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12-7 Lines of Best Fit Course 3 Warm Up Warm Up Lesson Presentation Lesson Presentation

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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? dashed above Course 3 12-7 Lines of Best Fit

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

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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: (x m, y m ) draw the line through (x m, y m ) that appears to best fit the data. estimate the coordinates of another point on the line. find the equation of the line. Course 3 12-7 Lines of Best Fit

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Plot the data and find a line of best fit. Additional Example 1: Finding a Line of Best Fit Plot the data points and find the mean of the x- and y-coordinates. x m = = 6 4 + 7 + 3 + 8 + 8 + 6 6 y m = = 4 4 + 5 + 2 + 6 + 7 + 4 6 2 3 x473886 y452674 2 3 (x m, y m )= 6, 4 Course 3 12-7 Lines of Best Fit

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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! Course 3 12-7 Lines of Best Fit

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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 Course 3 12-7 Lines of Best Fit

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

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Plot the data and find a line of best fit. Check It Out: Example 1 Plot the data points and find the mean of the x- and y-coordinates. x m = = 2 –1 + 0 + 2 + 6 + –3 + 8 6 y m = = 1 –1 + 0 + 3 + 7 + –7 + 4 6 x–1026–38 y–1037–74 (x m, y m ) = (2, 1) Course 3 12-7 Lines of Best Fit

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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. Course 3 12-7 Lines of Best Fit

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

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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. Example 2: Sports Application Let 1990 represent year 0. The first point is then (0, 98), and the last point is (12, 107). x m = = 5 0 + 2 + 4 + 7 + 12 5 Year19901992199419972002 Distance (ft)98101103106107 y m = = 103 98 + 101 + 103 + 106 + 107 5 (x m, y m ) = (5, 103) Course 3 12-7 Lines of Best Fit

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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. Course 3 12-7 Lines of Best Fit

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m = = 0.8 107 - 103 10 - 5 Find the slope. y – y 1 = m(x – x 1 )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. Additional Example 2 Continued Course 3 12-7 Lines of Best Fit Since 1990 represents year 0, 2006 represents year 16.

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

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Predict the winning weight lift in 2010. Check It Out: Example 2 Let 1990 represent year 0. The first point is then (0, 100), and the last point is (10, 170). x m = = 6 0 + 5 + 7 + 8 + 10 5 y m = = 132 100 + 120 + 130 + 140 + 170 5 Year19901995199719982000 Lift (lb)100120130140170 (x m, y m ) = (6, 132) Course 3 12-7 Lines of Best Fit

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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) 0 100 120 140 160 180 246810 200 Course 3 12-7 Lines of Best Fit

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m = = 4 140 – 132 7 – 5 Find the slope. y – y 1 = m(x – x 1 )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

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

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Lesson Quiz Plot the data to find the line of best fit. 1. 2. Insert Lesson Title Here Possible answer: y = 2x + 1 Possible answer: y = –10x + 9 Course 3 12-7 Lines of Best Fit

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