x 0 1 2 3 4
f(x) 12.5 13.75 15.125 16.638 18.301
If the change in x values and change in y values are constant then the data will be linear.
So, change in x from 0 to 1 is 1-0=1
Similarly change in x from 1 to 2 is 2-1=1
So, change in x's is constant in the given data.
Now let's check for change in y values.
Change in y from 12.5 to 13.75 is 13.75-12.5=1.25
Similarly change in y from 13.75 to 15.125 is 15.125-13.75=1.375
Changes in y's are not constant.
So, the given data is not linear.
Let's assume the exponential function is:
![y=a(b)^x](https://tex.z-dn.net/?f=%20y%3Da%28b%29%5Ex%20)
Let's take any two points from the data and plug into the above equation to get the equation. Let's plug in the first point (0, 12.5) in the above function. So,
![12.5=a(b)^0](https://tex.z-dn.net/?f=%2012.5%3Da%28b%29%5E0%20)
Since ![b^0=1](https://tex.z-dn.net/?f=%20b%5E0%3D1%20)
So, a=12.5
Next step is to plug in the second point (1, 13.75) and a=12.5 in the same equation to get the value of b. Hence,
![13.75=12.5(b)^1](https://tex.z-dn.net/?f=%2013.75%3D12.5%28b%29%5E1%20)
13.75=12.5b
Dividing each sides by 12.5.
So, b=1.1
Now we can plug in the value of a and b to get the exponential formula. Hence,
![y=12.5(1.1)^x](https://tex.z-dn.net/?f=%20y%3D12.5%281.1%29%5Ex%20)
So, the correct choice is a.