Example 3: y=14.14(1.41)^x Example 4: y=10.28(1.5)^x
Practice makes perfect
Let's look at the given table.
Year, x
1
2
3
4
5
6
7
Number of trampolines, y
15
23
40
52
80
105
140
We will repeat the steps in Examples 3 and 4 using the table above.
Example 3
We will begin by making a scatter plot of the data.
Looking at the graph, we can conclude that the data appear exponential. Therefore, the model that represents the data will be an exponential function which has the form of y=ab^x. To find the values of a and b, let's use the points ( 3, 40) and ( 5, 80). By substituting the points in the function, we can have a system.40=ab^3 & (I) 80=ab^5 & (II)
Let's solve the system!
Because base of an exponential function is positive, the value of b is about 1.41. Next, we will find the value of a by substituting b=1.41 in Equation I.
Therefore, the model can be written as below.
y=14.14(1.41)^x
Example 4
We will first create a table of data pairs (x,ln y).
Year, x
1
2
3
4
5
6
7
Number of trampolines, y
15
23
40
52
80
105
140
ln y
2.71
3.14
3.69
3.95
4.38
4.65
4.94
Next, we will plot the transformed points.
The points lie close to a line, so an exponential model should be a good fit for the original data. To write an exponential model y=ab^x, let's use the points ( 2, 3.14) and ( 4, 3.95) because they appear to be on the line. With this, we will write an equation in point-slope form.
ln y- ln y_1= m(x- x_1)
In this form, m is the slope and ( x_1, ln y_1) is a reference point that is on the line. Using the Slope Formula, we can substitute the chosen points and find the slope.
Now that we found the slope, we can write the equation in point-slope form. Let the point ( 2, 3.14) be our reference point.
ln y-3.14=0.405(x-2)
Next, we will isolate y to have the model in the form of y=ab^x.
