Question

Find the polynomial with the smallest degree that goes through the given points. $$(-3,3),(1,3) \text { and }(2,3)$$

Short answer

The polynomial is \( f(x) = 3 \).

Step-by-step solution

Determine the Polynomial Degree

First, observe the number of given points, which are three: (-3,3), (1,3), and (2,3). A polynomial that passes through n points and results uniquely is of degree at most n-1. Therefore, the polynomial we are looking for is of degree at most 2 because we have 3 points.

Examine the Y-Values

Notice that the y-values for all these points are 3. This means that all the points lie on a horizontal line. Hence, the polynomial doesn't actually change with x.

Identify the Polynomial

Since the y-values are constant (3 for all x-values), this implies that the polynomial is a constant function. The simplest polynomial that represents a horizontal line at y=3 for any x is the constant polynomial.

State the Polynomial

The constant polynomial that passes through all the given points is simply:\[ f(x) = 3 \]

Key concepts

Degree of a Polynomial

Polynomials are mathematical expressions involving a sum of powers of variables multiplied by coefficients. The "degree" of a polynomial is a crucial concept in defining its complexity and behavior. It is determined by the highest power of the variable in the polynomial. For example, in the polynomial expression \(7x^3 + 4x^2 - 2x + 9\), the highest power of \(x\) is 3, making it a polynomial of degree 3.

To determine the degree of a polynomial that passes through given points, one might initially consider the number of points. However, a polynomial that can satisfy \(n\) unique data points is typically of degree \(n-1\). This is because each point contributes to defining the polynomial. In our initial exercise, since there are three points, the polynomial's degree can be no more than 2, unless further simplification is possible due to the nature of the points themselves.

• A linear polynomial is one where the degree is 1.
• A quadratic polynomial has a degree of 2.
• The constant polynomial has a degree of 0, which leads us perfectly to the next section about constant polynomials.

Constant Polynomial

A constant polynomial is perhaps the simplest form of a polynomial. It has the form \(f(x) = c\), where \(c\) is a constant. Essentially, this means the output value of the polynomial is the same for any input \(x\). In mathematical terms, it doesn't "change" with \(x\); it forms a horizontal line on a graph.

In the given exercise, the solution resulted in a constant polynomial because every point provided \((-3,3), (1,3), (2,3)\) has the same y-value (3). This is indicative of a horizontal line at \(y=3\). Thus, the polynomial that satisfies these conditions is simply \(f(x) = 3\).

• Having a degree zero, it is unaffected by the x-input and regularly appears as \(y = c\).
• This constant nature makes it unique among polynomials, as it ignores any variance in x.

Polynomial Interpolation

Polynomial interpolation involves finding a polynomial that exactly passes through a given set of points. It is a fundamental tool in numerical analysis and computer graphics for constructing new data points within a range of known data points.

The exercise you encountered is a real-life example of polynomial interpolation, albeit a simple one. We identified the polynomial that passes through the points \((-3, 3), (1, 3), (2, 3)\). Since all these points lie on a line parallel to the x-axis at y = 3, polynomial interpolation was straightforward, leading to the constant polynomial \(f(x) = 3\).

In more complex scenarios, polynomial interpolation might involve quadratic or cubic equations, especially if the points are not aligned horizontally or vertically. It's pivotal:

• To ensure the degree of the polynomial correctly reflects the number of points minus one.
• To provide insight into the curve's behavior between the known points.

Understanding these basics of polynomial interpolation aids in solving problems efficiently, as seen with the constant solution for our exercise.