Question 285372
A trend line shows the average of a series of points over time.
We can assume the independent variable, age, is the x-axis.
That means height is a function of age, that is f(x), so it will be plotted on the y-axis.
The data points you have are:
(0,4)
(4,7)
(7,11.5)
.
To draw a line, you need two points.
(0,4) should be one of them.
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The slope-intercept form is:  y = mx+b
In this case, we know that when x=0, y=4, so the y-intercept has to be 4.
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The slope is the rise/run, which is the change in y divided by the change in x.
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From (0,4) to (4,7),
change in y = 7-4 = 3
change in x = 4-0 = 4
m = 3/4 = .75
.
From (4,7) to (7,11.5),
change in y = 11.5-7 = 4.5
change in x = 7-3 = 4
slope = 4.5/4 = 1.25
.
So the slope is getting steeper, which means a straight line will have to miss one or both of the points.  That means the trend line will be somewhere between them.
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We can pick the average value of 'x' and 'y' as the point that will be the second point on which we will base our trend line.
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The average value of 'y' = (7+11.5)/2 = 18.5/2 = 9.25
The average value of 'x' = (4+7)/2 = 11/2 = 5.5
So the point is (5.5,9.25)
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Based on the point (0,4) as our starting point...which reasonable since that is the at birth height...
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change in y = 9.25-4 = 5.25
change in x = 5.5-0 = 5.5
slope = 5.25/5.5 = .954545
.
Recall our second point will be (5.5,9.25)
The equation will be:
y = .954545*x + 4
.
{{{ drawing(500,500,-5,15,-5,15, 
grid(1),
graph(500,500,-5,15,-5,15,.95*x+4),
blue(circle(0, 4,.15)),
blue(circle(4, 7,.15)),
blue(circle(7, 11.5,.15))
) }}}