Problem 17 - College Algebra

Question

For the following exercises, find the domain, vertical asymptotes, and horizontal asymptotes of the functions. $$ f(x)=\frac{x+5}{x^{2}-25} $$

Step-by-Step Solution

Answer

Domain: \((-\infty, -5) \cup (-5, 5) \cup (5, \infty)\); Vertical asymptotes: \(x = 5\) and \(x = -5\); Horizontal asymptote: \(y = 0\).

Step 1: Identify the Denominator

In order to find the domain, we need to identify the denominator of the function. The denominator is given by \(x^2 - 25\).

Step 2: Determine Points Where Denominator is Zero

The function is undefined where the denominator equals zero. Set \(x^2 - 25 = 0\) and solve for \(x\):\[x^2 - 25 = 0\]\[(x - 5)(x + 5) = 0\]\(x = 5\) or \(x = -5\).These points \(x = 5\) and \(x = -5\) are where the function is not defined.

Step 3: Find the Domain

The domain of the function consists of all real numbers except where the function is undefined. Therefore, the domain is all real numbers except \(x = 5\) and \(x = -5\), or \( (-\infty, -5) \cup (-5, 5) \cup (5, \infty) \).

Step 4: Identify Vertical Asymptotes

Vertical asymptotes occur at the values of \(x\) where the denominator is zero, provided that these values don't cancel with the numerator. Here, factors \(x - 5\) and \(x + 5\) do not cancel with the numerator. Hence, vertical asymptotes occur at \(x = 5\) and \(x = -5\).

Step 5: Determine Horizontal Asymptotes

For rational functions of the form \(\frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials, we compare the degrees:- If the degree of the numerator is less than the degree of the denominator, \(y = 0\) is the horizontal asymptote.- Here, the degree of the numerator \((x+5)\) is 1 and the degree of the denominator \(x^2 - 25\) is 2, so the horizontal asymptote is \(y = 0\).

Key Concepts

Domain of a Function

The domain of a function refers to the set of all possible input values (usually represented as \(x\)) for which the function is defined. For rational functions, these are all real numbers except where the denominator is zero. When the denominator equals zero, the function becomes undefined at those points.
To determine the domain of the function \(f(x) = \frac{x+5}{x^2 - 25}\):

• Identify the denominator, which is \(x^2 - 25\).
• Set the denominator to zero: \(x^2 - 25 = 0\).
• Factor the equation: \((x - 5)(x + 5) = 0\), yielding solutions \(x = 5\) and \(x = -5\).

These solutions indicate the values of \(x\) that make the denominator zero, which are not part of the domain. Therefore, the domain of \(f(x)\) is all real numbers except \(x = 5\) and \(x = -5\). In interval notation, this is written as \((-\infty, -5) \cup (-5, 5) \cup (5, \infty)\).

Vertical Asymptotes

Vertical asymptotes are invisible lines on the graph of a function where the function approaches infinity, indicating that the function is undefined at those \(x\)-values. For rational functions, vertical asymptotes occur at values of \(x\) where the denominator is zero, provided that these zeros do not cancel with zeros in the numerator. To find the vertical asymptotes of \(f(x) = \frac{x+5}{x^2 - 25}\):

• Factor the denominator: \((x - 5)(x + 5) = 0\).
• Look for zeros in the numerator \(x + 5\), which does not cancel with the zeros of the denominator. So, the zeros \(x = 5\) and \(x = -5\) do not cancel.
• Vertical asymptotes are then present at \(x = 5\) and \(x = -5\).

In summary, vertical asymptotes are a result of division by zero that doesn't cancel out, causing the graph to spike upward or downward, infinitely close to these points.

Horizontal Asymptotes

Horizontal asymptotes reflect the behavior of a function as \(x\) approaches infinity. They provide a horizontal line that the function approaches as the input values go to positive or negative infinity but never actually reaches.
For rational functions, the location of horizontal asymptotes depends on the degrees of the numerator and denominator polynomials:

• If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \(y = 0\).
• For \(f(x) = \frac{x+5}{x^2 - 25}\), the numerator \(x+5\) has a degree of 1, while the denominator \(x^2 - 25\) has a degree of 2.

Since 1 is less than 2, the horizontal asymptote is at \(y = 0\). This means as \(x\) becomes very large positively or negatively, \(f(x)\) approaches 0.