Question

4 question project.

Help?

Answer 1

Iris has been studying an invasive population of snails. This particular snail has no local predators, so the population grows wildly. She has observed that the population follows an exponential rate of growth for fifteen years.

1. Create your own exponential function, f(x), which models the snail population. You will need to identify the principal population of the snails and the rate of growth each year. Explain to Iris how your function shows the principal population and the rate of growth, in complete sentences.


2. A local snail population grows according to the function g(x) = 200(1.03)2x. Demonstrate the steps to convert g(x) into an equivalent function with only x as the exponent. Then, explain to Iris how the key features of this local snail population compares to the key features of the invasive population.


3. Iris wants to graph the invasive snail population to show the city council. Justify what the appropriate domain and range would be for the function f(x), what the y-intercept would be, and if the function is increasing or decreasing.


4. In five years, a garden festival plans on using the park where Iris has been studying the invasive snails. Explain to the garden festival committee how to find the average rate of change for the snail population between years 2 and 5. Describe what this average rate of change represents.

Answer 2

Ahhh so many wordsssss :O(

Answer 3

Number 1 is pretty straight forward: `create your own`, do you know how to do that?

Answer 4

I know >_>

Answer 5

This is what i put

(1) The exponential equation with the population starting at 45 and doubling each year would be f(x)=45(2)^x where x is the year

Answer 6

Ooo very nice!

Answer 7

ye

Answer 8

So where we stuck? Number 2?

Answer 9

yes

Answer 10

\[\Large\rm g(x) = 200(1.03)^{2x}\]Hmm I guess they want us to use some exponent rules...

Answer 11

One of our exponent rules allows us to do this, yes?
\[\Large\rm g(x) = 200\left[(1.03)^{x}\right]^2\]

Answer 12

I'm not sure if that's the one we want to use though, hmm...

Answer 13

Our other option would be to use the addition rule.. thing...\[\Large\rm g(x) = 200(1.03)^{x+x}=200(1.03)^x\cdot(1.03)^x\]But then again, that just leads us to the square.

Answer 14

So what would we say has changed about the key features of g(x)? +_+ Hmmm

Answer 15

Something like.... how the growth rate is being squared?
So the population will increase much faster.

Answer 16

It doesn't change the initial population or anything like that.

Answer 17

Do you understand the exponent rule I used? :U

Answer 18

Sorry, I had to eat!

@zepdrix

Answer 19

I dont really understand any of this v_v

Answer 20

D:

Answer 21

I know, I'm sorry ;(

Answer 22

Exponent rule tells us:\[\LARGE\rm (a^{\color{orangered}{b}})^{\color{royalblue}{c}}=a^{\color{orangered}{b}\color{royalblue}{c}}\]We just multiply the exponents.
We want to use this rule in reverse,\[\LARGE\rm g(x) = 200(1.03)^{\color{orangered}{2}\color{royalblue}{x}}\]\[\LARGE\rm g(x) = 200\left[(1.03)^{\color{royalblue}{x}}\right]^{\color{orangered}{2}}\]

Answer 23

Ok

Answer 24

So with our function:\[\Large\rm g(x) = 200\left[(1.03)^{x}\right]^2\]This is the population:\[\Large\rm g(x) = \color{orangered}{200}\left[(1.03)^{x}\right]^2\]And this is what's making the population grow:\[\Large\rm g(x) = 200\color{orangered}{\left[(1.03)^{x}\right]^2}\]Soooo, for problem 2, maybe say something likeeeeee...
The growth rate is being squared, which causes it to grow much faster.
I dunno something like that :c

Answer 25

For problem 3:
The y-intercept is the `initial population`.
Notice that at the very start Iris said that the growth rate of these snails follows an exponential curve for a period of `15 years`. That tells us our domain, our x values that we should use. A good domain would be something like \(\Large\rm 0\le x\le 15\), yes?

Answer 26

The snails would `range` from the initial population up to whatever value they reach at 15 years.

Answer 27

Yes, Thank you so much!

Answer 28

And so number 4 I would just replace x with 5 in my equation?

Answer 29

Remember how to find `average rate of change` or `slope` ?\[\Large\rm m=\frac{y_2-y_1}{x_2-x_1}\]Our y_2 is the 5 year mark,
while our y_1 is the 2 year mark.

I'll rewrite the slope formula using function notation, because that's what's going to help us here:\[\Large\rm m=\frac{g(x_2)-g(x_1)}{x_2-x_1}\]Plugging in our 5 and 2 for x,\[\Large\rm m=\frac{g(5)-g(2)}{5-2}\]

Answer 30

So before you can calculate your slope/rate of change, you need to figure out the population at 5 years, at 2 years, and plug those in.

Answer 31

So m=g(3)/3

Answer 32

Our y_2 is the `population at the` 5 year mark,
blah i said that a little sloppy the first time.

Answer 33

No no no no no.

Answer 34

Or am I doing this wrong

Answer 35

\[\Large\rm g(5)-g(2)\ne g(3)\]I'm not sure what gave you the idea that you could do that :c

Answer 36

Ahh Sorry

Answer 37

Wait, thats what I put >_> I just put it over 3

Answer 38

Yes, do you see how I put a `not equal to` symbol between them? :o
That is not correct to do.

The 3 in the bottom is fine though.
\[\LARGE\rm g(\color{royalblue}{x}) = 200(1.03)^{2\cdot\color{royalblue}{x}}\]\[\LARGE\rm g(\color{royalblue}{5}) = 200(1.03)^{2\cdot\color{royalblue}{5}}\]

Answer 39

Here is how you would get your g(5) value.
Calculator should help out.

Answer 40

So what do you get for g(5)? +_+

Answer 41

2060

Answer 42

Right?

Answer 43

No not even close :c
Hmm let's make sure we're putting it into the calculator correctly.

Answer 44

\[\Large\rm 200(1.03)\text{^}(10)\]Put it in like that.

Answer 45

Yes, I did

Answer 46

OH WAIT

Answer 47

I had a blonde moment.. even though I'm not a blonde.. Wow, ok..

Exponent, duhh

Answer 48

lol :p

Answer 49

268.78

Answer 50

Mmm ya there we go! c:\[\Large\rm m=\frac{268.78-g(2)}{5-2}\]

Answer 51

How bout g(2)? Think you can figure that one out? c:

Answer 52

268.78-\[268.78 - 1.25\div5-2\]1.125

Answer 53

1.125 >_<

Answer 54

I cant do the equation thing.. lol

Answer 55

Your g(2) looks .... kinda crazy.

If g(0) represents the initial population.... 200 snails
and g(5) represents the snail population after 5 years .... 268 snails

do you think a population of 1 is about right after 2 years?

Answer 56

After 2 years, we would expect the population to be a little more than 200, since it's increasing.

So I dunno how you got that number >.<

Answer 57

\[\Large\rm g(2)=200(1.03)\text{^}(4)\]

Answer 58

I forgot to multiply the 200...

225.101

Answer 59

Oh ok.\[\Large\rm m=\frac{268.78-225.10}{5-2}\]Great, so wrap it up with one more calculation!

Answer 60

m=14.56

Answer 61

Yay good job!
That's telling us that over a 3 year period, the snail population was increasing at a rate of 14 and a half snails per year.