Question

A rare bacterial culture is being grown in a lab. As time passes, the cells multiply in a specific pattern.

After 1 day, there is only 1 cell.
After 2 days, there are 9 cells.
After 3 days, there are 20 cells.
After 4 days, there are 34 cells.
How many cells will there be after seven days?
A.149
B.120
C. 98
D.94

Answer 1

Let the growth function be the polynomial
f(x) = ax³ + bx² + cx + d

From the given information,
f(1)=1, therefore
a + b + c + d = 1 (1)
f(2) = 9, therefore
8a + 4b + 2c + d = 9 (2)
f(3) = 20, therefore
27a + 9b + 3c + d = 20 (3)
f(4) = 34, therefore
64a + 16b + 4c + d = 34 (4)

Assemble the four equations into the matrix equation (also shown in the figure)
| 1 1 1 1 | 1 |
| 8 4 2 1 | 9 |
|27 9 3 1 | 20 |
|64 16 4 1 | 34 |

Solve with the graphing calculator to obtain
a = 0
b = 1.5
c = 3.5
d = -4

Therefore
f(x) = 1.5x² + 3.5x - 4

After 7 days,
f(7) = 1.5*49 + 3.5*7 - 4 = 94

Answer: D. 94