Question

I am having such a hard time w/ this question, anyone? Helper will get a medal! Algebra:

Option 1 (amount in $) year 1 = 1300 year = 2 1600 year 3 = 1900
Option 2 ($) year 1 = 1120 year 2 = 1254.40 year 3 = 1404.93

Part A: What type of function, linear or exponential, can be used to describe the value of the investment after a fixed number of years using option 1 and option 2? Explain your answer.

Answer 1

Please ANYONE HELP ME!!!!! PLEASE!!!!!!

Answer 2

@pgpilot326

Answer 3

@AllTehMaffs please help me, I'm about to fail this test!

Answer 4

Well it keeps going up by 300 in the 1st one. So that one has to be linear.

Answer 5

Help, please? I'm really trying but I don't understand.

Answer 6

Okay I can try. So it would be option one is linear, option 2 is exponential. Because linear goes up by an even number. 300 is even if you can get to it by an even number.

Answer 7

Linear more specifically means that for the function that descirbes option 1, it can be expressed as
\[f(t) = 1000 + 300t\]
where t is in years

Answer 8

Ok, and Option 2 is um, .....I don't know. ;__;

Answer 9

Help?

Answer 10

That one is exponential, trying to come up with its equation.

Answer 11

So, is the question asking me which one is linear and which one is exponential?

Answer 12

yeah. I thought that you had to come up with their functions but apparently that's not true.

Answer 13

Oh. So, Option 1. is linear because the it is going evenly through out the 3 years, and Option 2. is exponential because it isn't going by a linear pattern.

Answer 14

Would you mind helping me w/ Part B.? I won't bother you again!

Answer 15

Yeah, I believe that suffices. And I'll try ^_^

Answer 16

Awesome, thank you! Okay, here it is:

Write one function for each option to describe the value of the investment f(n), in dollars, after n years.

Option 1. I already know is f(x) = 1,000x +300. Could you help me find the function for Option 2?

Answer 17

@AllTehMaffs

Answer 18

I'm trying, but my brain has apparently shut off.

First off, your first function is incorrect.

It's
\[f(t)=1,000 + 300t\]
or, as you wrote it
\[f(x) = 1,000 + 300x\]
where year 1 is assigned to t=1

Answer 19

it's increasing by 300 each time increment, so it's the 300 term that has the time variable attached to it.

Answer 20

Oh, whoops. Here f(t) = 1,000 + 300t

Answer 21

I understand, just made an error.

Answer 22

Now, what next oh great one? :)

Answer 23

Are you solving it? :3

Answer 24

Trying, my brain is not cooperating with me today.

Answer 25

Me too. Hmm...maybe I could help a little.

Answer 26

@mebmod any suggestions?

Answer 27

trying

Answer 28

yay, okay, so it fits the form

\[N = N_0(k^t)\]

Answer 29

Awesome!!.....uh....hm....I think I get it, but would you mind explaining it to me, just to make sure?

Answer 30

where k is some growth constant, and t is the time. For t=0 (year 1), we have our original population, but for t=1 (year 2)
\[1254.4 = 1120(k^1)\]
or just
\[1254.4 = 1120\times k\]
so what's k?

Answer 31

.......I don't know. So sorry!

Answer 32

WAIT

Answer 33

HOLD ON A SECOND, I THINK I AM UNDERSTANDING.

Answer 34

1.12?

Answer 35

So if you were given
\[1254.4=1120x\]
how would you find x?

Answer 36

I would divide both sides by 1,120.

Answer 37

And then what? :)

Answer 38

Then you just found k! ^_^

it's the same process

1254.4 = 1120k
\[k=\frac{1254.4}{1120} = ?\]

Answer 39

k = 1.12! :D

Answer 40

So, a review:
Option 1.)'s function would be f(t) = 1000 +300t. But here's the thing, it says it needs to be f(n) and then "n". What should I do?

Answer 41

yup yup ^_^

So the exponential function for any time, t is

\[f(t) = 1120k^t\]
you can check that it gives the correct output for t=2 (year3)

So first, I'm gonna make a correction, and say that at year 1, t=0, so the function turns into

\[f(n) = 1300 + 300n\]

All you're doing is adding 300 every year>1 so them saying f(n) is just a function where the thing that's going up is called n (maybe for numberof years).

there wasn't some defined value for n, was there?

Answer 42

n = years, and f(n) was dollars.

Answer 43

Then yeah

\[f(n) = $1200 + ($300/year)n\]
where n is years.

Answer 44

oops, sorry, should be 1300 for the first term

Answer 45

Ok, so Option 1. is f(n) = 1300 +300n. Option 2.) is 1,254.40 = 1,120(n)

Answer 46

hello?

Answer 47

Wait, when I do the math for the first options function, it gives me 1,900 for the second year. :(

Answer 48

For option 1, remember that n=0 means year 1, so for years 2, n=1

Also, not quite right for option 2

Remember that it's exponential, so the n has to be in the exponent! ^_^

It's also still a function of n, so 1254.4 is just an output of the function.

\[f(n) = 1120\times(1.12)^n\]
where, again,
n=0 corresponds to year 1,
n=1 is year 2
n=2 is year 3
etc.

Answer 49

I understand now! Awesome!

Answer 50

1 more question, what would be the value of the investment f(n) for each options function? Dumb, I know.

Answer 51

I don't understand the question. I'm bad with financial terms :/

Answer 52

value is like the gains on the original investment, right?

Answer 53

I don't know my friend, I don't know.

Answer 54

Then your guess is as good as mine ^_^ Sorry :P

Answer 55

It's cool, thank you so much for all the help!

Answer 56

Very welcome. Glad that I could help! ^_^