Problem 16 - College Algebra with Corequisite Support

Question

For the following exercises, find the domain, vertical asymptotes, and horizontal asymptotes of the functions. $$ f(x)=\frac{x^{2}-1}{x^{3}+9 x^{2}+14 x} $$

Step-by-Step Solution

Answer

Domain is all real numbers except \( x = 0, -7, -2 \); vertical asymptotes at \( x = 0, -7, -2 \); horizontal asymptote at \( y = 0 \).

Step 1: Find the Domain

To find the domain of the function \( f(x) = \frac{x^2 - 1}{x^3 + 9x^2 + 14x} \), we need to identify the values of \( x \) for which the denominator is zero, as these are the values that are excluded from the domain. Set the denominator equal to zero and solve for \( x \):\[ x^3 + 9x^2 + 14x = 0 \]Factor out an \( x \):\[ x(x^2 + 9x + 14) = 0 \]Next, factor the quadratic expression:\[ x(x + 7)(x + 2) = 0 \]The solutions to this equation are \( x = 0, -7, -2 \). Therefore, the domain of \( f(x) \) is all real numbers except \( x eq 0, -7, -2 \). So, the domain is \( (-\infty, -7) \cup (-7, -2) \cup (-2, 0) \cup (0, \infty) \).

Step 2: Find Vertical Asymptotes

Vertical asymptotes occur at the values of \( x \) that make the denominator zero but do not make the numerator zero simultaneously. From our earlier factorization:\[ x(x + 7)(x + 2) = 0 \]We see that the potential points for vertical asymptotes are at \( x = 0, -7, -2 \).Check if the numerator is zero at these points:\[ x^2 - 1 = (x - 1)(x + 1) \]- At \( x = 0 \): \( 0^2 - 1 eq 0 \), so \( x = 0 \) is a vertical asymptote.- At \( x = -7 \): \((-7)^2 - 1 = 48 eq 0 \), so \( x = -7 \) is a vertical asymptote.- At \( x = -2 \): \((-2)^2 - 1 = 3 eq 0 \), so \( x = -2 \) is a vertical asymptote.Thus, the vertical asymptotes are \( x = 0, -7, -2 \).

Step 3: Find Horizontal Asymptotes

Horizontal asymptotes are determined by comparing the degrees of the numerator and the denominator:- Degree of numerator: 2 (\( x^2 \))- Degree of denominator: 3 (\( x^3 \))Since the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at \( y = 0 \).This implies that as \( x \) approaches positive or negative infinity, \( f(x) \) approaches zero.

Key Concepts

Vertical Asymptotes

Vertical asymptotes are like invisible walls on a graph that a function approaches but never touches. These occur at values of \( x \) where the denominator of a rational function is zero and the numerator is not zero. These __points__ mean the function is undefined at these specific \( x \) values.
For example, in the function \( f(x) = \frac{x^2 - 1}{x^3 + 9x^2 + 14x} \):

• We must find when \( x^3 + 9x^2 + 14x = 0 \).
• This factors to \( x(x + 7)(x + 2) = 0 \).
• Thus, \( x = 0, -7, \text{ or } -2 \) make the denominator zero.

However, for these points to be considered vertical asymptotes, the numerator should not also be zero at these \( x \) values.

• Checking \( x = 0 \), \((-7)\), and \((-2)\), \( x^2 - 1 \) doesn't equal zero, confirming vertical asymptotes at these points.

In summary, the vertical asymptotes for our function are at \( x = 0, -7, \) and \( -2 \). These are places where the graph shoots up to infinity or downward, creating an important characteristic of the function's behavior.

Horizontal Asymptotes

A horizontal asymptote of a graph tells us the value that the function approaches as \( x \) approaches positive or negative infinity. It is a horizontal line that the graph may get closer to but never actually touch.
To determine the horizontal asymptote of \( f(x) = \frac{x^2 - 1}{x^3 + 9x^2 + 14x} \), consider the degrees of the polynomial:

• Numerator degree (top): \( 2 \) (since it is \( x^2 \))
• Denominator degree (bottom): \( 3 \) (since it is \( x^3 \))

When the degree of the denominator is greater than that of the numerator, the horizontal asymptote is \( y = 0 \).
This is because as \( x \) becomes very large (positive or negative), the denominator grows much faster than the numerator, forcing the value of the function to get very close to zero.
This horizontal asymptote means that no matter how large or small \( x \) gets, the function's output (or \( y \)), will approach zero, demonstrating the function's behavior at its extremes.

Domain of a Function

The domain of a function refers to the set of all possible input values (\( x \)) for which the function is defined, ensuring no division by zero occurs. For rational functions, such as our example \( f(x) = \frac{x^2 - 1}{x^3 + 9x^2 + 14x} \), the primary concern is the denominator.

To ensure the function is defined, exclude any \( x \) values making the denominator zero. Consider the factorized form:

• \( x(x + 7)(x + 2) = 0 \).

These values are \( x = 0, -7, \text{ and } -2 \). These are the values that cause division by zero and must be excluded from the domain.
The domain is thus all real numbers except these points. In interval notation, the domain is expressed as:

• \( (-\infty, -7) \cup (-7, -2) \cup (-2, 0) \cup (0, \infty) \).

This tells us that the function can accept any real number as input except for \( 0, -7, \text{ and } -2 \). The domain explains where and why the graph is valid and ensures proper evaluation without undefined regions.