If the ratios of consecutive y-values are equal, then the data can be modeled by an exponential function. If the difference of consecutive y-values is constant, then the data can be modeled by a linear function.
Exponential relationship of data: No. Example model: y=- 0.8x+66 Explanation: See solution.
Practice makes perfect
We want to determine whether the data show an exponential relationship. Then we will write a function that models the data. Let's do those things one at a time.
Determining the Type of the Model
If the ratios of consecutive y-values are equal, then the data can be modeled by an exponential function. If the difference of consecutive y-values is constant, then the data can be modeled by a linear function. Consider the given table.
x
0
10
20
30
40
50
60
y
66
58
48
42
31
26
21
Let's calculate the ratios of the consecutive y-values.
58/66 &≈ 0.879, 48/58 ≈ 0.828, [0.8em]
42/48&=0.875, 31/42 ≈ 0.738, [0.8em]
26/31 &≈ 0.839 , 21/26 ≈ 0.808
We can see that the ratios are not equal to each other, so the data cannot be modeled by an exponential function. Let's determine the differences between the consecutive y-values.
66-58&= 8, 58-48= 10,
48-42&=6, 42-31=11,
31-26&=5, 26-21=5
The difference of consecutive y-values is almost constant, around 8, so the data can be modeled by a linear function.
y=mx+b
Writing the Model
To find the values of m and b, we will use two of the ordered pairs given in the table. We will use (0,66) and (30,42). Let's start by substituting 0 for x and for 66.
Now that we know that b=66, we can partially write the equation.
y=mx+66
To find the value of m, we will substitute 30 for x and 42 for y in the above equation.
Now that we know that m=- 45, we can write the full equation that models the data in the given table.
y=- 4/5x+66
Writing 45 as 0.8, we can obtain our final answer.
y=- 4/5x+66 ⇔ y=- 0.8x+66
