Curve fitting (polyfit, fit) in MATLAB - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of polyfit polynomial coefficients

What is the output of the following MATLAB code?

x = [1 2 3 4 5]; y = [2.2 4.8 7.1 8.9 11.3]; p = polyfit(x, y, 1); disp(round(p,1))

MATLAB

x = [1 2 3 4 5];
y = [2.2 4.8 7.1 8.9 11.3];
p = polyfit(x, y, 1);
disp(round(p,1))

A [2.3 0.7]

B [1.8 2.2]

C [2.2 1.8]

D [2.2 0.2]

💡 Hint

Recall polyfit returns coefficients starting with highest degree term.

polyfit fits y = p1*x + p2. The slope is about 2.2 and intercept about 0.2 after rounding.

❓ Predict Output

intermediate

Result of fit with custom model

What is the value of the coefficient 'a' after running this MATLAB code?

x = (1:5)'; y = [2.2 4.8 7.1 8.9 11.3]'; ft = fittype('a*x+b'); f = fit(x,y,ft); disp(round(f.a,1))

MATLAB

x = (1:5)';
y = [2.2 4.8 7.1 8.9 11.3]';
ft = fittype('a*x+b');
f = fit(x,y,ft);
disp(round(f.a,1))

A 4.8

B 1.7

C 2.2

D 0.7

💡 Hint

Coefficient 'a' is the slope in the linear model.

The fit finds a slope close to 2.2 matching the data trend.

❓ Predict Output

advanced

Output of polynomial evaluation after polyfit

What is the output of this MATLAB code?

x = 1:4; y = [1 8 27 64]; p = polyfit(x,y,3); val = polyval(p, 2.5); disp(round(val,1))

MATLAB

x = 1:4;
y = [1 8 27 64];
p = polyfit(x,y,3);
val = polyval(p, 2.5);
disp(round(val,1))

A 15.6

B 14.5

C 16.0

D 15.0

💡 Hint

The data is cubic: y = x^3, so polyval should approximate 2.5^3.

2.5^3 = 15.625, rounded to 15.6 matches option A.

❓ Predict Output

advanced

Error type from incorrect fit model syntax

What error does this MATLAB code produce?

x = (1:3)'; y = [2 4 6]'; ft = fittype('a*x^2 + b'); f = fit(x,y,ft);

MATLAB

x = (1:3)';
y = [2 4 6]';
ft = fittype('a*x^2 + b');
f = fit(x,y,ft);

A Error: Undefined function or variable 'a'

B No error, fit succeeds

C Error: Invalid model expression

D Error: Fit cannot converge

💡 Hint

Check if the model string syntax is valid for fittype.

No error is produced; the expression 'a*x^2 + b' is valid because '^' is supported in fittype model strings.

🧠 Conceptual

expert

Choosing polynomial degree with polyfit

Given noisy data points, which polynomial degree chosen with polyfit is most likely to cause overfitting?

A Degree 10 (very high)

B Degree 2 (quadratic)

C Degree 1 (linear)

D Degree 3 (cubic)

💡 Hint

Higher degree polynomials can fit noise, not just trend.

A very high degree polynomial (like 10) fits noise and causes overfitting, losing generalization.

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