How Do I Derive Equation From a Graph?

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Homework Help Overview

The discussion centers around deriving an equation from a set of points plotted on a graph, specifically focusing on the relationship between x and y coordinates. The subject area involves concepts from precalculus and polynomial functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore various methods to derive an equation, including linear equations, piecewise functions, and polynomial fitting. Questions arise regarding the nature of the relationship represented by the points and the implications of fitting different types of functions.

Discussion Status

The discussion is active, with participants providing insights into different approaches to modeling the data. Some suggest using polynomial fitting while others emphasize the importance of understanding the underlying relationship between the points. There is an acknowledgment of the complexity involved in finding an appropriate function.

Contextual Notes

Participants note that the data may represent a curve rather than a simple linear relationship, and there are considerations regarding the degree of the polynomial needed to fit the points accurately. The presence of multiple data points complicates the equation derivation process.

Gordon Arnaut
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How can I derive the equation from a Graph?

The x coordinates are: 28,30,32,33,34,35

The Y coordinates are: 7,8,9,10,11,12


We could express this as a relation, r:

r{(7,28), (8,30), (9,32), (10,33), (11,34) (12,35)}


Regards,

Gordon,
 
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Are you familiar with the general form of a linear equation, y=mx+b? What are m and b?

Graph your points by hand, and you should be able to figure out m and b. Then, you should look in your book to see how you figure out m and b when you are given just two x,y points for a line in the general case...
 
BTW, I moved this thread to Homework Help, Precalculus Math. Please be sure to post all homework and coursework problems in the Homework Help forums, and not in the general forums.
 
Berkeman,

The relation describes a curve, not a line.

See the attached graph.


Regards,

Gordon
 

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Ah, I'd just glanced at the first few data points, and it looked like the delta y was 1 for each 2 delta x. But now I see that it's actually two straight lines that join in the middle...

So you should be able to write the equation as piecewise continuous straight lines over two x intervals...
 
Okay thanks, Berkeman.

Actually there are more data points, so it becomes more like a curve than two straight lines.

I'm going to try function transformations to see if I can come up with a formula that comes close to matching the curve.

Regards,

Gordon.
 
The most straightforward approach is to fit an (n-1) degree polynomial to the n points, in this case a degree 5 since you have 6 points. It means solving a 6 x 6 set of simultaneous linear equations to get the coefficients of the polynomial. The fit will be exact, but the function may be wiggly - may or may not be what you want.

If you are looking for an approximating function that is fit to the points in a least squares sense, that's more complicated. There is no way I know of to determine the best form of the equation to model the points (exhaustive trial and error will get you close). If the points came from experimental data, you might glean a hint from knowledge of the mathematical/physical nature of the experiment. Short of that, you can look at the points and make intelligent guesses as to functions to try (e.g. low order polynomials, exponential, etc.). It means trial and error, but that's a perfectly valid technique.
.
 
Last edited:
There exists an infinite number of graphs that will pass through any given finite set of points. What other conditions do you have? Are you looking for the polynomial of lowest degree that gives a graph through those points?
 
You can get a best fit line from a graphing calculator or Excel, seeing as you already have the data in there
 

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