Polynomial evaluation and roots in MATLAB - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to evaluate the polynomial p at x = 2.

MATLAB

p = [1 -3 2];
y = polyval(p, [1]);

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

A2

B3

Cx

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using the variable name instead of a number.

Forgetting to pass the evaluation point as the second argument.

The function polyval evaluates the polynomial p at the given value. Here, we want to evaluate at 2.

2fill in blank

medium

Complete the code to find the roots of the polynomial p.

MATLAB

p = [1 -3 2];
r = [1](p);

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

Apolyval

Broots

Csolve

Dfind

Attempts:

3 left

💡 Hint

Common Mistakes

Using polyval instead of roots.

Using find which is for arrays, not polynomials.

The roots function returns the roots of the polynomial defined by the vector p.

3fill in blank

hard

Fix the error in the code to correctly evaluate the polynomial p at x = 5.

MATLAB

p = [2 0 -1 3];
y = polyval([1], 5);

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

Ap

Bpolyval

C[2 0 -1 3]

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping the arguments of polyval.

Passing the number as the first argument.

The first argument to polyval must be the polynomial coefficients vector p, not the evaluation point.

4fill in blank

hard

Fill both blanks to create a polynomial vector for p(x) = 3x^2 - 4x + 1 and evaluate it at x = 3.

MATLAB

p = [[1] [2] 1];
y = polyval(p, 3);

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

A3

B-4

C4

D-3

Attempts:

3 left

💡 Hint

Common Mistakes

Using positive 4 instead of -4.

Swapping the order of coefficients.

The polynomial coefficients are in descending powers: 3 for x^2, -4 for x, and 1 constant.

5fill in blank

hard

Fill all three blanks to create a polynomial vector for p(x) = x^3 - 6x^2 + 11x - 6 and find its roots.

MATLAB

p = [[1] [2] [3] -6];
r = roots(p);

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

A1

B-6

C11

D6

Attempts:

3 left

💡 Hint

Common Mistakes

Using positive 6 instead of -6.

Mixing up the order of coefficients.

The polynomial coefficients are for x^3, x^2, and x terms respectively: 1, -6, and 11.

Practice

1. What does the MATLAB function polyval(p, x) do when p is a vector of polynomial coefficients and x is a number?

easy

A. Evaluates the polynomial at the value x.

B. Finds the roots of the polynomial p.

C. Plots the polynomial curve over x.

D. Calculates the derivative of the polynomial at x.

Solution

1. Step 1: Understand the function inputs

   The vector p contains polynomial coefficients, and x is the point where we want to evaluate the polynomial.

2. Step 2: Identify the function purpose

   polyval calculates the polynomial value at x using the coefficients in p.

3. Final Answer:

   Evaluates the polynomial at the value x. -> Option A

4. Quick Check:

   polyval = evaluate polynomial [OK]

Hint: polyval means evaluate polynomial at given points [OK]

Common Mistakes:

• Confusing polyval with roots function
• Thinking polyval finds zeros of polynomial
• Assuming polyval plots the polynomial

2. Which of the following is the correct syntax to find the roots of a polynomial with coefficients stored in vector p in MATLAB?

easy

A. polyval(p)

B. roots(p)

C. findroots(p)

D. solve(p)

Solution

1. Step 1: Recall MATLAB root-finding function

   MATLAB uses the function roots to find zeros of a polynomial given its coefficients.

2. Step 2: Check syntax correctness

   The correct syntax is roots(p), where p is the coefficient vector.

3. Final Answer:

   roots(p) -> Option B

4. Quick Check:

   roots function syntax = roots(p) [OK]

Hint: Use roots() to find polynomial zeros in MATLAB [OK]

Common Mistakes:

• Using polyval instead of roots
• Trying non-existent functions like findroots
• Passing wrong arguments to roots

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

p = [1 -3 2]; % Represents x^2 - 3x + 2
r = roots(p);

medium

A. [1 2]

B. [3 2]

C. [2 1]

D. [-1 -2]

Solution

1. Step 1: Identify the polynomial

   The vector p = [1 -3 2] represents the polynomial x^2 - 3x + 2.

2. Step 2: Find roots of the polynomial

   Factoring: (x-1)(x-2) = 0 so roots are 1 and 2. MATLAB returns a column vector [2; 1], matching option [2 1].

3. Final Answer:

   [2 1] -> Option C

4. Quick Check:

   Roots of x^2-3x+2 = 1, 2 [OK]

Hint: Roots are where polynomial equals zero [OK]

Common Mistakes:

• Mixing up coefficient order
• Forgetting roots are values of x
• Assuming roots are negative

4. The following MATLAB code is intended to evaluate the polynomial p = [2 -5 3] at x = 4. What is the error?

p = [2 -5 3];
y = roots(p, 4);

medium

A. The syntax is correct; no error.

B. The variable y is not defined before use.

C. The vector p should be transposed before calling roots.

D. roots does not take a second argument; use polyval instead.

Solution

1. Step 1: Understand function usage

   The function roots only takes one input: the coefficient vector p.

2. Step 2: Identify correct function for evaluation

   To evaluate polynomial at a point, use polyval(p, x), not roots.

3. Final Answer:

   roots does not take a second argument; use polyval instead. -> Option D

4. Quick Check:

   roots(p) finds zeros; polyval(p,x) evaluates [OK]

Hint: Use polyval to evaluate, roots only finds zeros [OK]

Common Mistakes:

• Passing extra arguments to roots
• Confusing roots and polyval functions
• Assuming roots evaluates polynomial

5. You have a polynomial p = [1 -6 11 -6] representing x^3 - 6x^2 + 11x - 6. Which MATLAB code correctly finds the roots and evaluates the polynomial at x = 2?

hard

A. r = roots(p); y = polyval(p, 2);

B. r = polyval(p, 2); y = roots(p);

C. r = roots(2); y = polyval(p);

D. r = roots(p, 2); y = polyval(p, 2);

Solution

1. Step 1: Find roots of polynomial

   Use roots(p) to get zeros of polynomial p.

2. Step 2: Evaluate polynomial at x=2

   Use polyval(p, 2) to find polynomial value at 2.

3. Final Answer:

   r = roots(p); y = polyval(p, 2); -> Option A

4. Quick Check:

   roots(p) finds zeros, polyval(p,x) evaluates [OK]

Hint: roots() finds zeros; polyval() evaluates at points [OK]

Common Mistakes:

• Swapping roots and polyval calls
• Passing wrong arguments to roots
• Using roots with two arguments