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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
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.Course 3
12-7 Lines of Best Fit
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.
xm = = 6 4 + 7 + 3 + 8 + 8 + 66
ym = = 4 4 + 5 + 2 + 6 + 7 + 46
23
x 4 7 3 8 8 6
y 4 5 2 6 7 4
23(xm, ym)= 6, 4
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!
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.
23
Course 3
12-7 Lines of Best Fit
Find the slope.
y – y1 = m(x – x1) Use point-slope form.
y – 4 = (x – 6)23
23 Substitute.
y – 4 = x – 423
23
23y = x +2
3
The equation of a line of best fit is .23y = x +23
Additional Example 1 Continued
23
13m = = =
6 – 4
8 – 6
1
223
Course 3
12-7 Lines of Best Fit
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.
xm = = 2 –1 + 0 + 2 + 6 + –3 + 8 6
ym = = 1 –1 + 0 + 3 + 7 + –7 + 46
x –1 0 2 6 –3 8
y –1 0 3 7 –7 4
(xm, ym) = (2, 1)
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.
Course 3
12-7 Lines of Best Fit
Find the slope.
y – y1 = m(x – x1) Use point-slope form.
y – 1 = (x – 2)98 Substitute.
y – 1 = x –98
94
The equation of a line of best fit is . y = x –98
54
Check It Out: Example 1 Continued
m = = 10 – 1 10 – 2
98
y = x –98
54
Course 3
12-7 Lines of Best Fit
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.
Additional Example 2: Sports Application
Let 1990 represent year 0. The first point is then (0, 98), and the last point is (12, 107).
xm = = 50 + 2 + 4 + 7 + 125
Year 1990 1992 1994 1997 2002
Distance (ft) 98 101 103 106 107
ym = = 103 98 + 101 + 103 + 106 + 107 5
(xm, ym) = (5, 103)
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.
Course 3
12-7 Lines of Best Fit
m = = 0.8 107 - 10310 - 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.
Additional Example 2 Continued
Course 3
12-7 Lines of Best Fit
Since 1990 represents year 0, 2006 represents year 16.
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.
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).
xm = = 60 + 5 + 7 + 8 + 105
ym = = 132100 + 120 + 130 + 140 + 170 5
Year 1990 1995 1997 1998 2000
Lift (lb) 100 120 130 140 170
(xm, ym) = (6, 132)
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
weig
ht
(lb)
0100
120
140
160
180
2 4 6 8 10
200
Course 3
12-7 Lines of Best Fit
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
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 193 lbs is a reasonable prediction.
Check It Out: Example 2 Continued
Course 3
12-7 Lines of Best Fit
