Problem 85 - Calculus Volume 1

Question

For the following exercises, 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. $$ f(x)=\frac{1}{2} x^{2}-1 $$

Step-by-Step Solution

Answer

Degree: 2, Zeros: \(\sqrt{2}, -\sqrt{2}\), y-intercept: \(-1\), End behavior: goes to infinity, Even function.

Step 1: Determine the Degree of the Polynomial

The degree of a polynomial is the highest power of the variable in the polynomial expression. In the given polynomial, \(f(x) = \frac{1}{2}x^2 - 1\), the highest power of \(x\) is 2. Therefore, the degree of the polynomial is 2.

Step 2: Find the Zeros of the Polynomial

To find the zeros, set the polynomial equal to zero and solve for \(x\):\[\frac{1}{2}x^2 - 1 = 0 \]Add 1 to both sides:\[\frac{1}{2}x^2 = 1\]Multiply both sides by 2:\[x^2 = 2\]Take the square root of both sides:\[x = \pm \sqrt{2}\]Thus, the zeros are \(x = \sqrt{2}\) and \(x = -\sqrt{2}\).

Step 3: Find the y-intercept

The \(y\)-intercept of a polynomial is the value of the function when \(x = 0\). Substitute \(x = 0\) in the polynomial:\[f(0) = \frac{1}{2}(0)^2 - 1 = -1\]Thus, the \(y\)-intercept is \(-1\).

Step 4: Determine the End Behavior Using the Leading Coefficient

The leading term of the polynomial is \(\frac{1}{2}x^2\). Since the degree is even and the leading coefficient \(\frac{1}{2}\) is positive, the end behavior is: as \(x\) approaches \(\pm\infty\), \(f(x)\) approaches \(+\infty\).

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

A polynomial function is even if \(f(-x) = f(x)\) and odd if \(f(-x) = -f(x)\). Let's check:1. Calculate \(f(-x)\):\[f(-x) = \frac{1}{2}(-x)^2 - 1 = \frac{1}{2}x^2 - 1 = f(x)\]This satisfies \(f(-x) = f(x)\), indicating the function is even.

Key Concepts

Degree of a Polynomial

When you look at a polynomial, the degree is the highest power of the variable within its terms.
This tells us a lot about the behavior and nature of the polynomial. The degree helps in predicting how many times the graph of the polynomial might intersect the x-axis, or in other terms, how many zeros the polynomial can have.

• If the polynomial is given as a single term, the degree is the exponent of that term.
• If the polynomial is a sum of terms, look for the highest exponent present.

For example, in the polynomial function \(f(x) = \frac{1}{2}x^2 - 1\), the degree is 2, since the highest power of \(x\) is 2. This makes it a quadratic polynomial that has specific characteristics, like a parabolic graph.

Zeros of a Polynomial

The zeros of a polynomial are the values of \(x\) that make the polynomial equal to zero.
Essentially, these are the points where the graph of the polynomial touches or crosses the x-axis.

• To find the zeros, set the polynomial equal to zero and solve for \(x\).
• This may involve factoring, using the quadratic formula, or other algebraic techniques.

In our example, solving \(\frac{1}{2}x^2 - 1 = 0\) gives us \(x = \pm \sqrt{2}\). Hence, these zeros, \(x = \sqrt{2}\) and \(x = -\sqrt{2}\), are where the polynomial intersects the x-axis.

End Behavior of Polynomials

The end behavior of a polynomial describes what happens to the polynomial's values as \(x\) approaches positive or negative infinity.
This is determined by two main factors:

• The degree of the polynomial (whether it's odd or even).
• The leading coefficient (whether it's positive or negative).

For an even-degree polynomial like \(f(x) = \frac{1}{2}x^2 - 1\), the graph tends to rise to infinity on both ends if the leading coefficient is positive.
In our case, the leading coefficient is \(\frac{1}{2}\), which is positive, suggesting that both ends of the polynomial graph will rise upwards, approaching \(+\infty\) as \(x\) moves towards \(+\infty\) or \(-\infty\).

Even and Odd Functions

Determining if a polynomial is even, odd, or neither can help us understand its symmetrical properties.
A function is said to be:

• Even if \(f(-x) = f(x)\) for all \(x\). This means the graph is symmetrical about the y-axis.
• Odd if \(f(-x) = -f(x)\) for all \(x\). This means the graph is symmetrical about the origin.
• Neither if it doesn't satisfy either condition.

For \(f(x) = \frac{1}{2}x^2 - 1\), calculating \(f(-x)\) gives us \(f(-x) = \frac{1}{2}x^2 - 1 = f(x)\).
This confirms that the polynomial is an even function because \(f(-x) = f(x)\), illustrating symmetry around the y-axis.