Problem 86 - Calculus Volume 1

Question

For each polynomial, a. find the degree; b. find the zeros, if any; c. find the \(y\) -intercept(s), if any; d. use the leading coefficient to determine the graph's end behavior; and e. determine algebraically whether the polynomial is even, odd, or neither. \(f(x)=x^{3}+3 x^{2}-x-3\)

Step-by-Step Solution

Answer

Degree: 3; Zeros: -1, 1, -3; Y-intercept: (0, -3); End behavior: Rises right, falls left; Neither even nor odd.

Step 1: Determine the Degree of the Polynomial

The degree of a polynomial is the highest power of the variable present in the polynomial. In this case, the highest power of the variable \( x \) is 3, so the degree of the polynomial is 3.

Step 2: Find the Zeros of the Polynomial

To find the zeros of the polynomial, set \( f(x) = 0 \):\[x^3 + 3x^2 - x - 3 = 0\]By using the Rational Root Theorem and synthetic division, we find that \( x = -1 \) is a root. Dividing the polynomial by \( x + 1 \) leaves \( x^2 + 2x - 3 \). Factoring it gives \( (x - 1)(x + 3) \). Thus, the zeros are \( x = -1, 1, -3 \).

Step 3: Find the Y-Intercept of the Polynomial

The y-intercept is found by evaluating the polynomial at \( x = 0 \). Therefore:\[f(0) = 0^3 + 3\cdot0^2 - 0 - 3 = -3\]Hence, the y-intercept is \( (0, -3) \).

Step 4: Determine the End Behavior Using the Leading Coefficient

The leading term is \( x^3 \) and its coefficient is positive. For polynomials with an odd degree and a positive leading coefficient, as \( x \to +fty \), \( f(x) \to +fty \), and as \( x \to -fty \), \( f(x) \to -fty \). So the graph will start low in the left and rise high on the right.

Step 5: Determine If the Polynomial is Even, Odd, or Neither

A polynomial is even if \( f(x) = f(-x) \) and odd if \( f(-x) = -f(x) \). Let's check:\[f(-x) = (-x)^3 + 3(-x)^2 -(-x) - 3 = -x^3 + 3x^2 + x - 3\]Since \( f(-x) eq f(x) \) and \( f(-x) eq -f(x) \), the polynomial is neither even nor odd.

Key Concepts

Degree of Polynomial

When we talk about the degree of a polynomial, we are referring to the highest power of the variable within that polynomial. This is a crucial concept because the degree gives us insight into the polynomial's shape and behavior. In our function, \(f(x) = x^{3} + 3x^{2} - x - 3\), we notice that the highest exponent of \(x\) is 3. Therefore, the degree of this polynomial is 3.
This tells us that it is a cubic polynomial. This is important since the degree influences the number of turning points the polynomial can have and its end behavior.

Zeros of Polynomial

Finding the zeros of a polynomial means finding values of \(x\) that make the polynomial equal to zero; these are also known as the roots. To find them, we set the polynomial equation to zero: \[x^3 + 3x^2 - x - 3 = 0\]
Using tools like the Rational Root Theorem and synthetic division, we can simplify this equation and discover its roots. For our specific function, we find that \(x = -1\) is a root. Dividing the polynomial by \(x + 1\) results in \(x^2 + 2x - 3\), which can be factored further into \((x - 1)(x + 3)\).
This means the zeros are \(x = -1, 1, \text{ and } -3\). These zeros indicate where the graph of the polynomial crosses the x-axis.

Leading Coefficient End Behavior

The leading coefficient and the degree of the polynomial together determine the end behavior of the graph. The leading coefficient is the constant multiplying the term with the highest power. In our polynomial \(f(x) = x^3 + 3x^2 - x - 3\), the leading term is \(x^3\) and the leading coefficient is 1, which is positive.
Since the degree is odd (3 in this case), and the leading coefficient is positive, the graph will descend as \(x\) approaches negative infinity and ascend as \(x\) goes to positive infinity. Simply put, the left end of the graph goes down, and the right end goes up.

Even and Odd Functions

A polynomial is classified as even if \(f(x) = f(-x)\) for all \(x\), and it is odd if \(f(-x) = -f(x)\). Checking these criteria helps to understand the symmetry of the function. In our example, substituting \(-x\) into the function \(f(x) = x^3 + 3x^2 - x - 3\) gives: \[f(-x) = (-x)^3 + 3(-x)^2 - (-x) - 3 = -x^3 + 3x^2 + x - 3\]
We can quickly observe that \(f(-x)\) is not equal to \(f(x)\) and \(f(-x)\) is not the negative of \(f(x)\), meaning it doesn't satisfy the conditions for being either even or odd. Hence, this polynomial is neither even nor odd. Understanding this aspect of the function helps us predict its symmetry and how it will graph.