Chapter 3: Problem 48
Find the values of \(k_{1}\) and \(k_{2}\) such that \((-1,0)\) and \((1,0)\) are \(x\) -intercepts for the graph of \(f(x)=k_{1} x^{4}-\) \(k_{2} x^{3}+x-4\).
Short Answer
Expert verified
The values are \(k_{1} = 4\) and \(k_{2} = 1\).
Step by step solution
01
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.
02
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 \(k_{1} + k_{2} = 5\).
03
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 \(k_{1} - k_{2} = 3\).
04
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 \(k_{1} = 4\).
05
Finding the value of k_{2}
Substitute \(k_{1} = 4\) back into the first equation: \(4 + k_{2} = 5\). Solving this gives \(k_{2} = 1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
Next, substitute \(k_1 = 4\) back into the first equation to solve for \(k_2\):
- Equation 1: \(k_1 + k_2 = 5\)
- Equation 2: \(k_1 - k_2 = 3\)
- \((k_1 + k_2) + (k_1 - k_2) = 5 + 3\)
- This simplifies to \(2k_1 = 8\).
Next, substitute \(k_1 = 4\) back into the first equation to solve for \(k_2\):
- \(4 + k_2 = 5\)
- Solve to find \(k_2 = 1\).
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.
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.
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:
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 \(x = 1\): \(k_{1}(1)^4 - k_{2}(1)^3 + 1 - 4 = 0\)