Problem 68 - College Algebra with Corequisite Support

Question

For the following exercises, construct a polynomial function of least degree possible using the given information. Real roots: \(-2, \frac{1}{2}\) (with multiplicity 2) and \((-3, f(-3))=(-3,5)\)

Step-by-Step Solution

Answer

The polynomial is \( f(x) = -\frac{20}{49}(x + 2)(x - \frac{1}{2})^2 \).

Step 1: Identifying the Roots

The roots given are - \(x = -2\),- \(x = \frac{1}{2}\) with a multiplicity of 2.The polynomial will have a factor of each form for every root. For the root \(x = -2\), the factor is \((x + 2)\). For \(x = \frac{1}{2}\) with multiplicity 2, the factor is \((x - \frac{1}{2})^2\).

Step 2: Constructing the Polynomial Using the Known Roots

The polynomial can be constructed from these factors:\[ f(x) = a(x + 2)(x - \frac{1}{2})^2 \]Where \(a\) is a constant to be determined.

Step 3: Using the Point (-3, 5) to Find 'a'

To determine the value of \(a\), use the given point \((-3, f(-3)) = (-3, 5)\).Substitute \(x = -3\) and \(f(-3) = 5\) into the polynomial:\[ 5 = a(-3 + 2)(-3 - \frac{1}{2})^2 \]

Step 4: Simplifying to Solve for 'a'

Calculate the expressions within the equation:\[5 = a(-1)(-\frac{7}{2})^2\]This simplifies to:\[5 = a(-1)(\frac{49}{4})\]\[5 = -\frac{49a}{4}\]Multiply through by \(-4/49\) to solve for \(a\):\[a = -\frac{20}{49}\]

Step 5: Final Polynomial Function

Substitute \(a = -\frac{20}{49}\) back into the polynomial:\[f(x) = -\frac{20}{49}(x + 2)(x - \frac{1}{2})^2\]Expand if necessary to present the polynomial as a standard form expression.

Key Concepts

Real Roots

Real roots of a polynomial refer to the values of x for which the polynomial equals zero. Put simply, if you plug the root into the polynomial, the result will be zero. In our exercise, the real roots provided are

• \(x = -2\)
• \(x = \frac{1}{2}\) with a multiplicity of 2

This means there are actual points on the x-axis, where the polynomial function will intersect. Each of these roots represents a factor of the polynomial:

• A root of \(x = -2\) contributes the factor \((x+2)\)
• A root of \(x = \frac{1}{2}\) contributes the factor \((x-\frac{1}{2})\)

Multiplicity of Roots

When we say a root has a 'multiplicity', we're talking about how many times that root appears as a factor of the polynomial. This also influences the shape of the graph at the point of the root.

• For a simple root, the graph will cross the x-axis.
• For a root with even multiplicity, the graph will touch and "bounce" off the x-axis.

In this exercise, \(x = \frac{1}{2}\) is given a multiplicity of 2.
This means the factor \((x - \frac{1}{2})\) appears twice in the polynomial: \((x - \frac{1}{2})^2\).
The graph will just touch the x-axis at \(x = \frac{1}{2}\).
Understanding multiplicities helps us predict the behavior of the polynomial graph.

Constant Determination

In building a polynomial, identifying the roots only brings us part of the way. The constant factor \(a\) in the polynomial equation is crucial as it impacts the stretch and orientation of the graph.
To determine this constant, a known point on the polynomial graph, other than the roots, must be used.
In our example, the point \((-3, 5)\) is used. We plug this into the polynomial \[f(x) = a (x + 2)(x - \frac{1}{2})^2\] substituting \(x = -3\) and \(f(x) = 5\) to solve for \(a\):

• Substitute the values into the factorized form.
• Calculate to find \(a\).

This gives us the specific constant needed to tailor the polynomial to this particular set of data.

Polynomial Construction

The construction of a polynomial based on given real roots is a methodical process. The roots and their factors provide the basic framework of the polynomial.
In this case, with roots \(-2\) and \(\frac{1}{2}\), the polynomial starts with the factors

• \((x + 2)\)
• \((x - \frac{1}{2})\)^2

Combining these, the polynomial initially looks like \[f(x) = a(x + 2)(x - \frac{1}{2})^2\]
With the constant \(a\) previously determined as \(-\frac{20}{49}\), the final polynomial is \[f(x) = -\frac{20}{49}(x + 2)(x - \frac{1}{2})^2\].This helps in crafting the polynomial that not only has the correct roots but also aligns perfectly with any additional conditions or points given in the problem.