Curve fitting (polyfit, fit) in MATLAB - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to fit a polynomial of degree 2 to data points.

MATLAB

p = polyfit(x, y, [1]);

Drag options to blanks, or click blank then click option'

A1

B2

C0

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Using degree 1 fits a straight line, not a curve.

Using degree 0 fits a constant, which is not a curve.

The third argument to polyfit specifies the degree of the polynomial. Here, degree 2 fits a quadratic curve.

2fill in blank

medium

Complete the code to evaluate the polynomial fit at points x_new.

MATLAB

y_fit = polyval(p, [1]);

Drag options to blanks, or click blank then click option'

Ax_new

By_new

Cp

Dx

Attempts:

3 left

💡 Hint

Common Mistakes

Using y-values instead of x-values as input to polyval.

Passing the coefficients vector again instead of points.

polyval evaluates the polynomial coefficients p at the points given by x_new.

3fill in blank

hard

Fix the error in the code to create a fit object for a linear fit.

MATLAB

f = fit(x, y, [1]);

Drag options to blanks, or click blank then click option'

A'poly2'

B'linear'

C'exp1'

D'poly1'

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'linear' which is not a valid model name for fit.

Using 'poly2' which fits a quadratic, not linear.

To fit a linear polynomial using fit, use the model name 'poly1'.

4fill in blank

hard

Fill both blanks to create a quadratic fit and plot it.

MATLAB

f = fit(x, y, [1]);
plot(f, [2]);

Drag options to blanks, or click blank then click option'

A'poly2'

Bx

Cx_new

D'poly1'

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'poly1' instead of 'poly2' for quadratic fit.

Plotting against original x instead of new x-values.

'poly2' fits a quadratic polynomial. To plot the fit against new x-values, use x_new.

5fill in blank

hard

Fill all three blanks to create a cubic polynomial fit, evaluate it, and plot the results.

MATLAB

p = polyfit(x, y, [1]);
y_fit = polyval(p, [2]);
plot([3], y_fit, '-r');

Drag options to blanks, or click blank then click option'

A3

Bx_new

Dx

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong degree for cubic fit.

Plotting against original x instead of new x-values.

Degree 3 fits a cubic polynomial. Evaluate at x_new and plot y_fit against x_new.

Practice

1. What does the MATLAB function polyfit(x, y, n) do?

easy

A. It returns a fit object that can be used to predict new values.

B. It plots the data points x and y on a graph.

C. It finds the coefficients of a polynomial of degree n that fits the data points x and y.

D. It calculates the correlation coefficient between x and y.

Solution

1. Step 1: Understand the purpose of polyfit

   polyfit calculates polynomial coefficients that best fit the data points given by x and y.

2. Step 2: Differentiate from other functions

   Unlike fit, polyfit returns just coefficients, not a fit object or plot.

3. Final Answer:

   It finds the coefficients of a polynomial of degree n that fits the data points x and y. -> Option C

4. Quick Check:

   polyfit = polynomial coefficients [OK]

Hint: Remember: polyfit returns coefficients, fit returns a fit object [OK]

Common Mistakes:

• Confusing polyfit with fit function
• Thinking polyfit plots data
• Assuming polyfit returns a fit object

2. Which of the following is the correct syntax to fit a quadratic polynomial to data vectors x and y using polyfit?

easy

A. p = polyfit(x, y, 2);

B. p = polyfit(y, x, 2);

C. p = polyfit(x, y, 3);

D. p = polyfit(x, 2, y);

Solution

1. Step 1: Identify correct argument order

   The syntax is polyfit(x, y, n) where n is the polynomial degree.

2. Step 2: Confirm degree for quadratic

   Quadratic means degree 2, so the third argument must be 2.

3. Final Answer:

   p = polyfit(x, y, 2); -> Option A

4. Quick Check:

   polyfit(x, y, 2) fits quadratic [OK]

Hint: Remember order: x, y, degree in polyfit [OK]

Common Mistakes:

• Swapping x and y arguments
• Using wrong polynomial degree
• Incorrect argument order

3. What is the output of the following MATLAB code?

x = [1 2 3 4 5];
y = [2.2 4.1 6.0 8.1 10.2];
p = polyfit(x, y, 1);
y_fit = polyval(p, 6);

medium

A. y_fit = 12.1

B. y_fit = 12.3

C. y_fit = 13.0

D. y_fit = 10.2

Solution

1. Step 1: Fit a line to data

   Using polyfit with degree 1 fits a line y = mx + b to points.

2. Step 2: Calculate predicted value at x=6

   Using polyval, we evaluate the fitted line at x=6. The slope and intercept from polyfit give approximately y_fit = 12.1.

3. Final Answer:

   y_fit = 12.1 -> Option A

4. Quick Check:

   polyval(polyfit(x,y,1),6) ≈ 12.1 [OK]

Hint: polyval evaluates polynomial at given x [OK]

Common Mistakes:

• Confusing polyfit output with predicted value
• Using wrong degree in polyfit
• Calculating y_fit without polyval

4. Identify the error in this MATLAB code snippet for fitting data with fit:

x = (1:5)';
y = [2 4 6 8 10]';
f = fit(x, y, 'poly2');

medium

A. The fit type string should be 'poly1' for linear fit, not 'poly2'.

B. The fit function requires the Curve Fitting Toolbox to be installed.

C. The input vectors x and y must be row vectors, not column vectors.

D. The syntax is correct; there is no error.

Solution

1. Step 1: Check syntax and input shapes

   Vectors x and y as column vectors are valid inputs for fit.

2. Step 2: Verify toolbox requirement

   The fit function requires the Curve Fitting Toolbox; without it, the code will error.

3. Final Answer:

   The fit function requires the Curve Fitting Toolbox to be installed. -> Option B

4. Quick Check:

   fit needs toolbox installed [OK]

Hint: fit needs Curve Fitting Toolbox installed [OK]

Common Mistakes:

• Assuming fit works without toolbox
• Confusing poly1 and poly2 as errors
• Thinking vector orientation causes error

5. You have noisy data vectors x and y. You want to fit a cubic polynomial and then predict the value at x = 7. Which sequence of MATLAB commands correctly does this?

hard

A. p = polyfit(x, y, 3); y_pred = p(7);

B. f = fit(x, y, 'poly2'); y_pred = feval(f, 7);

C. p = polyfit(y, x, 3); y_pred = polyval(p, 7);

D. f = fit(x, y, 'poly3'); y_pred = feval(f, 7);

Solution

1. Step 1: Choose correct fit function and degree

   Using fit with 'poly3' fits a cubic polynomial to x and y.

2. Step 2: Predict value at x=7 using fit object

   Use feval(f, 7) or f(7) to get predicted y value from the fit object.

3. Final Answer:

   f = fit(x, y, 'poly3'); y_pred = feval(f, 7); -> Option D

4. Quick Check:

   fit with 'poly3' and feval predicts y at 7 [OK]

Hint: Use fit with 'poly3' and feval to predict [OK]

Common Mistakes:

• Swapping x and y in polyfit
• Using wrong polynomial degree
• Using poly2 instead of poly3 for cubic fit