Math Problem Statement

Find a polynomial whose graph passes through the points whose coordinates are 

(−2, −16),   (0, 2),   (1, −1),

 and 

(2, −4).

Solution

To find the polynomial whose graph passes through the given points, we use Lagrange interpolation or solve a system of equations for a polynomial of degree 3, as four points determine a cubic polynomial.

General Form of a Cubic Polynomial

The general form of a cubic polynomial is:

P(x)=ax3+bx2+cx+dP(x) = ax^3 + bx^2 + cx + d

Step 1: Set up equations for the given points

Using the coordinates of the points, substitute xx and yy into the polynomial equation P(x)=yP(x) = y:

  1. For (2,16)(-2, -16): 16=a(2)3+b(2)2+c(2)+d-16 = a(-2)^3 + b(-2)^2 + c(-2) + d 16=8a+4b2c+d-16 = -8a + 4b - 2c + d

  2. For (0,2)(0, 2): 2=a(0)3+b(0)2+c(0)+d2 = a(0)^3 + b(0)^2 + c(0) + d 2=d2 = d

  3. For (1,1)(1, -1): 1=a(1)3+b(1)2+c(1)+d-1 = a(1)^3 + b(1)^2 + c(1) + d 1=a+b+c+d-1 = a + b + c + d

  4. For (2,4)(2, -4): 4=a(2)3+b(2)2+c(2)+d-4 = a(2)^3 + b(2)^2 + c(2) + d 4=8a+4b+2c+d-4 = 8a + 4b + 2c + d

Step 2: Substitute d=2d = 2 into the equations

Using d=2d = 2, the equations become:

  1. 16=8a+4b2c+2-16 = -8a + 4b - 2c + 2: 18=8a+4b2c-18 = -8a + 4b - 2c

  2. 1=a+b+c+2-1 = a + b + c + 2: 3=a+b+c-3 = a + b + c

  3. 4=8a+4b+2c+2-4 = 8a + 4b + 2c + 2: 6=8a+4b+2c-6 = 8a + 4b + 2c

Step 3: Solve the system of equations

We now solve the system of three equations:

  1. 8a+4b2c=18-8a + 4b - 2c = -18
  2. a+b+c=3a + b + c = -3
  3. 8a+4b+2c=68a + 4b + 2c = -6

Simplify the equations:

  1. Divide the first equation by 2: 4a+2bc=9-4a + 2b - c = -9

  2. Leave the second equation as is: a+b+c=3a + b + c = -3

  3. Divide the third equation by 2: 4a+2b+c=34a + 2b + c = -3

Add equations (1) and (3):

(4a+2bc)+(4a+2b+c)=9+(3)(-4a + 2b - c) + (4a + 2b + c) = -9 + (-3) 4b=12    b=34b = -12 \implies b = -3

Substitute b=3b = -3 into equation (2):

a3+c=3a - 3 + c = -3 a+c=0    c=aa + c = 0 \implies c = -a

Substitute b=3b = -3 and c=ac = -a into equation (1):

4a+2(3)(a)=9-4a + 2(-3) - (-a) = -9 4a6+a=9-4a - 6 + a = -9 3a=3    a=1-3a = -3 \implies a = 1

Find cc:

c=a    c=1c = -a \implies c = -1

Step 4: Write the polynomial

Now that we know a=1a = 1, b=3b = -3, c=1c = -1, and d=2d = 2, the polynomial is:

P(x)=x33x2x+2P(x) = x^3 - 3x^2 - x + 2

Final Answer:

P(x)=x33x2x+2P(x) = x^3 - 3x^2 - x + 2

Would you like to see the detailed graph or verification of the solution? Let me know!


5 Follow-Up Questions:

  1. How can Lagrange interpolation be used to derive the same polynomial?
  2. Can we verify this polynomial by plugging in the given points?
  3. What happens if we add an extra point to this problem?
  4. How does the degree of a polynomial relate to the number of points it passes through?
  5. What are the applications of polynomial interpolation in real-world problems?

Tip:

When solving systems of equations, always simplify each step to avoid algebraic errors, especially when substituting values!

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Math Problem Analysis

Mathematical Concepts

Polynomial Interpolation
Cubic Polynomials
Systems of Linear Equations

Formulas

General form of a cubic polynomial: P(x) = ax^3 + bx^2 + cx + d
Equation substitution for polynomial points: P(x) = y
Solving linear equations: substitution and elimination

Theorems

Lagrange Interpolation (optional method)
Fundamental Theorem of Algebra (roots and polynomial structure)

Suitable Grade Level

Grades 10-12