Problem 16 - College Algebra

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: \( x \neq 0, -7, -2 \); Vertical asymptotes: \( x = 0, -7, -2 \); Horizontal asymptote: \( y = 0 \).

Step 1: Find the Domain

To find the domain of the function \( f(x)=\frac{x^{2}-1}{x^{3}+9 x^{2}+14 x} \), we need to identify the values of \( x \) that make the denominator zero, as these will be excluded from the domain. Set the denominator equal to zero:\[x^3 + 9x^2 + 14x = 0\]Factor out \( x \):\[x(x^2 + 9x + 14) = 0\]This gives one factor, \( x = 0 \). Now factor \( x^2 + 9x + 14 \) further:\[x(x + 7)(x + 2) = 0\]So, the values of \( x \) that make the denominator zero are \( x = 0 \), \( x = -7 \), and \( x = -2 \). Therefore, the domain is all real numbers except \( x = 0 \), \( x = -7 \), and \( x = -2 \).

Step 2: Identify Vertical Asymptotes

Vertical asymptotes occur at the values of \( x \) that make the denominator of the reduced form of the function zero, provided that they don't cancel with zeros in the numerator. Since the numerator \( x^2 - 1 \) factors to \( (x-1)(x+1) \), and none of these factors cancel with \( (x)(x+7)(x+2) \), the vertical asymptotes are directly at the roots we found. Therefore, vertical asymptotes occur at \( x = 0 \), \( x = -7 \), and \( x = -2 \).

Step 3: Find Horizontal Asymptotes

The degrees of the numerator and the denominator will help us find horizontal asymptotes. The numerator \( x^2 - 1 \) is of degree 2, and the denominator \( x^3 + 9x^2 + 14x \) is of degree 3. Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is \( y = 0 \).

Key Concepts

Domain of a Function

The domain of a function consists of all the possible input values (usually represented by \( x \)) for which the function is defined. When dealing with rational functions like \( f(x)=\frac{x^{2}-1}{x^{3}+9x^{2}+14x} \), the primary concern is that the denominator does not become zero, as this would make the function undefined.
To find the domain, set the denominator equal to zero: \[x^3 + 9x^2 + 14x = 0\]
Then solve the equation to find the values of \( x \) that are excluded from the domain:

• Factor out \( x \): \( x(x^2 + 9x + 14) = 0 \).
• This yields one solution, \( x = 0 \).
• Next, factor \( x^2 + 9x + 14 \) to get: \((x + 7)(x + 2) = 0\).

From these factors, the solutions \( x = -7 \) and \( x = -2 \) are obtained. Thus, the complete domain of the function includes all real numbers except for these three values. Therefore, the domain is all real numbers except \( x = 0 \), \( x = -7 \), and \( x = -2 \).

Vertical Asymptotes

Vertical asymptotes are vertical lines \( x = a \) where a function approaches infinity or negative infinity as \( x \) gets arbitrarily close to \( a \). For rational functions, vertical asymptotes occur at values where the denominator is zero, provided that these zeros do not cancel out with zeros in the numerator.
The numerator of our function, \( x^2 - 1 \), can be factored as \((x-1)(x+1)\), and these factors do not cancel with any factor of the denominator \( x(x+7)(x+2)\). Thus, the values of \( x \) that make the denominator zero give us vertical asymptotes:

• \( x = 0 \)
• \( x = -7 \)
• \( x = -2 \)

These values are where the function has vertical asymptotes. As \( x \) approaches any of these values, the function's value trends toward infinity or negative infinity.

Horizontal Asymptotes

Horizontal asymptotes occur when the output value of a function approaches a fixed number as the input \( x \) grows infinitely large in the positive or negative direction. For rational functions, the degrees of the polynomials in the numerator and the denominator of the function determine the horizontal asymptote.
In our function \( f(x)=\frac{x^{2}-1}{x^{3}+9x^{2}+14x} \), the degree of the numerator (2) is less than the degree of the denominator (3). This implies that as \( x\) approaches positive or negative infinity, the value of \( f(x) \) approaches 0.

• When the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \( y = 0 \).

Therefore, for this function, the horizontal asymptote is \( y = 0 \), indicating that \( f(x) \) levels off to 0 as \( x \) becomes very large or very small.