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(PRE-CALCULUS) Create a polynomial that goes through the following points

(PRE-CALCULUS) Create a polynomial that goes through the following points-example-1
User Andi Droid
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The polynomial y = - 3 · x³ - 2 · x² + x is the equation that passes through given points.

How to determine the polynomial equation that passes through a given set of points

According to algebra, a polynomial of grade n can be found by knowning n + 1 distinct points of the form (x, y). Since we need to determine a polynomial grade 3, then we need four distinct points and definition of polynomial:

y = a · x³ + b · x² + c · x + d, where a, b, c, d are real coefficients.

If we know that (8, - 1656), (- 2, 14), (9, - 2340) and (10, - 3190), then the coefficients of the polynomials are:

a · 8³ + b · 8² + c · 8 + d = - 1656

a · (- 2)³ + b · (- 2)² + c · (- 2) + d = 14

a · 9³ + b · 9² + c · 9 + d = - 2340

a · 10³ + b · 10² + c · 10 + d = - 3190

The resulting system of linear equations is:

512 · a + 64 · b + 8 · c + d = - 1656

- 8 · a + 4 · b - 2 · c + d = 14

729 · a + 81 · b + 9 · c + d = - 2340

1000 · a + 100 · b + 10 · c + d = - 3190

Then, the resulting coefficients of the polynomial are:

(a, b, c, d) = (- 3, - 2, 1, 0)

The polynomial that passes through given points is y = - 3 · x³ - 2 · x² + x.

User Grant Trevor
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