Ask yourself: does this function look like a straight line, a simple exponential of the form $P_0\cdot a^t$, or something else? Which option assumes a straight-line (constant) rate of change?

As $t$ becomes large, what happens to $e^{-0.6t}$? What does that do to the denominator $1+4e^{-0.6t}$ and to the overall value of $P(t)$?

Use $P(0)$ and $P(0.6)$ from the formula. If the population really doubled every 0.6 months, what should $P(0.6)$ be compared with $P(0)$, and does the function produce that result?

Enter the function as y = 820/(1+4e^(-0.6x)). Use $x$ in place of $t$ since Desmos uses $x$ for the horizontal axis by default.

Look at the point where the graph crosses the $y$-axis (when $x=0$). Read that $y$-value from the graph or from a table (add a table and set the first $x$-value to 0) to see the modeled starting number of fish.

In the table, include $x=0$ and $x=0.6$. Compare the $y$-values at these two $x$-values. If the population doubled every 0.6 months, the $y$-value at $0.6$ would be exactly twice the $y$-value at 0.

Look at the shape of the graph. A constant rate of change would produce a straight line. Notice whether this graph is straight or curves and levels off, and use that to judge any statements about a constant rate.

Zoom out along the positive $x$-axis (to large $x$-values like 20, 30, or more) and watch what $y$-value the graph gets closer to as $x$ increases. Check whether the graph ever rises above that $y$-value, and compare this behavior with the remaining answer choice.

To find the initial population, plug $t=0$ into the model:

So the initial population is 164 fish, not 820 fish. This rules out the statement that the initial population was 820 fish.

A constant rate of change (like "0.6 fish per month") would come from a linear function of the form $P(t) = mt + b$, whose graph is a straight line.

But this model is

which contains an exponential expression $e^{-0.6t}$ in the denominator. That means the rate of change is not constant; it changes over time. Therefore, the statement that the population increases at a constant rate of 0.6 fish per month is not supported by this model.

If a population doubles every fixed time interval (like every 0.6 months), the model usually looks like an exponential growth function, for example $P(t) = P_0 \cdot 2^{t/0.6}$, where each interval multiplies the population by 2.

Here, the function has the form

which is a different type of function (logistic) and has a numerator and a denominator. To see it does not double every 0.6 months, compare values:

• From step 1, $P(0) = 164$.
• Now compute $P(0.6)$:

If the population doubled every 0.6 months, $P(0.6)$ would have to equal $2\cdot 164 = 328$. A calculator shows $P(0.6)$ is not 328, so the population does not double every 0.6 months.

Focus on the exponential part $e^{-0.6t}$ in the denominator. As $t$ becomes very large, $-0.6t$ becomes a large negative number, so

Then the denominator $1 + 4e^{-0.6t}$ approaches $1 + 4\cdot 0 = 1$, so $P(t)$ approaches

Also, because $e^{-0.6t} > 0$ for all real $t$, the denominator $1 + 4e^{-0.6t}$ is always greater than 1, so $P(t)$ is always less than 820 for all $t \ge 0$.

This means the model predicts that the fish population levels off and gets closer and closer to 820 fish but never becomes larger than 820 fish. That matches the conclusion: The population will never exceed 820 fish.