Problem 27 - Algebra Form and Function

Question

Find a polynomial of degree 4 that has zeros at \(x=\) \(-2, x=-1, x=2,\) and \(x=3\) and whose graph contains the point (0,6) .

Step-by-Step Solution

Answer

Answer: The polynomial function is \(f(x) = \frac{1}{2}(x + 2)(x + 1)(x - 2)(x - 3)\).

Step 1: Write down the general form of the polynomial

Since we know that the polynomial has zeros at \(x = -2, -1, 2,\) and \(3\), we can write it in the general form: \(f(x) = A(x + 2)(x + 1)(x - 2)(x - 3)\) where A is a constant to be determined.

Step 2: Plug in the given point (0,6)

We're given that the graph of the polynomial passes through the point (0,6). This means that \(f(0) = 6\). Plug in x = 0 into the general form of the polynomial: \(6 = A(0 + 2)(0 + 1)(0 - 2)(0 - 3)\).

Step 3: Solve for the constant A

We have \(6 = A(2)(1)(-2)(-3)\), which simplifies to \(6 = A(12)\). Divide both sides by 12 to get \(A = \frac{1}{2}\).

Step 4: Write the final polynomial

Now that we know the value of A, substitute it back into the general form of the polynomial to get the final polynomial: \(f(x) = \frac{1}{2}(x + 2)(x + 1)(x - 2)(x - 3)\).

Key Concepts

Zeros of Polynomial

To understand what it means for a polynomial to have zeros, imagine you're identifying key points where the graph of the polynomial touches or crosses the x-axis. These points are called "zeros" because this is where the polynomial equals zero. For a polynomial with degree 4, like in our exercise, you might find up to four zeros, depending on their multiplicity. When you have a list of zeros, say

• \(x = -2\),
• \(x = -1\),
• \(x = 2\), and
• \(x = 3\),

these values can be used to construct factors of the polynomial. This leads us to form the technique: - Each zero \(x = a\) corresponds to a factor \((x - a)\). Thus, given the zeros, our polynomial can be expressed in terms of \[f(x) = A(x+2)(x+1)(x-2)(x-3)\] where \(A\) is a constant that we will determine later.

Degree of Polynomial

The degree of a polynomial is an indicator of the "highest power" present in the polynomial expression. It can tell us a lot about the polynomial's behavior. In this case, we knew we were searching for a polynomial of degree 4.The degree is crucial because:

• It determines the general shape of the graph; defining how many turning points are possible.
• It predicts the number of zeros you should look out for.

When each zero comes from a distinct linear factor (as in our example with the zeros), this matches the degree directly.In simple terms:- The degree informed us that we should construct a polynomial that has exactly four linear factors multiplying together (each corresponding to one zero). This is how we arrived at \(f(x) = A(x+2)(x+1)(x-2)(x-3)\). Every time you work through polynomial problems, remember the degree will guide your expectations and the number of zeros you may find.

Constant Determination

Finding the constant \(A\) in a polynomial is all about identifying how the polynomial functions return specific values. In the given exercise, we were provided with the point (0,6), which directly helped us compute the constant. Here's a simple thought process to understand it:1. **Substitute the Point Given**: We use the point (0,6) because it tells us that when \(x = 0\), the polynomial should equal 6. Substituting into our polynomial \(6 = A(0+2)(0+1)(0-2)(0-3)\).2. **Simplifications**: When you plug in 0 into the factors, they simplify to 2, 1, -2, and -3. Therefore: \(6 = A \times 2 \times 1 \times (-2) \times (-3)\). 3. **Solve for A**: After multiplication inside the brackets: \(6 = 12A\). To find \(A\), divide both sides by 12, giving \(A = \frac{1}{2}\).This constant scales the polynomial so it fits perfectly with any additional points given. Completing this allows for accurately charting the correct polynomial graph.