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11-7

11-7. Lines of Best Fit. Course 3. Warm Up. Problem of the Day. Lesson Presentation. 11-7. Lines of Best Fit. Course 3. Warm Up Answer the questions about the inequality 5 x + 10 y > 30. 1. Would you use a solid or dashed boundary line?

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11-7

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  1. 11-7 Lines of Best Fit Course 3 Warm Up Problem of the Day Lesson Presentation

  2. 11-7 Lines of Best Fit Course 3 Warm Up Answer the questions about the inequality 5x + 10y > 30. 1.Would you use a solid or dashed boundary line? 2. Would you shade above or below the boundary line? 3. What are the intercepts of the graph? dashed above (0, 3) and (6, 0)

  3. 11-7 Lines of Best Fit Course 3 Problem of the Day Write an inequality whose positive solutions form a triangular region with an area of 8 square units. (Hint: Sketch such a region on a coordinate plane.) Possible answer: y < –x + 4

  4. 11-7 Lines of Best Fit Course 3 Learn to recognize relationships in data and find the equation of a line of best fit.

  5. 11-7 Lines of Best Fit Course 3 When data show a correlation, you can estimate and draw a line of best fit that approximates a trend for a set of data and use it to make predictions. • To estimate the equation of a line of best fit: • calculate the means of the x-coordinates and y-coordinates: (xm, ym) • draw the line through (xm, ym) that appears to best fit the data. • estimate the coordinates of another point on the line. • find the equation of the line.

  6. 11-7 Lines of Best Fit 2 3 2 3 xm = = 6 ym = = 4 4 + 5 + 2 + 6 + 7 + 4 4 + 7 + 3 + 8 + 8 + 6 6 6 (xm, ym)= 6, 4 Course 3 Additional Example 1: Finding a Line of Best Fit Plot the data and find a line of best fit. Plot the data points and find the mean of the x- and y-coordinates.

  7. 11-7 Lines of Best Fit Remember! The line of best fit is the line that comes closest to all the points on a scatter plot. Try to draw the line so that about the same number of points are above the line as below the line. Course 3

  8. 11-7 Lines of Best Fit Draw a line through 6, 4 that best represents the data. Estimate and plot the coordinates of another point on that line, such as (8, 6). Find the equation of the line. 2 3 Course 3 Additional Example 1 Continued

  9. 11-7 Lines of Best Fit 6 – 4 1 m = = = 8 – 6 2 y – 4 = (x – 6) y – 4 = x – 4 1 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 2 2 y = x + y = x + 3 3 The equation of a line of best fit is . Course 3 Additional Example 1 Continued Find the slope. y – y1 = m(x – x1) Use point-slope form. Substitute.

  10. 11-7 Lines of Best Fit xm = = 2 ym = = 1 –1 + 0 + 2 + 6 + –3 + 8 –1 + 0 + 3 + 7 + –7 + 4 6 6 Course 3 Try This: Example 1 Plot the data and find a line of best fit. Plot the data points and find the mean of the x- and y-coordinates. (xm, ym) = (2, 1)

  11. 11-7 Lines of Best Fit Course 3 Try This: Example 1 Continued Draw a line through (2, 1) that best represents the data. Estimate and plot the coordinates of another point on that line, such as (10, 10). Find the equation of the line.

  12. 11-7 Lines of Best Fit 9 y – 1 = (x – 2) 8 9 9 9 5 9 5 y – 1 = x – 8 8 8 4 4 4 9 10 – 1 m = = 8 y = x – 10 – 2 The equation of a line of best fit is . y = x – Course 3 Try This: Example 1 Continued Find the slope. y – y1 = m(x – x1) Use point-slope form. Substitute.

  13. 11-7 Lines of Best Fit xm = = 5 0 + 2 + 4 + 7 + 12 98 + 101 + 103 + 106 + 107 5 5 ym = = 103 Course 3 Additional Example 2: Sports Application Find a line of best fit for the Main Street Elementary annual softball toss. Use the equation of the line to predict the winning distance in 2006. Let 1990 represent year 0. The first point is then (0, 98), and the last point is (12, 107). (xm, ym) = (5, 103)

  14. 11-7 Lines of Best Fit Course 3 Additional Example 2 Continued Draw a line through (5, 103)that best represents the data. Estimate and plot the coordinates of another point on that line, such as (10, 107). Find the equation of the line.

  15. 11-7 Lines of Best Fit 107 - 103 m = = 0.8 10 - 5 Course 3 Additional Example 2 Continued Find the slope. y – y1 = m(x – x1) Use point-slope form. y – 103 = 0.8(x – 5) Substitute. y – 103 = 0.8x – 4 y = 0.8x + 99 The equation of a line of best fit is y = 0.8x + 99. Since 1990 represents year 0, 2006 represents year 16.

  16. 11-7 Lines of Best Fit Course 3 Additional Example 2 Continued y = 0.8(16) + 99 Substitute. y = 12.8 + 99 y = 111.8 The equation predicts a winning distance of about 112 feet for the year 2006.

  17. 11-7 Lines of Best Fit xm = = 6 0 + 5 + 7 + 8 + 10 100 + 120 + 130 + 140 + 170 5 5 ym = = 132 Course 3 Try This: Example 2 Predict the winning weight lift in 2010. Let 1990 represent year 0. The first point is then (0, 100), and the last point is (10, 170). (xm, ym) = (6, 132)

  18. 11-7 Lines of Best Fit 200 180 160 weight (lb) 140 120 100 2 4 6 8 10 0 Years since 1990 Course 3 Try This: Example 2 Continued Draw a line through (5, 132)the best represents the data. Estimate and plot the coordinates of another point on that line, such as (7, 140). Find the equation of the line.

  19. 11-7 Lines of Best Fit 140 – 132 m = = 4 7 – 5 Course 3 Try This: Example 2 Continued Find the slope. y – y1 = m(x – x1) Use point-slope form. y – 132 = 4(x – 5) Substitute. y – 132 = 4x – 20 y = 4x + 112 The equation of a line of best fit is y = 4x + 112. Since 1990 represents year 0, 2010 represents year 20.

  20. 11-7 Lines of Best Fit Course 3 Try This: Example 2 Continued y = 4(20) + 112 Substitute. y = 192 The equation predicts a winning lift of about 192 lb for the year 2010.

  21. 11-7 Lines of Best Fit Course 3 Insert Lesson Title Here Lesson Quiz Plot the data to find the line of best fit. 1. 2. Possible answer: y = 2x + 1 Possible answer: y = –10x + 9

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