Course 3 12-7 Lines of Best Fit Learn to recognize relationships in data and find the equation of a line of best fit.

Course 3 12-7 Lines of Best Fit 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.

Additional Example 1: Finding a Line of Best Fit Course 3 12-7 Lines of Best Fit Additional Example 1: Finding a Line of Best Fit Plot the data and find a line of best fit. x 4 7 3 8 6 y 5 2 Plot the data points and find the mean of the x- and y-coordinates. xm = = 6 4 + 7 + 3 + 8 + 8 + 6 6 2 3 (xm, ym)= 6, 4 ym = = 4 4 + 5 + 2 + 6 + 7 + 4 6 2 3

Lines of Best Fit 12-7 Remember! Course 3 12-7 Lines of Best Fit A line of best fit is a line that comes close 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. Remember!

Additional Example 1 Continued Course 3 12-7 Lines of Best Fit Additional Example 1 Continued 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

Additional Example 1 Continued Course 3 12-7 Lines of Best Fit Additional Example 1 Continued 2 3 1 m = = = 6 – 4 8 – 6 Find the slope. y – y1 = m(x – x1) Use point-slope form. y – 4 = (x – 6) 2 3 Substitute. y – 4 = x – 4 2 3 2 3 y = x + The equation of a line of best fit is . 2 3 y = x +

Lines of Best Fit 12-7 Check It Out: Example 1 Course 3 12-7 Lines of Best Fit Check It Out: Example 1 Plot the data and find a line of best fit. x –1 2 6 –3 8 y 3 7 –7 4 Plot the data points and find the mean of the x- and y-coordinates. xm = = 2 –1 + 0 + 2 + 6 + –3 + 8 6 (xm, ym) = (2, 1) ym = = 1 –1 + 0 + 3 + 7 + –7 + 4 6

Check It Out: Example 1 Continued Course 3 12-7 Lines of Best Fit Check It Out: 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.

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

Additional Example 2: Sports Application Course 3 12-7 Lines of Best Fit 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. Is it reasonable to make this prediction? Explain. Year 1990 1992 1994 1997 2002 Distance (ft) 98 101 103 106 107 Let 1990 represent year 0. The first point is then (0, 98), and the last point is (12, 107). xm = = 5 0 + 2 + 4 + 7 + 12 5 (xm, ym) = (5, 103) ym = = 103 98 + 101 + 103 + 106 + 107 5

Additional Example 2 Continued Course 3 12-7 Lines of Best Fit 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.

Additional Example 2 Continued Course 3 12-7 Lines of Best Fit Additional Example 2 Continued m = = 0.8 107 - 103 10 - 5 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.

Additional Example 2 Continued Course 3 12-7 Lines of Best Fit Additional Example 2 Continued y = 0.8(16) + 99 Substitute. y = 12.8 + 99 Add to find the distance. y = 111.8 The equation predicts a winning distance of about 112 feet for the year 2006. A toss of about 112 feet is a reasonable prediction.

Lines of Best Fit 12-7 Check It Out: Example 2 Course 3 12-7 Lines of Best Fit Check It Out: Example 2 Predict the winning weight lift in 2010. Year 1990 1995 1997 1998 2000 Lift (lb) 100 120 130 140 170 Let 1990 represent year 0. The first point is then (0, 100), and the last point is (10, 170). xm = = 6 0 + 5 + 7 + 8 + 10 5 (xm, ym) = (6, 132) ym = = 132 100 + 120 + 130 + 140 + 170 5

Check It Out: Example 2 Continued Course 3 12-7 Lines of Best Fit Check It Out: 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. Years since 1990 weight (lb) 100 120 140 160 180 2 4 6 8 10 200

Check It Out: Example 2 Continued Course 3 12-7 Lines of Best Fit Check It Out: Example 2 Continued m = = 4 140 – 132 7 – 5 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.

Check It Out: Example 2 Continued Course 3 12-7 Lines of Best Fit Check It Out: Example 2 Continued y = 4(20) + 112 Substitute and add to find the winning weight lift. y = 192 The equation predicts a winning weight lift of about 192 lb for the year 2010. A weight lift of 193 lbs is a reasonable prediction.