### Video Transcript

A population that grows over time π‘ to a fixed positive limit may be modeled by a transformed exponential function π of π‘ is equal to π plus π times π raised to the power ππ‘ for suitable π, π, and π. The following is a graph of this. If the initial and limiting populations are π΄ and πΏ, respectively, which would be a suitable function? (A) πΏ plus πΏ plus π΄ multiplied by π raised to the power of five π‘. (B) πΏ minus πΏ minus π΄ multiplied by π raised to the power five π‘. (C) πΏ plus πΏ minus π΄ multiplied by π raised to the power negative five π‘. (D) πΏ minus πΏ minus π΄ multiplied by π raised to the power negative five π‘. Or (E) πΏ minus πΏ plus π΄ multiplied by π raised to the power negative five π‘.

Weβre given that a population grows over time from an initial population π΄, that it can be modeled by a transformed exponential function, and that it grows to a fixed positive limit πΏ. With this information and the graph shown, weβre asked to determine which of the given functions is a suitable model for this behavior.

We first recall that in the modified or transformed exponential function π of π‘ is equal to π plus π times π raised to the power ππ‘, since we know that our population levels out after some time, our constant π in the exponent must be negative. This is because with negative π, as π‘ tends to β, or for very large values of π‘, π raised to the power ππ‘ tends to zero. If, on the other hand, π was positive, our second term would tend to β as π‘ tends to β, which we know is not the case. This means that as π‘ tends to β, the function π of π‘ tends to the constant π plus zero. For our population, we know that the limiting value is πΏ so that the constant π must be equal to πΏ.

So far then, we have π of π‘ is equal to πΏ plus lowercase π multiplied by π raised to the power ππ‘. Weβre also given the initial value of π΄. This means that at π‘ is equal to zero, our population is πΏ plus lowercase π multiplied by π raised to the power π times zero, and thatβs equal to our initial value π΄. Now, π times zero is equal to zero. And since we know that π raised to the power zero is equal to one, we have that π at π‘ is equal to zero is equal to πΏ plus π multiplied by one, which is equal to our initial value uppercase π΄. π΄ multiplied by one is simply π΄. And now subtracting πΏ from both sides, we have the constant π is equal to π΄, our initial value, minus πΏ.

Now, making a note of the values weβve found at π is equal to πΏ and π is π΄ minus πΏ, we can write the modified function as π of π‘ is equal to πΏ plus π΄ minus πΏ multiplied by π raised to the power ππ‘. Now, this doesnβt quite match any of the given functions yet. But if we multiply our parentheses by negative one, we have πΏ minus πΏ minus π΄ π raised to the power ππ‘. So this could correspond to either option (B) or option (D). But remember, our constant π has to be less than zero. And only option (D) of the two remaining options has a negative constant in the exponent. Hence, if we make the choice that π is equal to negative five, then π of π‘ is πΏ minus πΏ minus π΄ multiplied by π raised to the power negative five π‘. And this corresponds to option (D).