(x) = x3 + 5, write the rule for g(x) = f(x – 1) – 2 and sketch its graph. g(x) = (x – 1)3 + 3 2. Write a function that reflects f(x) = 2x3 + 1 across the x-axis and shifts it 3 units down. h(x) = –2x3 – 4

  3. 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.

  4. Example 1A: Using Finite Differences to Determine Degree Use finite differences to determine the degree of the polynomial that best describes the data. The x-values increase by a constant 2. Find the differences of the y-values. 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.

  5. Example 1B: Using Finite Differences to Determine Degree Use finite differences to determine the degree of the polynomial that best describes the data. The x-values increase by a constant 3. Find the differences of the y-values. 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.

  6. Check It Out! Example 1 Use finite differences to determine the degree of the polynomial that best describes the data. The x-values increase by a constant 3. Find the differences of the y-values. 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.

  7. 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. 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

  8. 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) ≈ 104.58x3 – 283.85x2 + 1098.34x + 4266.79

  9. 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. 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

  10. 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

  11. 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.

  12. 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. 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.

  13. Example 3 Continued Step 2Write the polynomial model. The data can be modeled by f(x) = 32.23x4 – 339.13x3 + 1069.59x2 – 858.99x + 693.88 Step 3Find 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.

  14. 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. 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.

  15. Check It Out! Example 3 Continued Step 2Write the polynomial model. The data can be modeled by f(x) = 19.09x4 – 377.90x3 + 2153.24x2 – 2183.29x + 3871.46 Step 3Find 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.

  16. Lesson Quiz: Part I 1.Use finite differences to determine the degree of the polynomial that best describes the data. cubic

  17. 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. f(x) = 7.08x4 – 126.92x3 + 595.95x2 – 241.81x + 2780.54; about $3003.50

More Related