Problem 12 - Algebra Form and Function

Question

A polynomial \(p(x)\) can be written in two forms: I. \(p(x)=\left(x^{2}+4\right)\left(4-x^{2}\right)\) II. \(\quad p(x)=16-x^{4}\) Which form most readily shows (a) The number of zeros of \(p(x) ?\) Find them. (b) The vertical intercept? What is it? (c) The sign of \(p(x)\) as \(x\) gets large, either positive or negative. What are the signs?

Step-by-Step Solution

Answer

Question: Determine the number of zeros, the vertical intercept, and the sign of the polynomial \(p(x)\) when \(x\) goes to either positive or negative infinity for the following polynomial: Form I: \(p(x) = (x^2 + 4)(4 - x^2)\) Form II: \(p(x) = 16 - x^4\) Answer: The polynomial \(p(x)\) has 2 zeros at \(x = 2\) and \(x = -2\). The vertical intercept is \((0, 16)\), and as \(x \to \pm \infty\), \(p(x)\) tends to negative values.

Step 1: Find the zeros of the polynomial

In order to find the number of zeros of the polynomial \(p(x)\), we will first compare the two forms, Form I and Form II, to determine which form is more suitable. Form I: \(p(x) = (x^2 + 4)(4 - x^2)\) Form II: \(p(x) = 16 - x^4\) Form I is more useful because it is in factored form, which is helpful when identifying the zeros of a polynomial. Now let's find the zeros of the polynomial p(x) from Form I: Set the factors equal to zero: \(x^2 + 4 = 0\) \(4 - x^2 = 0\) Solve each equation for x: \(x^2 = -4\) - No real solutions \(x^2 = 4\) And, solving for x gives: \(x = \pm \sqrt{4} = \pm 2\) The polynomial \(p(x)\) has 2 zeros: \(x = 2\) and \(x = -2\).

Step 2: Find the vertical intercept

To find the vertical intercept, we will find the value of the polynomial \(p(x)\) when \(x=0\). In this case, it is easier to use Form II because it has fewer terms and no additional factors. Evaluate Form II of the polynomial at \(x=0\): \(p(0) = 16 - (0)^4\) \(p(0) = 16\) The vertical intercept is \((0, 16)\).

Step 3: Determine the sign of p(x) when x goes to either positive or negative infinity

To find the sign of \(p(x)\) when \(x\) goes to either positive or negative infinity, we need to analyze the behavior of the polynomial when \(x\) becomes very large. For this purpose, we will use Form II because the highest degree term controls the behavior when \(x\) has extremely large values. As \(x \to \pm \infty\), the dominant term of \(p(x)\) is \(-x^4\). This term is always non-positive since any exponent of a power of 4 is always square and thus non-negative. Moreover, the sign of \(p(x)\) depends on whether the other terms in the polynomial, in this case, 16, can change the overall sign. When \(x\) gets large enough, the value of \(x^4\) will be much greater than 16, and \(p(x)\) will be negative. Therefore, as \(x \to \pm \infty\), \(p(x)\) tends to negative values.

Key Concepts

Zeros of Polynomial

The zeros of a polynomial are the values of \(x\) for which the polynomial equals zero. To find them, we often write the polynomial in factored form because it simplifies the process of setting each factor to zero. This transformation is helpful because it break downs complex expressions into manageable parts. In our exercise, the polynomial \(p(x) = (x^2 + 4)(4 - x^2)\) from Form I is suitable for finding zeros. Here’s why:

• The expression \(x^2 + 4 = 0\) does not yield real solutions because no real number squared will result in a negative number.
• The other expression \(4 - x^2 = 0\) simplifies to \(x^2 = 4\), giving us the solutions \(x = \pm 2\).

Therefore, the polynomial \(p(x)\) has two real zeros: \(x = 2\) and \(x = -2\). Understanding these solutions tell us where the graph of the polynomial intersects the x-axis. In a learning context, identifying zeros is crucial as they represent the roots of the polynomial.

Vertical Intercept

The vertical intercept of a polynomial is the point where the graph intersects the y-axis. This occurs when \(x\) is equal to zero. To find this point, substitute \(x = 0\) in the polynomial and solve for \(p(x)\). Taking our polynomial's Form II, \(p(x) = 16 - x^4\), makes it straightforward:

• Calculate \(p(0) = 16 - (0)^4 = 16\).

Thus, the vertical intercept is at \((0, 16)\). This point is significant because it gives us an initial value of the polynomial when graphed. Knowing how to locate and interpret the vertical intercept can help in graphing or planning the polynomial's behavior without plotting numerous other points.

End Behavior of Polynomials

The end behavior of a polynomial refers to what happens to \(p(x)\) as \(x\) goes to very large positive or negative values. This behavior is primarily influenced by the term with the highest degree in the polynomial. For \(p(x) = 16 - x^4\), Form II clearly indicates that \(-x^4\) dominates as \(x\) grows very large:

• This term \(-x^4\) becomes overwhelmingly larger and negative compared to the constant 16.

Therefore, as \(x\) approaches infinity or negative infinity, \(p(x)\) trends towards negative infinity. Recognizing how the highest degree terms dominate the end behavior is essential in sketching the polynomial's graph and understanding its overall shape. This insight is invaluable in calculus and mathematical modeling, providing a more comprehensive view of how functions behave beyond immediate observation.