Question

medal and fan!

Answer 1

Belinda wants to invest $1000. The table below shows the value of her investment under two different options for three different years:


Number of years 1 2 3
Option 1 (amount in dollars) 1100 1200 1300
Option 2 (amount in dollars) 1100 1210 1331


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. (2 points)

Part B: Write one function for each option to describe the value of the investment f(n), in dollars, after n years. (4 points)

Part C: Belinda wants to invest in an option that would help to increase her investment value by the greatest amount in 20 years. Will there be any significant difference in the value of Belinda's investment after 20 years if she uses option 2 over option 1? Explain your answer, and show the investment value after 20 years for each option. (4 points)

Answer 2

option A can be described by a linear function, namely
\(y=ax+b\)

Answer 3

oops.. I meant option 1

Answer 4

okay and option2

Answer 5

would be exponential?

Answer 6

yes! by exclusion, option 2 can be described by an exponential function

Answer 7

is that all you have to do in Part a

Answer 8

I think it is better to write the corresponding equations for both function, lnear and exponential

Answer 9

functions*

Answer 10

@Michele_Laino i have to go to the bath room can i mg you wen i com back. i should be around 2 min.

Answer 11

ok!

Answer 12

thank you! brb

Answer 13

@Michele_Laino

Answer 14

for linear function I use the data you provided, so I replace \(x=1\) and \(y=1100\), so I get:
\(1100=a \cdot 1+b=a+b\)
please solve for \(b\)

Answer 15

please suppose that \(a\) is known

Answer 16

what does that mean?

Answer 17

for example if I subtract \(a\) from both sides, I get:
\[1100 - a = a + b - a\]
please simplify

Answer 18

a = 1100-b?

Answer 19

hint:
we have \(a+b-a=b\)
so we get \(b=1100-a\)

Answer 20

next, I replace \(x=2\) and \(y=1200\) into the formula \(y=ax+b\), so I get:
\(1200=2a+b\)

Answer 21

then I subsitute \(b=1100-a\), so I write this:
\(1200=2a+1100-a\)
please solve for \(a\)

Answer 22

a =100

Answer 23

that's right!
so I can write this:
\[b = 1100 - a = 1100 - 100 = ...?\]

Answer 24

im strugling

Answer 25

please, it is simple, what is \(1100-100...?\)

Answer 26

1000

Answer 27

perfect! SO the linear function can be written as below:
\(f(N) = 100N+1000\)

Answer 28

oops... So*

Answer 29

wow! so this is all for just option1?

Answer 30

partA?

Answer 31

no, since we have to write an equation for the exponential function

Answer 32

no i mean like what we just did is for option 1

Answer 33

?

Answer 34

answer to part A is:
option 1 can be descibed by a \(linear\) function, since its values increase linearly
and option 2, by exclusion, can be described by an exponential function

now, we are writing the equations for both functions, so we are answering to part B

Answer 35

oh okay! im sorry i dont mean to frustrate you.

Answer 36

no problem :)

Answer 37

now we have to write the general equation for an exponential function

Answer 38

how do we do that?

Answer 39

I'm thinking...

Answer 40

the general formula is:
\(\huge y=a \cdot b^x\)

Answer 41

now I replace \(x=1\) and \(y=1100\), so I get:
\[\Large 1100 = a \cdot {b^1} = ab\]

Answer 42

so, we get \(ab=1100\)
next I replace \(x=2\) and \(y=1210\), so I can write this:
\[\Large 1210 = a \cdot {b^2} = ab \cdot b = 1100 \cdot b\]
please find \(b\)

Answer 43

okay i will try

Answer 44

hint:
\[b = \frac{{1210}}{{1100}} = ...?\]

Answer 45

b=11/10

Answer 46

yes! It is \(b=1.1\)
now, please find \(a\):
\[a = \frac{{1100}}{b} = \frac{{1100}}{{1.1}} = ...?\]

Answer 47

a=1000

Answer 48

correct!
So, the exponential function, is:
\[\huge g\left( N \right) = 1000 \cdot {1.1^N}\]

Answer 49

so for part B it is

Option1: G(n)=1000x1.1^n

Answer 50

answer to part B, is:
option 1 linear function \(f(N)=100N+1000\)
option 2, exponential function \(gN)=1000 \cdot 1.1^N\)

Answer 51

oh okay!

Answer 52

oops.. \(g(N)=1000 \cdot 1.1^N\)

Answer 53

for part C, we have to compute these quantities:
\[\begin{gathered}
f\left( {20} \right) = 100 \cdot 20 + 1000 = ... \hfill \\
\hfill \\
g\left( {20} \right) = 1000 \cdot {1.1^{20}} = ... \hfill \\
\end{gathered} \]

Answer 54

\[\huge \begin{gathered}
f\left( {20} \right) = 100 \cdot 20 + 1000 = ... \hfill \\
\hfill \\
g\left( {20} \right) = 1000 \cdot {1.1^{20}} = ... \hfill \\
\end{gathered} \]

Answer 55

f(20)=3000?

Answer 56

yes!

Answer 57

G(20) = 6727.49994932

Answer 58

correct! We can round that value to \(6,728\)

Answer 59

now, please compare such values

Answer 60

what do you mean?

Answer 61

you have to establish which value is the greatest value between them

Answer 62

the second one?

Answer 63

yes! That's right!

Answer 64

okay so how would i write that in part c?

Answer 65

answer to part C, can be this:
investment after 20 years for option 1 is \(\$3000\)
investment after 20 years for option 2 is \(\$6728\)
so, we get more profit using option 2

Answer 66

okay thank you so much @Michele_Laino unlike others you take the time to help people and go through whatever problem with them and i admire that about you. i wish i could give you a million medals right now!

Answer 67

thanks! :D