Question Video: Determining Which Exponential Function Corresponds to a Particular Graph | Nagwa Question Video: Determining Which Exponential Function Corresponds to a Particular Graph | Nagwa

# Question Video: Determining Which Exponential Function Corresponds to a Particular Graph Mathematics • Second Year of Secondary School

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A population that grows over time π‘ to a fixed positive limit may be modeled by a transformed exponential function π(π‘) = π + ππ^(ππ‘) for suitable π, π, and π. The following is a graph of this. If the initial and limiting populations are π΄ and πΏ, respectively, which would be a suitable function?

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### 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 ππ‘ for suitable π, π, and π. The following is a graph of this. If the initial and limiting populations are π΄ and πΏ, respectively, which would be a suitable function? (A) πΏ plus πΏ plus π΄ multiplied by π raised to the power of five π‘. (B) πΏ minus πΏ minus π΄ multiplied by π raised to the power five π‘. (C) πΏ plus πΏ minus π΄ multiplied by π raised to the power negative five π‘. (D) πΏ minus πΏ minus π΄ multiplied by π raised to the power negative five π‘. Or (E) πΏ minus πΏ plus π΄ multiplied by π raised to the power negative five π‘.

Weβre 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 πΏ. With this information and the graph shown, weβre 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 ππ‘, 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 ππ‘ 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 πΏ so that the constant π must be equal to πΏ.

So far then, we have π of π‘ is equal to πΏ plus lowercase π multiplied by π raised to the power ππ‘. Weβre also given the initial value of π΄. This means that at π‘ is equal to zero, our population is πΏ plus lowercase π multiplied by π raised to the power π times zero, and thatβs 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 πΏ plus π multiplied by one, which is equal to our initial value uppercase π΄. π΄ multiplied by one is simply π΄. And now subtracting πΏ from both sides, we have the constant π is equal to π΄, our initial value, minus πΏ.

Now, making a note of the values weβve found at π is equal to πΏ and π is π΄ minus πΏ, we can write the modified function as π of π‘ is equal to πΏ plus π΄ minus πΏ multiplied by π raised to the power ππ‘. Now, this doesnβt quite match any of the given functions yet. But if we multiply our parentheses by negative one, we have πΏ minus πΏ minus π΄ π raised to the power ππ‘. 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 πΏ minus πΏ minus π΄ multiplied by π raised to the power negative five π‘. And this corresponds to option (D).

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