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## Algebra 2 (Eureka Math/EngageNY)

### Unit 3: Lesson 11

Topic D: Lessons 25-28: Modeling with exponential functions- Interpreting change in exponential models
- Interpret change in exponential models
- Interpreting time in exponential models
- Interpret time in exponential models
- Constructing exponential models
- Constructing exponential models: half life
- Constructing exponential models: percent change
- Construct exponential models
- Interpreting change in exponential models: with manipulation
- Interpret change in exponential models: with manipulation
- Interpreting change in exponential models: changing units
- Interpret change in exponential models: changing units

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# Constructing exponential models

CCSS.Math: , , ,

Sal models the distribution of a chain letter using an exponential function.

## Want to join the conversation?

- cant we just write a function for the rate of change per week instead of every three weeks? I find it easier to write a function in case t is not a multiple of 3, i.e. f(t)=40*1.3^t. Since it grows 90% every three weeks, it would grow 30% every week, i.e. 1.3.(13 votes)
- How do I find what "A" is if I try to convert 40*1.9^(t/3) to 40*A^t.

My goal is to find "A" and know the weekly rate of email chain growth.(4 votes)- You have 40 * 1.9^(t/3) and want it in the form 40 * A^t.

Note that 1.9^(t/3) = [ 1.9^(1/3) ]^t

and then use the calculator to find the value of 1.9^(1/3) which equals A.

Then interpret the value of A to give you the weekly rate of e-mail chain growth.

OK?(5 votes)

- Shouldn't the answer be 40.((1.9)^[t/3]) rather than 40.((1.9)^(t/3))? (greatest integer function)(6 votes)
- With (t/3) you can get an answer for any week, while [t/3] rounds up to answers for multiples of 3 weeks.(2 votes)

- What would you do with decay functions? Would you still add the +1 like he did for the 9/10?(3 votes)
- The 1 represents that you start with 100%. For a decay problem, the 100% decreases, so we subtract.

So, you would use: 1 - decay percentage.

Hope this helps.(4 votes)

- Why is it P(t) and not simply just P in the exponential function and table?(1 vote)
- The problems asks you to "write a function". Functions are written using function notation, which is why Sal uses P(t) rather than just P.(5 votes)

- How come at2:26he is multiplying by 1.9. Aren't you supposed to multiply it by 0.9 or 90% ?(2 votes)
- 10 * 1.9 = 19 - 90% increase

10 * 0.9 = 9 - 10% reduction(2 votes)

- At1:36, why is it that 40+9/10*40 equal to 40(1+9/10)?(1 vote)
- He used the distributive property to factor out a common factor of 40. If you are unfamiliar with factoring using the distributive property, see this link: https://www.khanacademy.org/math/algebra/polynomial-factorization#factoring-polynomials-1-common-factors(2 votes)

- The problem says that the recipients of letters increased every 3 weeks, but it does not day it increased by any other periods. If we use the answer given in the video, the recipient number increases in 1 week, 2 weeks so forth. Are we supposed to assume the number increases between 3 weeks' intervals?(1 vote)
- I don't understand how he goes from 40 * (9/10) + 40 to 40(1 + (9/10)) ? Like, I thought it was factoring but how do you factor a 40 out of 9/10 ?

Here's what I tried:

(40)(40 + (9/10)) // factor out a 40 from each set of parens

(40)(1)(1 + ?) // if 9/10 was divisible by 40, here's where you'd put (9/10) / 40, but it's not so I dunno what I'm missing

40(1 + (9/10)) // this is what Sal ends up with

Btw, I did the math and confirmed that 40 * (9/10) + 40 does equal 40(1 + (9/10)), so my only conclusion is that I'm missing something about how to factor stuff?(1 vote)- You changed "40 times 9/10" into "40 plus 9/10". This is why your version isn't working.

40*9/10, you can factor out the 40 and end up with 9/10.

Hope this helps.(1 vote)

- so is P(t)=40 x 1.9^t/3 the answer? I'm sooo confused TvT(0 votes)

## Video transcript

- [Voiceover] Derek sent a
chain letter to his friends, asking them to forward the
letter to more friends. The group of people who
receive the email gains 9/10 of its size every three weeks, and can be modeled by a function, P, which depends on the
amount of time, t in weeks. Derek initially sent the
chain letter to 40 friends. Write a function that
models the group of people who receive the email t weeks since Derek initially
sent the chain letter. So pause the video if you
wanna have a go at this. All right, now the way I'd
like to think about this, let's just gonna table with values for t and our function P which is a function of t. For some values that we can just pull out of the description here. So when t is zero, when
it's been zero weeks since Derek initially
sent the chain letter, how many people have gotten it? Well, they tell us. He initially, Derek initially sent the chain letter to 40 friends. So t equals zero. P of t or P of zero is 40. Now, what's an interesting time period? It says that the email,
the number of people who receive the email gains 9/10 or increases by 9/10 every three weeks, every three weeks. So after three weeks, so three weeks have gone by. So I'm just adding three to t. What is P of t going to be? Well, they tell us it's going
to gain 9/10 of its size. So it's going to be 40 plus 9/10 times 40 which is going to be equal to, what? Well, that's equal to 40. We factor out our 40. 40 times one plus 9/10 or you could say this
is equal to 40 times, whoops, 40 times 1.9. Another way of thinking about it, after three weeks, we've grown 90%. That's another way of saying
that the number of people who receive the email
gains 9/10 of its size. You could say the group of
people who receive the email grows 90% every three weeks. And so if we go another three weeks, so plus another three weeks, I could say, well, let me just write this, this is six weeks. Well, how many people would
have received the email? It was going to be this number. It's going to be grown another 90%. So we're gonna multiply
it times 1.9 again. So it's going to be 40 times 1.9 times 1.9. We're gonna grow by another 9/10. Growing by 9/10 is the
same thing as multiplying by one and 9/10. The one is what you already are. Then you're growing by another 9/10. So this is the same thing as 40 times 1.9 squared. We go another three weeks, nine weeks. We're gonna grow another 90%. So you're gonna multiply it by, you're gonna take this number and multiply by 1.9 again which is going to be
1.9 to the third power. And so what's going on over here? Well, we can see some
exponential function. We have our initial value. And every three weeks,
we're multiplying by 1.9. So 1.9 would be our common ratio. So we could say that P of t is
equal to our initial value 40 times our common ratio 1.9. And we multiply by 1.9 every three weeks. So we could just say, how many three week
periods have passed by? Well, we will take t
and divide it by three. T divided by three is a
number of three week periods that have gone by. And there you have it. And notice, t equals zero, 1.9 to the zeroth power is one so 40 times one. T equals three, that's gonna
be 1.9 to the first power, three over three. And so we're gonna grow by
90% and so on and so forth. So feeling pretty good about this.