Problem 67 - College Algebra

Question

For the following exercises, construct a polynomial function of least degree possible using the given information. Real roots: \(-1,1\) (with multiplicity 2 and 1\()\) and \((2, f(2))=(2,4)\)

Step-by-Step Solution

Answer

The polynomial is \(f(x) = \frac{4}{9}(x+1)^2(x-1)\).

Step 1: Identify Known Roots

We are given the polynomial has real roots at \(-1\) (with multiplicity 2) and \(1\) (with multiplicity 1). These roots can be expressed as the factors \((x + 1)^2\) and \((x - 1)\).

Step 2: Write the General Form of the Polynomial

Combine the factors to form a general polynomial function: \(f(x) = a(x+1)^2(x-1)\), where \(a\) is a constant that needs to be determined.

Step 3: Use Given Function Value to Solve for Constant

The polynomial passes through the point \((2, 4)\), so substitute \(x = 2\) and \(f(x) = 4\) into the polynomial to solve for \(a\):\[4 = a(2+1)^2(2-1)\]\[4 = a(3)^2(1)\]\[4 = 9a\]\[a = \frac{4}{9}\]

Step 4: Write the Polynomial Function

Substitute the value of \(a\) back into the general polynomial equation to obtain the final polynomial:\[f(x) = \frac{4}{9}(x+1)^2(x-1)\]

Key Concepts

Real Roots

In polynomial functions, 'real roots' are specific values of the variable, often denoted as \(x\), where the polynomial equals zero. These points are crucial because they indicate where the graph of the polynomial will intersect the x-axis. For the polynomial to touch or cross the x-axis at these points, the function value at these roots must be zero. In the given exercise, real roots are identified as \(-1\) and \(1\). Understanding real roots is vital as it helps in forming the polynomial, where each real root can be expressed as a factor. For example, a root at \(-1\) translates to the factor \((x + 1)\), and at \(1\) to \((x - 1)\). Understanding these factor forms will help reconstruct the polynomial step-by-step.

Multiplicity

Multiplicity refers to the number of times a particular root is repeated in a polynomial equation. It highlights the root's role in how the polynomial behaves at that point on the graph. If a root has a multiplicity greater than 1, the graph will touch the x-axis but not cross it. In the given question, the real root at \(-1\) has a multiplicity of 2, meaning \((x + 1)^2\) is a factor, causing the graph to merely "touch" but not cross the x-axis at \(-1\). On the other hand, the root at \(1\), with a multiplicity of 1, forms the factor \((x - 1)\), where the graph will cross the x-axis. Recognizing the multiplicity helps in understanding the overall shape and behavior of the graph of the polynomial function.

Polynomial Degree

The degree of a polynomial is a crucial aspect that indicates the highest power of the variable present in the polynomial expression. It also determines the maximum number of roots and the shape of the graph. In constructing a polynomial, we aim for the lowest degree, especially given the roots and their multiplicities. To find the degree, sum the multiplicities of all distinct roots. In this exercise, we have two roots, \(-1\) with a multiplicity of 2 and \(1\) with a multiplicity of 1. This sums up to a degree of 3, indicating a cubic polynomial. Polynomials with odd degrees have different end behavior going to infinity or negative infinity, which is important for sketching the graph accurately.

Function Value

The 'function value' at a particular given point provides more information needed to pin down the exact polynomial function. It is often used to find unknown coefficients. For this exercise, we use the fact that the polynomial passes through the point \((2, 4)\). This means when \(x = 2\), the function value \(f(x) = 4\). Such values allow for determining any constants or coefficients that are not clear from roots alone. By substituting \(x = 2\) into our polynomial with unidentified constant \(a\), we calculate \(a\) such that when \(x = 2\), the polynomial indeed evaluates to \(4\). Solving this gives the coefficient \(a = \frac{4}{9}\), allowing the final polynomial to be accurately expressed as \(f(x) = \frac{4}{9}(x+1)^2(x-1)\).