Problem 13 - A Graphical Approach to Precalculus with Limits

Question

Give the equations of any vertical, horizontal, or oblique asymptotes for the graph of each rational function. State the domain of \(f .\) $$f(x)=\frac{x^{2}-1}{x+3}$$

Step-by-Step Solution

Answer

Vertical asymptote: \(x = -3\); Oblique asymptote: \(y = x - 3\). Domain: \(x \neq -3\).

Step 1: Identify Vertical Asymptote

To find vertical asymptotes of the function \(f(x) = \frac{x^2 - 1}{x + 3}\), set the denominator equal to zero and solve for \(x\). This ensures the function is undefined at these values. \(x + 3 = 0\) gives \(x = -3\). Thus, there is a vertical asymptote at \(x = -3\).

Step 2: Simplify the Function

The next step is to check if the factor causing the vertical asymptote also cancels with the numerator. Factor the numerator: \(x^2 - 1 = (x - 1)(x + 1)\). Since the factors in the numerator do not cancel with the denominator, the vertical asymptote remains unchanged at \(x = -3\).

Step 3: Identify Horizontal or Oblique Asymptote

To determine if there is a horizontal or oblique asymptote, compare the degrees of the numerator and denominator. The numerator \(x^2 - 1\) has degree 2, and the denominator \(x + 3\) has degree 1. Since the degree of the numerator is one higher than the denominator, there is an oblique asymptote. To find it, perform polynomial long division of \(x^2 - 1\) by \(x + 3\).

Step 4: Perform Polynomial Long Division

Divide \(x^2 - 1\) by \(x + 3\). The first term of the quotient is \(x\), since \(x\times(x + 3) = x^2 + 3x\). Subtract \(x^2 + 3x\) from \(x^2 - 1\) to get \(-3x - 1\). Divide \(-3x\) by \(x\) to get \(-3\). Multiply \(-3\times(x + 3) = -3x - 9\), and subtract this product from \(-3x - 1\) to finish with a remainder of 8. Thus, the division gives \(x - 3\) with a remainder, which means the oblique asymptote is \(y = x - 3\).

Step 5: State the Domain

The domain of the function consists of all real numbers except where the function is undefined. The function is undefined where the denominator is zero, so the domain is \(x eq -3\).

Key Concepts

Vertical Asymptotes

Vertical asymptotes occur in rational functions when the denominator equals zero, making the function undefined. For \(f(x) = \frac{x^2 - 1}{x + 3}\), to find vertical asymptotes, we set the denominator \(x + 3\) to zero. Solving \(x + 3 = 0\) gives us \(x = -3\). Here's what happens:

• Identify the denominator: \(x + 3\)
• Set it to zero: \(x + 3 = 0\)
• Solve for \(x\): \(x = -3\)

This process shows that the vertical asymptote is at \(x = -3\), which is where the function gets classified as undefined. Remember, factors in the numerator that don't cancel out with the denominator confirm that the vertical asymptote persists, as seen in the factorization of \(x^2 - 1 = (x - 1)(x + 1)\). It does not match with \(x + 3\), so the asymptote stays.

Horizontal Asymptotes

Horizontal asymptotes provide insights into a function's behavior as it approaches infinity or negative infinity. They appear when the degree of the numerator is less than or equal to the degree of the denominator in rational functions. For \(f(x) = \frac{x^2 - 1}{x + 3}\), the degree of the numerator is 2 and the degree of the denominator is 1. Since the numerator's degree is greater, there isn’t a horizontal asymptote to consider. In general, horizontal asymptotes follow these rules:

• If the degrees are equal, the asymptote is \(y = \frac{a}{b}\), where \(a\) and \(b\) are the leading coefficients.
• If the degree of the numerator is less, the asymptote is \(y = 0\).
• If the numerator's degree is greater, as in this function, we consider oblique asymptotes instead.

So, checking degree can quickly tell us whether horizontal asymptotes exist, and help in understanding the function's behavior at extremes.

Oblique Asymptotes

Oblique, or slant, asymptotes appear in rational functions when the degree of the numerator is exactly one higher than the degree of the denominator. For \(f(x) = \frac{x^2 - 1}{x + 3}\), the numerator's degree is 2, and the denominator's is 1. This signals the presence of an oblique asymptote. To find it, we perform polynomial long division:

• Divide the leading term of the numerator by the leading term of the denominator: \(x^2 \div x = x\).
• Multiply back and subtract: \(x \times (x + 3) = x^2 + 3x\), leaving \(-3x - 1\).
• Repeat the process: divide \(-3x\) by \(x\) to get \(-3\), subtract, leaving a remainder.

Once completed, the quotient \(x - 3\) represents the oblique asymptote, \(y = x - 3\). This line describes how the function behaves near infinity and negative infinity.

Polynomial Long Division

Polynomial long division is a method used to determine more complex aspects of rational functions, like oblique asymptotes. It's similar to numerical long division, yet here, we deal with polynomials.Working through \(f(x) = \frac{x^2 - 1}{x + 3}\):

• Divide \(x^2\) by \(x\) to get \(x\).
• Multiply \(x\) by \(x + 3\), which equals \(x^2 + 3x\).
• Subtract from original, results in \(-3x - 1\).
• Divide \(-3x\) by \(x\) to get \(-3\).
• Multiply \(-3\) with \(x + 3\) giving \(-3x - 9\).
• Subtract to get a remainder of \(8\).

After completing division, the main interest is the quotient \(x - 3\), indicating an oblique asymptote \(y = x - 3\). This practice helps reveal the asymptotic behavior beyond the domain restriction.