If the population doubles every 6 hours, then after 6 hours it should be $1000$, after 12 hours it should be $2000$, and so on. Your model should use a factor of $2$ for each 6-hour period.

How many 6-hour periods are in $t$ hours? That number should be the exponent on the base of the exponential function. Look for an exponent involving $t/6$, not just $6$ or $t$ alone.

In Desmos, enter each option as a separate function using $x$ for time (hours):

• $y_1 = 500(2x)^6$
• $y_2 = 500(1/2)^{x/6}$
• $y_3 = 2\cdot 500^{x/6}$
• $y_4 = 500\cdot 2^{x/6}$

For each function, either tap to create a table or just substitute $x=0$ and see the $y$-value. The correct model must have $y=500$ when $x=0$.

For the functions that have $y(0)=500$, evaluate $y$ at $x=6$ (and optionally $x=12$). The correct model will have the output double every 6 hours (e.g., $500$ at $x=0$, $1000$ at $x=6$, $2000$ at $x=12$). The function that shows this pattern is the right choice.

At noon (when $t=0$), the culture has 500 bacteria, so the function must satisfy $P(0)=500$.

The population doubles every 6 hours, which means:

• After 6 hours, the amount is multiplied by $2$.
• After 12 hours, it is multiplied by $2$ again (so total factor $4$), and so on.

If $t$ is the number of hours after noon, then the number of 6-hour periods that have passed is $t/6$.

So after $t$ hours, the population has been doubled $t/6$ times.

For exponential growth with doubling:

• Start with the initial amount (here, $500$).
• Multiply by the doubling factor $2$ raised to the number of doubling periods.

So the model must have the form:

• "initial value $500$" times
• "$2$ raised to the power $t/6$".

Look for the answer choice that matches this structure exactly.

Putting it all together, the function is

This matches choice D.