Chapter 1: Problem 14
Find the polynomial with the smallest degree that goes through the given points. $$(-3,10),(-1,2),(1,2) \text { and }(2,5)$$
Short Answer
Expert verified
A cubic polynomial passes through the points: \(f(x) = x^3 - 2x^2 - 2x + 2\).
Step by step solution
01
Understand the problem
We need to find a polynomial of the smallest degree that goes through all the given points: \((-3,10), (-1,2), (1,2), (2,5)\). A polynomial of degree \(n\) can pass through exactly \(n+1\) points.
02
Determine the polynomial degree
The polynomial must pass through 4 points. Since a polynomial of degree \(n\) passes through exactly \(n+1\) distinct points, a quadratic (degree 3) polynomial can pass through exactly 4 points. Hence, we'll start with a cubic polynomial.
03
Construct the general cubic polynomial
A general cubic polynomial can be written as \(f(x) = ax^3 + bx^2 + cx + d\). We need to find the coefficients \(a, b, c,\) and \(d\) such that the polynomial satisfies all given points.
04
Set up equations using the points
Use each point to create an equation by substituting the \(x\) and \(y\) values into the polynomial. The equations are: \[ \begin{align*} 27a - 9b - 3c + d &= 10, \ -a + b - c + d &= 2, \ a + b + c + d &= 2, \ 8a + 4b + 2c + d &= 5. \end{align*} \]
05
Solve the system of equations
Solve the system of four equations with four unknowns to find the coefficients \(a, b, c,\) and \(d\). This typically involves methods like substitution or elimination.
06
Verify the polynomial
Once the coefficients are found, verify the polynomial by substituting back the points to check that they all satisfy the polynomial equation. This ensures the polynomial is correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cubic Polynomial
A cubic polynomial is a type of polynomial that has a degree of three. This means that the highest exponent of the variable in the equation is three. A general form of a cubic polynomial can be written as \( f(x) = ax^3 + bx^2 + cx + d \), where \( a, b, c, \) and \( d \) are coefficients and \( a eq 0 \).
A cubic polynomial can have a wide variety of shapes, including one or two turns, because it can intersect the x-axis up to three times, depending on its roots. In the context of polynomial interpolation, cubic polynomials are often used since they can fit exactly four points. This is because a polynomial of degree \( n \) can fit \( n+1 \) points. Therefore, with a cubic polynomial (degree 3), you can fit exactly four distinct points.
The reason why cubic polynomials are frequently employed in interpolation is due to their balance between flexibility and simplicity. They offer sufficient complexity to accurately capture the pattern across multiple points without becoming overly complicated.
A cubic polynomial can have a wide variety of shapes, including one or two turns, because it can intersect the x-axis up to three times, depending on its roots. In the context of polynomial interpolation, cubic polynomials are often used since they can fit exactly four points. This is because a polynomial of degree \( n \) can fit \( n+1 \) points. Therefore, with a cubic polynomial (degree 3), you can fit exactly four distinct points.
The reason why cubic polynomials are frequently employed in interpolation is due to their balance between flexibility and simplicity. They offer sufficient complexity to accurately capture the pattern across multiple points without becoming overly complicated.
System of Equations
A system of equations is a set of equations with multiple variables which need to be solved together.
Each equation adds constraints and collectively they help to pinpoint specific solutions. In the process of determining a cubic polynomial through interpolation, as in the provided exercise, a system of equations is derived from the given points. These equations represent the constraints that the polynomial must satisfy for each given point.
For the points \((-3,10), (-1,2), (1,2), (2,5)\), each point provides one equation that collectively forms a system:
Each equation adds constraints and collectively they help to pinpoint specific solutions. In the process of determining a cubic polynomial through interpolation, as in the provided exercise, a system of equations is derived from the given points. These equations represent the constraints that the polynomial must satisfy for each given point.
For the points \((-3,10), (-1,2), (1,2), (2,5)\), each point provides one equation that collectively forms a system:
- \(27a - 9b - 3c + d = 10\)
- \(-a + b - c + d = 2\)
- \(a + b + c + d = 2\)
- \(8a + 4b + 2c + d = 5\)
Degree of a Polynomial
The degree of a polynomial is the highest power of the variable in the polynomial expression. It determines the number of roots the polynomial can have and the polynomial's shape when graphed. The degree of a polynomial is an important concept because it tells us about the behavior and characteristics of the polynomial. For example, the degree will tell us how many points it can interpolate.
In polynomial interpolation, the degree indicates how many points the polynomial can potentially fit. For instance, a linear polynomial (degree 1) can fit two points, a quadratic polynomial (degree 2) can fit three points, and a cubic polynomial (degree 3) can fit four points, as seen in our exercise.
This understanding helps when choosing the type of polynomial needed to interpolate given data points. In the context of this exercise, we needed a polynomial of at least degree 3 to fit through 4 points. By choosing the smallest possible degree, solutions remain simple and efficient, avoiding unnecessary complexity.
In polynomial interpolation, the degree indicates how many points the polynomial can potentially fit. For instance, a linear polynomial (degree 1) can fit two points, a quadratic polynomial (degree 2) can fit three points, and a cubic polynomial (degree 3) can fit four points, as seen in our exercise.
This understanding helps when choosing the type of polynomial needed to interpolate given data points. In the context of this exercise, we needed a polynomial of at least degree 3 to fit through 4 points. By choosing the smallest possible degree, solutions remain simple and efficient, avoiding unnecessary complexity.