Problem 48 - Precalculus with Calculus Previews

Question

Find the values of \(k_{1}\) and \(k_{2}\) such that \((-1,0)\) and \((1,0)\) are \(x\) -intercepts for the graph of \(f(x)=k_{1} x^{4}-\) \(k_{2} x^{3}+x-4\).

Step-by-Step Solution

Answer

The values are \(k_{1} = 4\) and \(k_{2} = 1\).

Step 1: Understanding x-intercepts

The x-intercepts of a graph are the values of \(x\) for which \(f(x) = 0\). Given the points \((-1,0)\) and \((1,0)\) as x-intercepts, we substitute these into the function to set up equations.

Step 2: Substituting the first intercept

Substitute \(x = -1\) into the function: \(f(-1) = k_{1}(-1)^4 - k_{2}(-1)^3 + (-1) - 4 = 0\). Simplify to get \(k_{1} + k_{2} - 1 - 4 = 0\), which simplifies to \(k_{1} + k_{2} = 5\).

Step 3: Substituting the second intercept

Substitute \(x = 1\) into the function: \(f(1) = k_{1}(1)^4 - k_{2}(1)^3 + 1 - 4 = 0\). Simplify to get \(k_{1} - k_{2} + 1 - 4 = 0\), which simplifies to \(k_{1} - k_{2} = 3\).

Step 4: Solving the system of equations

We now have a system of equations: \(k_{1} + k_{2} = 5\) and \(k_{1} - k_{2} = 3\). Add these equations: \((k_{1} + k_{2}) + (k_{1} - k_{2}) = 5 + 3\), simplifying to \(2k_{1} = 8\), so \(k_{1} = 4\).

Step 5: Finding the value of k_{2}

Substitute \(k_{1} = 4\) back into the first equation: \(4 + k_{2} = 5\). Solving this gives \(k_{2} = 1\).

Key Concepts

Solving Systems of Equations

Understanding how to solve systems of equations is a foundational skill in algebra. It involves finding the values of variables that satisfy multiple equations. In this instance, we have two linear equations with variables \(k_1\) and \(k_2\), derived from the x-intercepts of the polynomial:

• Equation 1: \(k_1 + k_2 = 5\)
• Equation 2: \(k_1 - k_2 = 3\)

To solve this system, we can use the method of addition (also known as the elimination method). Here's how it works: add the two equations together to eliminate one of the variables:

• \((k_1 + k_2) + (k_1 - k_2) = 5 + 3\)
• This simplifies to \(2k_1 = 8\).

Solve for \(k_1\) by dividing both sides by 2, yielding \(k_1 = 4\).
Next, substitute \(k_1 = 4\) back into the first equation to solve for \(k_2\):

• \(4 + k_2 = 5\)
• Solve to find \(k_2 = 1\).

This process highlights how solving equations provides us with the exact values that align with specific conditions—in this case, making sure the polynomial meets its x-intercept requirements.

Polynomial Functions

Polynomial functions are a type of mathematical expression that involve sums of powers of variables with coefficients. These functions are expressed in the form \(f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0\), where \(a_n, a_{n-1}, ..., a_0\) are constants.
The degree of the polynomial is determined by the highest power of \(x\) in the expression. Here, the polynomial \(f(x)=k_{1}x^{4}-k_{2}x^{3}+x-4\) is a fourth-degree polynomial because the highest power is 4.

• Each term in a polynomial function is composed of a coefficient and a variable raised to an exponent.
• The coefficients dictate the stretch and orientation of the function, while the exponents determine the curve's shape.

Polynomials are continuous and smooth, which means that they have no breaks or sharp corners, making them a vital tool for modeling real-world scenarios.

Zeros of Functions

The zeros of a function, often referred to as roots or x-intercepts, are the inputs \(x\) that make the output of the function zero, i.e., \(f(x) = 0\). Identifying zeros is crucial when graphing and analyzing functions.
For the function \(f(x)=k_{1}x^{4}-k_{2}x^{3}+x-4\), we know that the points \((-1,0)\) and \((1,0)\) should be zeros of the function. This means substituting these 'x' values into the function yields zero, helping to form equations:

• At \(x = -1\): \(k_{1}(-1)^4 - k_{2}(-1)^3 + (-1) - 4 = 0\)
• At \(x = 1\): \(k_{1}(1)^4 - k_{2}(1)^3 + 1 - 4 = 0\)

By solving the system of equations derived from these zeros, we ensure that our polynomial reflects the necessary intercept behavior. Understanding zeros helps us factor and sketch functions, crucially linking algebraic expressions with their graphical representations.