Question

Which of the following exponential regression equations best fits the data shown below

Answer 1

(D)
`8.46\cdot 3.51^x`

Explanation:

One could run this on a computer and verify the best fit through brute force. The more elegant way is, as usual, to think: What are special values of x for an exponential function? Zero, for starters - anything to the power of zero is 1. The function value for x=0 in the table is 10. Which choices A through D are close to 10 for x=0? Well, (C) and (D), the rest is too far. Next, what is the function value for the next easy one, x=1? The table says 30. Which of (C) and (D) is close to 30 for x = 1. It turns out we can safely exclude (C) because 10.84*1.77 is about 19 and that's way too far from 30. Let's check (D): 8.46*3.51=29.7 - that's quite close. Since there is no other candidate left, I bet my money on (D). Feel free to verify closeness for the other values of x if you are unconvinced yet.

Answer 2

`\text{D. }y = 8.46(3.51)^(x)`

Explanation:

The quickest way is probably to do some direct substitution.

First, let's try f(0).

The values of y predicted by the four equations are, respectively, 3.14, 5.32. 10.84, and 8.46.

It looks like C and D are the closest to the observed data.

Now, let's try f(4) on C and D.

C: 10.84(1.77)⁴ = 10.84 × 9.815 = 106.4

D: 8.46(3.51)⁴ = 8.46 × 151.4 = 1284

The best curve is
`y = 8.46(3.51)^(x).`