Problem 54 - Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

Domain: All real numbers. Range: \(-2 < y < 0\). Symmetry: Even.

Step 1: Understand the Function

The given function is \( y = \frac{-2}{x^{2} + 1} \). This function is a rational function where the numerator is a constant (-2) and the denominator is a quadratic expression \( x^2 + 1 \).

Step 2: Determine Domain

For rational functions like \( y = \frac{-2}{x^{2} + 1} \), the domain includes all values for which the denominator is not zero. Since \( x^2 + 1 \geq 1 \) for all real numbers, \( x^2 + 1 \) is never zero. Therefore, the domain of the function is all real numbers.

Step 3: Analyze the Range

For a function \( y = \frac{-2}{x^{2} + 1} \), identify how \( y \) changes with different values of \( x \). The expression \( x^2 + 1 \) is always positive, ranging from 1 to infinity, hence the entire \( \frac{-2}{x^2 + 1} \) is negative and its values range from \( -2 \) to 0.

Step 4: Check for Symmetry

The function is even since replacing \( x \) with \( -x \) does not change the function: \( y(-x) = \frac{-2}{(-x)^2 + 1} = \frac{-2}{x^2 + 1} = y(x) \). Thus, the function is symmetric with respect to the y-axis.

Step 5: Find Intercepts

The y-intercept occurs when \( x = 0 \). Substituting \( x = 0 \) into the function, \( y = \frac{-2}{0^2 + 1} = -2 \). There are no x-intercepts as the fraction never equals zero.

Step 6: Identify Asymptotes

There are no vertical asymptotes since the denominator \( x^2 + 1 \) is never zero. There is a horizontal asymptote at \( y = 0 \) because as \( x \to \pm \infty \), \( y = \frac{-2}{x^2 + 1} \to 0 \).

Key Concepts

Domain of a Function

The domain of a function is the set of all possible input values, usually denoted as "x", for which the function is defined. For rational functions, you need to check where the denominator is zero because division by zero is undefined.
In the function \( y = \frac{-2}{x^2 + 1} \), the denominator is \( x^2 + 1 \). It is important to note that \( x^2 + 1 \) is always greater than or equal to 1 for all real numbers because the square of any real number \( x^2 \) is always non-negative, and adding one makes it strictly positive. Thus, the denominator never becomes zero.
This means the domain of the function is all real numbers, or \( (-\infty, \infty) \). It's quite simple because this particular rational function doesn't exclude any real numbers from its domain.

Range of a Function

To determine the range of a function, we must identify all possible output values it can produce. In the function \( y = \frac{-2}{x^2 + 1} \), the expression \( x^2 + 1 \) gives values starting from 1 to infinity since \( x^2 \) is always non-negative.
When we divide -2 by these positive values, the output \( y \) is always negative and falls between -2 and 0.
Think of it this way:

• If \( x^2 + 1 = 1 \), the smallest possible value, \( y = -2 \).
• If \( x^2 + 1 \) becomes very large, approaching infinity, \( y \) approaches 0 but is always less than 0.

Thus, the range of the function is \( (-2, 0] \). This helps understand that the function's outputs are limited to negative values close to zero and -2.

Symmetry of Functions

A function's symmetry helps in understanding its behavior without calculating every single point. The function \( y = \frac{-2}{x^2 + 1} \) is symmetric because it's even. To check for even symmetry, you substitute \( x \) with \( -x \) and see if the function remains unchanged.
For this function:

• Substituting \( -x \) yields \( y(-x) = \frac{-2}{(-x)^2 + 1} = \frac{-2}{x^2 + 1} \).
• This is identical to the original \( y(x) = \frac{-2}{x^2 + 1} \).

Since \( y(-x) = y(x) \), the function has even symmetry, meaning it is symmetric with respect to the y-axis. Knowing a function's symmetry is key in graphing and understanding its overall shape.

Intercepts of Functions

Intercepts are points where a graph crosses the x-axis or y-axis. They provide fixed reference points for understanding a function's behavior.
For the rational function \( y = \frac{-2}{x^2 + 1} \):

• The y-intercept occurs where \( x = 0 \). Substituting \( x = 0 \) into the function, we find \( y = \frac{-2}{0^2 + 1} = -2 \). Thus, the y-intercept is at the point (0, -2).
• For x-intercepts, we look for when \( y = 0 \). However, the fraction \( \frac{-2}{x^2 + 1} \) is never zero since the numerator is a constant non-zero value (-2). Consequently, there are no x-intercepts.

Intercepts are essential as they provide starting points for sketching the graph and understanding the function’s intersection with the axes.

Asymptotes of Functions

Asymptotes are lines that a graph approaches but never touches. They reveal how a function behaves at extreme values and are crucial for understanding the long-term behavior of functions.
For the function \( y = \frac{-2}{x^2 + 1} \), asymptotes are determined as follows:

• Vertical asymptotes occur when the denominator is zero. Here, \( x^2 + 1 \) is never zero, so there are no vertical asymptotes.
• Horizontal asymptotes describe what happens as \( x \to \pm \infty \). For this function, as \( x^2 + 1 \) becomes very large, \( y \) gets closer to zero. Therefore, there is a horizontal asymptote at \( y = 0 \).

Understanding asymptotes helps to sketch rational functions, indicating the directions the curves will head toward as they extend indefinitely.