Contributor: Michelle Haver. Lesson ID: 13646
Certain situations can be modeled by linear functions while others can be modeled by exponential functions. Discover how to determine if it's linear or exponential.
You decide you'd like to start saving up some money but aren't entirely sure of the best way to do so. Your parents, acting as the bank, provide you with two options.
Option 1: Your parents will put aside $700 immediately and add $100 once a year.
Option 2: Your parents will put aside $100 immediately and double the amount set aside each year.
Look at both options to decide which is best!
Option 1 may seem great since you get $700 immediately! Let's see how much money you would save over five years.
In this option, you start with $700 and add $100 annually. You can lay this out with a table.
Contribution | Total Savings | |
initial contribution | $700 | |
after 1 year | $700 + $100 = $800 | |
after 2 years | $800 + $100 = $900 | |
after 3 years | $900 + $100 = $1000 | |
after 4 years | $1000 + $100 = $1100 | |
after 5 years | $1100 + $100 = $1200 |
You can see that, after five years, you will have saved $1,200 with your parents' help!
Suppose you want to know how much money you would have saved after 10 years. You could continue to use the table, but that would take a chunk of time to do!
In this case, you have a starting value ($700) and will increase the amount saved by an additional $100 yearly. This constant addition ($100) to the independent variable ($700) leads to a linear function.
The slope is the yearly increase, and the y-intercept is the starting value. So you can write the function.
s(x) = 100x + 700.
This function gives the amount of money saved x years after the initial investment with option 1.
Now, consider option 2 and see how this compares.
This one may not seem as good right off the bat because you only start with $100, but this time you will double the amount saved each year.
Look at this using a table.
Contribution | Total Savings | |
initial contribution | $100 | |
after 1 year | $100 x 2 = $200 | |
after 2 years | $200 x 2 = $400 | |
after 3 years | $400 x 2 = $800 | |
after 4 years | $800 x 2 = $1600 | |
after 5 years | $1600 x 2 = $3200 |
You can see that, after five years, you will have saved $3,200 with your parents' help! This is more than the previous option.
However, initially, there are several times where option 1 will yield more money. Again, suppose you want to know how much money you would have saved after 10 years.
In this case, you have a starting value ($100) and will multiply by 2 each year. However, the amount multiplied by 2 will not be the same each year but will instead depend on how many years have passed.
Since you are doubling the amount each year, that is represented by 2 to the power of how many years the money has been saved. This makes the scenario an exponential function because you multiply the independent variable by an exponent.
You can write a function that gives the amount of money saved x years after the initial investment with option 2.
m(x) = 100(2^{x})
This is loads more money than you would have saved with option 1! It's over 60 times more!
Explore the main differences a bit more.
In Option 1, you add the same number for each increase in x. This leads you to use a linear function!
These functions do not grow as quickly as the exponential functions used in Option 2. In Option 2, you are multiplying the same number for each increase in x.
To see some more examples showcasing the differences between situations modeled by linear functions and situations modeled by exponential functions, check out the following video.
You may have noticed some differences in phrasing, which can help you determine if a model is linear or exponential.
For exponential models, you want to look for terms such as "increases by a factor of" or "decreases by 5%".
For linear models, you want to look for situations with constant increase or decrease.
Now that wyouve seen the differences between linear and exponential models go ahead to the Got It? section to check your understanding!