If the population triples, are you multiplying by a number bigger than 1 or smaller than 1? What number represents "tripling" as a multiplication factor?

In $h$ hours, how many 6-hour intervals have passed? That count should be the exponent in your exponential expression.

For each option, try plugging in $h=0$ and $h=6$. Which equation gives 500 at $h=0$ and 3 times as many cells (1500) at $h=6$?

In Desmos, let the horizontal axis be $h$ (you can use h instead of x). Type each option as a separate function, for example P1(h)=500*(1/3)^(h/6), P2(h)=500*(1/6)^(h/3), P3(h)=500*3^(h/6), and P4(h)=500*6^(h/3).

For each function, evaluate it at $h=0$ (for example, type P1(0), P2(0), etc.). The correct model must give an initial value of 500 cells at $h=0$.

Now evaluate each function at $h=6$ (type P1(6), P2(6), etc.). The correct equation is the one whose value at $h=6$ is exactly three times its value at $h=0$, because the population triples every 6 hours.

The problem says there are 500 cells initially and the culture triples every 6 hours.

• "Initially" (when $h=0$) means $P(0)=500$.
• "Tripled" means we multiply by 3 each time the fixed time period (6 hours) passes. So the model should start with 500 and then repeatedly multiply by 3 as time goes on.

Exponential growth with a fixed period uses the idea:

• Base = growth factor per period
• Exponent = number of periods that have passed

Here, one period is 6 hours, and the growth factor per period is 3.

• In $h$ hours, the number of 6-hour periods is $\dfrac{h}{6}$. So the exponent in our exponential expression should be $\dfrac{h}{6}$, because it counts how many times we multiply by the growth factor.

Now combine everything:

• Initial amount: 500 (a factor out front)
• Growth factor per 6-hour period: 3 (the base of the exponent)
• Number of 6-hour periods in $h$ hours: $\dfrac{h}{6}$ (the exponent)

So the function for the number of cells after $h$ hours is

This matches choice C.