Problem 488 - Precalculus

Question

For the following exercises, find the slant asymptote of the functions. $$f(x)=\frac{x^{2}+5 x+4}{x-1}$$

Step-by-Step Solution

Answer

The slant asymptote is \(y = x + 6\).

Step 1: Check for Slant Asymptote Condition

A function has a slant asymptote if the degree of the numerator is exactly one more than the degree of the denominator. Here, the numerator is \(x^2 + 5x + 4\), which is degree 2, and the denominator is \(x - 1\), which is degree 1. The condition for a slant asymptote is satisfied.

Step 2: Perform Polynomial Long Division

Perform polynomial long division of the numerator \(x^2 + 5x + 4\) by the denominator \(x - 1\). Divide the first term of the numerator by the first term of the denominator: \(\frac{x^2}{x} = x\). Multiply \(x\) by \(x - 1\) to get \(x^2 - x\). Subtract \(x^2 - x\) from \(x^2 + 5x + 4\), resulting in \(6x + 4\). Divide \(6x\) by \(x\): \(\frac{6x}{x} = 6\). Multiply \(6\) by \(x - 1\) to get \(6x - 6\). Subtract \(6x - 6\) from \(6x + 4\) to get a remainder of 10.

Step 3: Determine the Slant Asymptote Equation

The slant asymptote is derived from the quotient obtained in the polynomial division. The quotient is \(x + 6\), so the slant asymptote is \(y = x + 6\). The remainder does not affect the slant asymptote.

Key Concepts

Polynomial Long Division

Polynomial long division is analogous to the regular long division technique used with numbers, but it's applied to polynomials. This method is useful for dividing a polynomial by another polynomial of lower degree.
In our example, we have a numerator of degree 2, which is \(x^2 + 5x + 4\), and we are dividing it by a denominator of degree 1, which is \(x - 1\). The goal is to simplify this into a form that gives us insights into the function's behavior, particularly any asymptotes.

• First, divide the leading term of the numerator by the leading term of the denominator: \(\frac{x^2}{x} = x\).
• Multiply this result by the entire divisor: \(x \cdot (x-1) = x^2 - x\).
• Subtract this from the original polynomial: \((x^2 + 5x + 4) - (x^2 - x) = 6x + 4\).
• Repeat the process with the new polynomial: \(\frac{6x}{x} = 6\).
• Multiply and subtract again: \((6x + 4) - (6x - 6) = 10\).

This leaves a quotient of \(x + 6\) and a remainder of 10. This quotient helps us learn more about the function's behavior at infinity.

Rational Functions

Rational functions are expressions of the form \(\frac{P(x)}{Q(x)}\) where both \(P(x)\) and \(Q(x)\) are polynomials. These functions are significant because they often have interesting and informative asymptotic behavior.
This behavior can tell us how the function behaves at extreme values of \(x\) (both very large positive and negative), which is critical for understanding functions in calculus and algebra. In our exercise, the function \(f(x)=\frac{x^{2}+5 x+4}{x-1}\) is an example of a rational function.

• The degree of the numerator \(P(x) = x^2 + 5x + 4\) is 2.
• The degree of the denominator \(Q(x) = x - 1\) is 1.

By analyzing these degrees, one can determine the presence and type of any asymptotes.

Asymptote Conditions

Asymptotes are lines that a graph approaches but never quite reaches. For rational functions, asymptotes can be vertical, horizontal, or slant. The conditions under which these asymptotes appear depend on the degrees of the numerator and denominator.
Slant (or oblique) asymptotes occur when the degree of the numerator is exactly one greater than the degree of the denominator. This is the case in our example function \(f(x) = \frac{x^2 + 5x + 4}{x - 1}\), where the degrees are 2 and 1, respectively.
To find the slant asymptote, you need to perform polynomial long division. The resulting quotient gives us the asymptote's equation, without considering the remainder.

• In our example, the quotient from polynomial long division is \(x + 6\), which means the slant asymptote is \(y = x + 6\).
• Vertical asymptotes, in contrast, occur when \(Q(x) = 0\), which in our case is \(x = 1\).

The remainder in the division, while interesting for finding the exact remainder form, does not affect the location of the slant asymptote.