Question Video: Determining Which Exponential Function Corresponds to a Particular Graph Mathematics • 9th Grade

Video Transcript

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

We’re given that a population grows over time from an initial population 𝐴, that it can be modeled by a transformed exponential function, and that it grows to a fixed positive limit 𝐿. With this information and the graph shown, we’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 𝑃 of 𝑡 is equal to 𝑐 plus 𝑎 times 𝑒 raised to the power 𝑏𝑡, since we know that our population levels out after some time, our constant 𝑏 in the exponent must be negative. This is because with negative 𝑏, as 𝑡 tends to ∞, or for very large values of 𝑡, 𝑒 raised to the power 𝑏𝑡 tends to zero. If, on the other hand, 𝑏 was positive, our second term would tend to ∞ as 𝑡 tends to ∞, which we know is not the case. This means that as 𝑡 tends to ∞, the function 𝑃 of 𝑡 tends to the constant 𝑐 plus zero. For our population, we know that the limiting value is 𝐿 so that the constant 𝑐 must be equal to 𝐿.

So far then, we have 𝑃 of 𝑡 is equal to 𝐿 plus lowercase 𝑎 multiplied by 𝑒 raised to the power 𝑏𝑡. We’re also given the initial value of 𝐴. This means that at 𝑡 is equal to zero, our population is 𝐿 plus lowercase 𝑎 multiplied by 𝑒 raised to the power 𝑏 times zero, and that’s equal to our initial value 𝐴. Now, 𝑏 times zero is equal to zero. And since we know that 𝑒 raised to the power zero is equal to one, we have that 𝑃 at 𝑡 is equal to zero is equal to 𝐿 plus 𝑎 multiplied by one, which is equal to our initial value uppercase 𝐴. 𝐴 multiplied by one is simply 𝐴. And now subtracting 𝐿 from both sides, we have the constant 𝑎 is equal to 𝐴, our initial value, minus 𝐿.

Now, making a note of the values we’ve found at 𝑐 is equal to 𝐿 and 𝑎 is 𝐴 minus 𝐿, we can write the modified function as 𝑃 of 𝑡 is equal to 𝐿 plus 𝐴 minus 𝐿 multiplied by 𝑒 raised to the power 𝑏𝑡. Now, this doesn’t quite match any of the given functions yet. But if we multiply our parentheses by negative one, we have 𝐿 minus 𝐿 minus 𝐴 𝑒 raised to the power 𝑏𝑡. So this could correspond to either option (B) or option (D). But remember, our constant 𝑏 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 𝑏 is equal to negative five, then 𝑃 of 𝑡 is 𝐿 minus 𝐿 minus 𝐴 multiplied by 𝑒 raised to the power negative five 𝑡. And this corresponds to option (D).