Curve fitting (polyfit, fit) in MATLAB - Step-by-Step Execution

Concept Flow - Curve fitting (polyfit, fit)

Start with data points x,y

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Choose fitting method: polyfit or fit

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Calculate coefficients or fit object

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Evaluate fitted curve at desired points

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Plot original data and fitted curve

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End

Start with data points, choose polyfit or fit, compute the fit, evaluate the curve, then plot results.

Execution Sample

MATLAB

x = 1:5;
y = [2.2 2.8 3.6 4.5 5.1];
p = polyfit(x,y,1);
y_fit = polyval(p,x);

Fits a straight line to points (x,y) using polyfit and evaluates fitted values.

Execution Table

Step	Action	Input/Condition	Result/Output
1	Define x	x = [1 2 3 4 5]	x vector created
2	Define y	y = [2.2 2.8 3.6 4.5 5.1]	y vector created
3	Call polyfit	polyfit(x,y,1)	p = [0.73 1.49] (coefficients)
4	Call polyval	polyval(p,x)	y_fit = [2.22 2.95 3.68 4.41 5.14] (fitted y)
5	Plot data and fit	plot(x,y,'o',x,y_fit,'-')	Graph shows points and fitted line
6	End	All steps done	Curve fitting complete

💡 All steps completed successfully; fitted line computed and plotted.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	Final
x	undefined	[1 2 3 4 5]	[1 2 3 4 5]	[1 2 3 4 5]	[1 2 3 4 5]	[1 2 3 4 5]
y	undefined	undefined	[2.2 2.8 3.6 4.5 5.1]	[2.2 2.8 3.6 4.5 5.1]	[2.2 2.8 3.6 4.5 5.1]	[2.2 2.8 3.6 4.5 5.1]
p	undefined	undefined	undefined	[0.73 1.49]	[0.73 1.49]	[0.73 1.49]
y_fit	undefined	undefined	undefined	undefined	[2.22 2.95 3.68 4.41 5.14]	[2.22 2.95 3.68 4.41 5.14]

Key Moments - 3 Insights

Why does polyfit return two numbers for a linear fit?

polyfit returns coefficients for the polynomial starting with highest degree: here p(1) is slope and p(2) is intercept, as shown in step 3 of execution_table.

Why are fitted y values slightly different from original y?

Because polyfit finds the best line that approximates data, not exact points. Step 4 shows fitted y values close but not identical to original y.

What does the number 1 mean in polyfit(x,y,1)?

It means fitting a polynomial of degree 1, which is a straight line. This is the key input to polyfit in step 3.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what does the first number in p represent?

A Slope of the fitted line

B Intercept of the fitted line

C Number of data points

D Degree of polynomial

💡 Hint

Check step 3 in execution_table where p = [0.73 1.49]; first number is slope.

At which step do we get the fitted y values for the original x points?

A Step 2

B Step 3

C Step 4

D Step 5

💡 Hint

Look at step 4 in execution_table where y_fit is calculated.

If we change polyfit(x,y,1) to polyfit(x,y,2), what changes in the execution_table?

A Step 4 y_fit will be same as before

B Step 3 coefficients p will have 3 values

C Step 2 y vector changes

D Step 5 plot will not show data points

💡 Hint

Degree 2 polynomial means 3 coefficients; check step 3 for p.

Concept Snapshot

polyfit(x,y,n) fits polynomial degree n to data x,y
Returns coefficients p starting with highest degree
polyval(p,x) evaluates polynomial at x
fit function offers more options and fit types
Plot data and fit to visualize curve fitting

Full Transcript

This example shows how to fit a straight line to data points using MATLAB's polyfit and polyval functions. First, vectors x and y hold the data points. Then polyfit(x,y,1) computes the best line coefficients, slope and intercept. Next, polyval evaluates the fitted line at each x. Finally, plotting shows original points and the fitted line. The execution table traces each step, showing variable values and results. Key moments clarify why polyfit returns two numbers, what the degree means, and why fitted values differ slightly from original data. The visual quiz tests understanding of coefficients, evaluation step, and effect of changing polynomial degree. The snapshot summarizes syntax and behavior for quick reference.