Question

Number of years 1 2 3
Option 1 (amount in dollars) 1300 1690 2197
Option 2 (amount in dollars) 1300 1600 1900
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.

Answer 1

you cn see from the first 3 values for years 1,2,and 3, Option 1 increases in value faster than Option 2

Answer 2

Option 2 is linear , with a slope of 300 dollars per year

Option 1 is exponential

Answer 3

For Option 1, use the general form

y = a*b^x

Answer 4

use a point (x,y) = (1,1300)

1300 = a*b^1

Another point (x1,y1) = (2,1690)

1690 = a*b^2

2 equations, solvable for constants a and b

Answer 5

does that form y = a*b^x look familiar from class ?

Answer 6

no not really

Answer 7

Ok, lets start with the easier one, option 2.....do you see that each year the value increases by the same amount, 300 dollars?

Answer 8

yes

Answer 9

Since the value increases at the same amount each year, the graph will be a line, we can use 2 of the points from the table to find the slope of the line. In general, the line will be in the form:
y = m*x

where m is the slope.

Answer 10

using points (1,1300) and (2,1600) the slope m, is the difference in y over differene in x

m = (1600-1300)/(2-1) = 300/1 = 300

The line has a slope of 300 dollars per year.

Answer 11

oh ok

Answer 12

and starts out at year 1 with 1300 dollars, so the equation of the line is , using the point- slope formula for a line, point (1,1300) and slope m=300

y - 1300 = 300(x-1)

y - 1300 = 300x - 300

Answer 13

The equation for OPTION 2 is:

\[\large\color{blue}{ y = 300x + 1000}\]

Answer 14

where x is the year, and y is the money value

Answer 15

ohh okayy thanks so much

Answer 16

After 20 years, OPTION 2 will be worth y = 300(20) + 1000
\[\large\color{blue}{OPTION~2 ~at ~20~years = $7000 }\]

Answer 17

That was for the linear option 2

For the OPTION 1, we have to find the Exponential equation

Answer 18

okay

Answer 19

The general form for an exponential equation is

\[\large\color{blue}{ y = C*a^x}\]

we need to use points from the table to figure out what C and a will be..you follow?

Answer 20

yes

Answer 21

Here are 2 points from OPTION 1: (x,y)

(1,1300) and (2,1690)

Answer 22

Lets use the first point in the general form above (x,y) = (1,1300)

\[1300 = C*a^1\]

Now use the second point in the same general equation: (x,y) = (2,1690)

\[1690 = C*a^2\]

see that ?

Answer 23

We have 2 equations, with 2 unknown variables 'C' and 'a'

\[\large\color{blue}{1300 = C*a^1~~~and~~~1690 = C*a^2 }\]

Answer 24

Solving the first equation for C gives,

\[C = \frac{ 1300 }{ a }\]

Putting that C value into the second equation gives,
\[1690 = C*a^2 = \frac{ 1300 }{ a }a^2\]

Answer 25

Now you can solve for what 'a' is:

1690 = (1300/a)*a^2

1690 = 1300a

a = 13/10

Answer 26

Now that 'a' is known , we can find C...

1300 = C*(13/10)^1

1300*10 = 13C
C= 1000

Answer 27

Overall the exponential equation for the OPTION 1 is given by:

\[y = 1000*(\frac{ 13 }{ 10 })^x\]

where y is the value in dollars, and x is the time in years

Answer 28

As a test, let x=3 for the third year:

y = 1000*(13/10)^3 = 2197

That is true and the equation works, at year 3, the value is 2197 dollars

Answer 29

So at year 20, (x=20), the value of OPTION 1 is:

y = 1000*(13/10)^20 = 190050 dollars

Answer 30

at the 20 year Mark:

OPTION 1 = $190,050
OPTION 2 = $7000

Big difference.