Chapter 3: Problem 2
Find a polynomial that passes through the four points $$ (0,0),(1,2),(4,-3), \text { and }(10,-1) $$
Short Answer
Expert verified
The polynomial is \(p(x) = \frac{7}{60}x^3 - \frac{23}{60}x^2 + \frac{41}{30}x\).
Step by step solution
01
Understand the Requirements
We need to find a polynomial that passes through four given points: \((0,0)\), \((1,2)\), \((4,-3)\), and \((10,-1)\). This means we need to find a polynomial \(p(x)\) such that \(p(0) = 0\), \(p(1) = 2\), \(p(4) = -3\), and \(p(10) = -1\). Since we have four points, we'll look for a cubic polynomial (degree 3): \(p(x) = ax^3 + bx^2 + cx + d\).
02
Set Up Equations from Points
Plug each point into the polynomial equation \(p(x) = ax^3 + bx^2 + cx + d\):1. For \((0,0)\): \(0 = a(0)^3 + b(0)^2 + c(0) + d = d\) so, \(d = 0\).2. For \((1,2)\): \(2 = a(1)^3 + b(1)^2 + c(1) + d\).3. For \((4,-3)\): \(-3 = a(4)^3 + b(4)^2 + c(4) + d\).4. For \((10,-1)\): \(-1 = a(10)^3 + b(10)^2 + c(10) + d\).
03
Substitute and Simplify Equations
Substitute \(d = 0\) in the remaining equations:1. \(2 = a + b + c\)2. \(-3 = 64a + 16b + 4c\)3. \(-1 = 1000a + 100b + 10c\)
04
Solve the System of Equations
To find the coefficients \(a\), \(b\), and \(c\), solve the system:1. From \(a + b + c = 2\)2. From \(64a + 16b + 4c = -3\)3. From \(1000a + 100b + 10c = -1\)By solving these using substitution or any preferred method, you'll find:\[a = \frac{7}{60}, b = -\frac{23}{60}, c = \frac{41}{30}, d = 0\]
05
Write the Polynomial
With the coefficients found, write the polynomial:\[p(x) = \frac{7}{60}x^3 - \frac{23}{60}x^2 + \frac{41}{30}x\]
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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 is characterized by having a degree of three. This means the highest power of the variable, usually denoted as \(x\), is three. The general form of a cubic polynomial is \(p(x) = ax^3 + bx^2 + cx + d\). Here:
- \(a\), \(b\), \(c\), and \(d\) are constants known as coefficients.
- \(x^3\) is the term that gives the polynomial its cubic nature, determining the polynomial's highest degree.
- If \(a\) is non-zero, then the polynomial maintains its cubic form.
System of Equations
A system of equations is essentially a set of two or more equations that we deal with together. In the context of finding a cubic polynomial, we use a system of equations to determine the polynomial coefficients that will make the polynomial curve go through the given points. Each point provides a specific equation when plugged into the polynomial formula. For instance, given the original points:
- \((0,0)\) leads to the equation \(d = 0\).
- \((1,2)\) gives us \(a + b + c + d = 2\).
- \((4,-3)\) provides \(64a + 16b + 4c + d = -3\).
- \((10,-1)\) results in \(1000a + 100b + 10c + d = -1\).
Polynomial Coefficients
The coefficients of a polynomial are crucial as they dictate the polynomial's shape and position. For a cubic polynomial in the form \(p(x) = ax^3 + bx^2 + cx + d\), the coefficients \(a\), \(b\), \(c\), and \(d\) play significant roles:
- \(a\) influences the leading behavior. A positive \(a\) will make the curve rise to the right, while a negative \(a\) causes it to fall.
- \(b\) and \(c\) account for how the curve twists and bends.
- \(d\), the constant term, shifts the curve up or down without affecting its shape.