Problem 41 - Precalculus with Calculus Previews

Question

In Problems 41 and 42 , find a cubic polynomial function \(f\) that satisfies the given conditions. rational zeros 1 and \(2, f(0)=1\) and \(f(-1)=4\)

Step-by-Step Solution

Answer

The cubic polynomial is \( f(x) = -\frac{7}{6}(x - 1)(x - 2)(x + \frac{7}{3}) \).

Step 1: Identify the Zeros and Form the Polynomial

Given the rational zeros are 1 and 2. A cubic polynomial can be expressed as \( f(x) = a(x - 1)(x - 2)(x - r) \), where \( r \) is another zero and \( a \) is a leading coefficient. Since only two zeros are given, \( r \) will be determined such that \( f(0) = 1 \).

Step 2: Use the Condition f(0)=1

Substitute \( x = 0 \) into the polynomial: \[ f(0) = a(0 - 1)(0 - 2)(0 - r) = a(-1)(-2)(-r) = 2ar \]Since \( f(0) = 1 \), we set up the equation:\[ 2ar = 1 \]

Step 3: Use the Condition f(-1)=4

Substitute \( x = -1 \) into the polynomial: \[ f(-1) = a(-1 - 1)(-1 - 2)(-1 - r) = a(-2)(-3)(-1 - r) \]We know \( f(-1) = 4 \), so:\[ -6a(1 + r) = 4 \]\[ -6a - 6ar = 4 \]

Step 4: Solve the Equations Simultaneously

From step 2, \( ar = \frac{1}{2} \).From step 3, substitute \( ar = \frac{1}{2} \):\[ -6a - 3 = 4 \]Solving for \( a \), \[ -6a = 7 \]\[ a = -\frac{7}{6} \]Substitute \( a \) into \( ar = \frac{1}{2} \):\[ -\frac{7}{6}r = \frac{1}{2} \]\[ r = -\frac{7}{3} \]

Step 5: Construct the Final Polynomial

Substitute the values of \( a \) and \( r \) into the polynomial expression:\[ f(x) = -\frac{7}{6}(x - 1)(x - 2)(x + \frac{7}{3}) \]Expand this to find the polynomial function.

Key Concepts

Rational Zeros

When working with polynomial functions, understanding rational zeros is crucial. Rational zeros are the roots of a polynomial that can be expressed as fractions or whole numbers. In our exercise, we are given the rational zeros 1 and 2. These zeros help us construct the factors of the polynomial. For instance, if 1 and 2 are zeros, the polynomial will have corresponding factors

• (x - 1)
• (x - 2)

These factors are important as they directly influence the form of the polynomial. Identifying rational zeros is a step that reduces complexity and guides you towards solving a polynomial equation more efficiently. Knowing their role aids in constructing the initial structure of a polynomial function.

Leading Coefficient

The leading coefficient of a polynomial is another vital concept. It is the coefficient of the term with the highest degree. In polynomial expressions such as cubic ones, it significantly impacts the shape and behavior of the graph. For example, consider a polynomial defined as \( f(x) = a(x - 1)(x - 2)(x - r) \).

• The leading coefficient here is "a"
• It determines how steep or flat the graph of the polynomial is

In our exercise, the value of the leading coefficient "a" was determined to be \(-\frac{7}{6}\). The sign and magnitude of 'a' can tell us whether the polynomial will open upwards or downwards. A negative leading coefficient like \(-\frac{7}{6}\) indicates that the graph of the polynomial will fall from left to right when moving along the x-axis. It's important to correctly establish the leading coefficient as it affects the polynomial's overall behavior and characteristics.

Polynomial Expansion

Expanding a polynomial is the process of expressing it as a sum of terms. It involves multiplying out the factors to reveal the entire polynomial structure. From our exercise, we have a polynomial written as \( f(x) = -\frac{7}{6}(x - 1)(x - 2)(x + \frac{7}{3}) \).

• Each pair of terms within the factors is multiplied together individually
• The results are summed up to develop the polynomial expression

This expansion is necessary to translate the factored form into a form that can be easily analyzed for its coefficients and other properties. Understanding polynomial expansion allows us to manifest the complete polynomial equation, which is pivotal for solving equations or graphing functions. This profound step helps to fully express the polynomial's power and intricacies in both professional and academic settings.