Question

Find the polynomial with the smallest degree that goes through the given points. $$(-1,-8),(1,-2) \text { and }(3,4)$$

Short answer

The polynomial is \(y = 3x - 5\).

Step-by-step solution

Understand the Problem

We need to find the polynomial of smallest degree which passes through all the given points: \((-1,-8)\), \((1,-2)\), and \((3,4)\). The smallest polynomial that can pass through three points is a quadratic polynomial, which is of the form \(y = ax^2 + bx + c\).

Set Up the System of Equations

Plug each given point into the quadratic polynomial equation \(y = ax^2 + bx + c\) to form a system of equations:1. For \((-1, -8)\): \[-8 = a(-1)^2 + b(-1) + c\] This simplifies to: \[a - b + c = -8\]2. For \((1, -2)\): \[-2 = a(1)^2 + b(1) + c\] This simplifies to: \[a + b + c = -2\]3. For \((3, 4)\): \[4 = a(3)^2 + b(3) + c\] This simplifies to: \[9a + 3b + c = 4\]

Solve the System of Equations

We now have three equations:1. \(a - b + c = -8\)2. \(a + b + c = -2\)3. \(9a + 3b + c = 4\)Subtract the first equation from the second to find \(b\):\[(a + b + c) - (a - b + c) = -2 + 8\]\[2b = 6 \Rightarrow b = 3\]Substitute \(b = 3\) into the first equation:\[a - 3 + c = -8 \Rightarrow a + c = -5\]Substitute \(b = 3\) into the third equation:\[9a + 3(3) + c = 4 \Rightarrow 9a + 9 + c = 4\]\[9a + c = -5\]Now we solve the equations \(a + c = -5\) and \(9a + c = -5\):

Solve for a, b, and c

Subtract \(a + c = -5\) from \(9a + c = -5\):\[(9a + c) - (a + c) = -5 + 5\]\[8a = 0 \Rightarrow a = 0\]Substitute \(a = 0\) into \(a + c = -5\):\[0 + c = -5\Rightarrow c = -5\]Our values are \(a = 0\), \(b = 3\), and \(c = -5\). Thus, the polynomial is \(y = 3x - 5\).

Verify the Solution

Plug each of the points back into the polynomial \(y = 3x - 5\) to ensure it passes through each point:1. For \((-1, -8)\): \[y = 3(-1) - 5 = -3 - 5 = -8\] (Correct)2. For \((1, -2)\): \[y = 3(1) - 5 = 3 - 5 = -2\] (Correct)3. For \((3, 4)\): \[y = 3(3) - 5 = 9 - 5 = 4\] (Correct)All points are correctly satisfied by the polynomial \(y = 3x - 5\).

Key concepts

Quadratic Polynomial

When trying to fit a polynomial through given points, we aim to find the polynomial equation that matches the specified points. For three points in the plane, the polynomial with the smallest possible degree is a quadratic polynomial. A quadratic polynomial is generally expressed as:
\[y = ax^2 + bx + c \]
where \(a\), \(b\), and \(c\) are constants. The presence of \(x^2\) shows us that it's a quadratic (second degree) equation. The task is to determine these constants so that the polynomial passes exactly through the given points, essentially forming a curve that touches each point.

System of Equations

To find the constants in the quadratic polynomial \(y = ax^2 + bx + c\), we need to use the given points to form a system of equations. Each point is substituted into the polynomial equation, giving us a distinct equation.
For example, given the points

• \((-1, -8)\), you plug into \(y = ax^2 + bx + c\) to obtain: \(-8 = a(-1)^2 + b(-1) + c\), simplifying to: \(a - b + c = -8\).
• \((1, -2)\), results in: \(-2 = a(1)^2 + b(1) + c\), simplifying to: \(a + b + c = -2\).
• \((3, 4)\), results in: \(4 = a(3)^2 + b(3) + c\), simplifying to: \(9a + 3b + c = 4\).

You will have as many equations as there are points. With three equations from three points, you can solve for the unknowns \(a\), \(b\), and \(c\).

Solving Equations

Solving the system of equations means finding the values for \(a\), \(b\), and \(c\) that satisfy all equations simultaneously.
You start by using elimination or substitution methods. From the system:

• Equation 1: \(a - b + c = -8\)
• Equation 2: \(a + b + c = -2\)
• Equation 3: \(9a + 3b + c = 4\)

First, subtract Equation 1 from Equation 2 to find \(b\):
\(2b = 6\), giving \(b = 3\).
Substitute \(b = 3\) into other equations and solve for \(a\) and \(c\) through similar elimination:
With \(a + c = -5\) and \(9a + c = -5\), you simplify to solve for \(a\) and find that \(a = 0\).
Finally substitute back to solve for \(c\), finding \(c = -5\). After solving, the final polynomial forms, \(y = 3x - 5\).

Verifying Solutions

Verification ensures that the polynomial found actually passes through all provided points. Each point is substituted back into the derived polynomial equation to confirm satisfaction.
Use the polynomial \(y = 3x - 5\):

• For \((-1, -8)\), substitute: \(y = 3(-1) - 5 = -3 - 5 = -8\), which is correct.
• For \((1, -2)\), substitute: \(y = 3(1) - 5 = 3 - 5 = -2\), which is correct.
• For \((3, 4)\), substitute: \(y = 3(3) - 5 = 9 - 5 = 4\), which is correct.

With all points satisfying the equation, the derived polynomial is verified. This cross-checking step is crucial as it validates the correctness of our calculations.