Solving for Car Depreciation: Linear, Quadratic, and Exponential Models

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SUMMARY

The discussion focuses on modeling car depreciation using linear, quadratic, and exponential equations based on provided data points. The linear model is represented as y = mx + b, the quadratic model as y = ax^2 + bx + c, and the exponential model as y = ab^x. The participants emphasize that the exponential model best fits the data, particularly for predicting the car's value after 10 years. They also mention the importance of least squares curve fitting for determining the most accurate models.

PREREQUISITES
  • Understanding of linear equations (y = mx + b)
  • Familiarity with quadratic equations (y = ax^2 + bx + c)
  • Knowledge of exponential functions (y = ab^x)
  • Basic principles of least squares curve fitting
NEXT STEPS
  • Learn how to implement least squares curve fitting using Excel
  • Explore the derivation of exponential decay models in real-world applications
  • Study the implications of different mathematical models in predicting trends
  • Investigate the use of statistical software for regression analysis
USEFUL FOR

Students in mathematical modeling courses, data analysts, and anyone interested in understanding depreciation modeling and curve fitting techniques.

Adam2987
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Homework Statement



A car depreciates over time.

x(years) y ($)
1 14000
2 9100
3 6200
4 4000
5 3000

i) determine the linear model (1 mark)
ii) determine the quadratic model (1 mark)
iii) determine the exponential model (1mark)

b) using all three models determine the amount the car is worth after 10 years
c) which model decribes this best?

Homework Equations





The Attempt at a Solution



If I only knew where to begin. This is a correspondance course, last question of the course, and the textbook discussed this for 3 pages and only gave us the exponential equation in an example. I have asked several teachers, university students and nobody can figure this out. I can handle b and c if I could only figure out a). Any help would be greatly appreciated.
 
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by the way, the ideas I have for a) are:

i) y=mx+b

ii) y=ax^2+bx+c

iii) y=ab^x

just not sure how to go about substituting in the variables to get y
 
As x increases, y decreases.
I can't figure out how y is decreasing, but once you have that, you can do this:

y=14000(b)^x

b = the decreasing factor
x=years

I believe it's like that.
 
iRaid said:
As x increases, y decreases.
I can't figure out how y is decreasing, but once you have that, you can do this:

y=14000(b)^x

b = the decreasing factor
x=years

I believe it's like that.

that would work if the data is linear, but it isn't. This is definitely an exponential curve when the data is graphed. The unit is mathematical modelling, so the linear and quadratic models are going to be off, the exponential equation should be dead on or closest in respect to the graphed data. I appreciate your time, I just don't think that is what they're after.
 
I need to keep in mind also that for part b), I need each equation to give me the answer when I plug in 10 for x. With non-linear data that will be difficult. I've got pages and pages of attempts, and I just can't grasp what it is that is so simple to make each model worth only one mark. I think I'm missing something simplistic. Please any help or impput would be appreciated, I've been trying to solve this question since Monday.
 
Have you covered least squares curve fitting? Here's a link to a wiki article - http://en.wikipedia.org/wiki/Least_squares

That article might be heavy going, but there are links to related topics that might be helpful.

The basic idea in least squares curve fitting is to find the line/parabola/exponential function that best fits the given data. For the line of best fit, you want the line y = ax + b so that the sum of the squares of the vertical distances between the line and the data is smallest.

For the parabola of best fit, you want the parabola y = ax^2 + bx + c, and for the exponential function, I think you want the function y = ae^(bx). Spreadsheets such as Excel can do this curve fitting.
 
Mark44 said:
Have you covered least squares curve fitting? Here's a link to a wiki article - http://en.wikipedia.org/wiki/Least_squares

That article might be heavy going, but there are links to related topics that might be helpful.

The basic idea in least squares curve fitting is to find the line/parabola/exponential function that best fits the given data. For the line of best fit, you want the line y = ax + b so that the sum of the squares of the vertical distances between the line and the data is smallest.

For the parabola of best fit, you want the parabola y = ax^2 + bx + c, and for the exponential function, I think you want the function y = ae^(bx). Spreadsheets such as Excel can do this curve fitting.

I've never even heard most of these terms before, and after almost completing this course, if I needed to I would have by now I assume. There's only 3 pages on this in my text, without any mentioning of least squares curve fitting. I don't understand how there can be a parabola in such a graph. I've stumped so many people with this question, I'm honestly starting to think there was a misprint in the question. I totally understand what you're trying to tell me, but again, basing what is needed on the marking scheme, I just don't think this is it. If somebody can provide me with an example of any of the three models needed, I should be able to take it from there. I just have no idea. So confused.
 
Can you give us a summary of the example in the book? Maybe we can work out from that example how you need to proceed.
 
the example they used had to do with planets and their distance form the sun and using Astronomical Units as the y-axis and years as the x-axis. They also mentioned Kepplar's Third Law which seemed randomly inserted.

They used the compound interest formula of A=P(1+i)^n where they somehow decided that i was 0.017 and never explained why. n was changed to x-p where p equals some year in the table. Nothing was ever really explained. They also added the the value of y at the end of the equation. For example A=130(1.017)^x-1900 +0.63. I couldn't make any sense out of this. And the data table is long and very complex and only a form of the exponential equation was used.

The only other example I had is where I had 6 tables of data, 6 equations and I had to math the equation to the data. If the example provided was in any way feasible I would have attempted this question around that. The sad fact is, this textbook taught me nothing about this question.
 
  • #10
Ok I think I found the quadratic model.

600x^2-6400x+20000

This is roughly what i need for the other two now lol.
 
  • #11
As for how I got that, I don't know but I sure do love excel
 
  • #12
exponential I got 21538.46(0.65^x) Can anyone verify these?
 
  • #13
If you have a calculator you can verify them yourself. Or use Excel to calculate them.
 

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