Problem 2 - Algebra 2

Question

Determine the equations of any vertical asymptotes and the values of \(x\) for any holes in the graph of each rational function. $$ f(x)=\frac{x-1}{x^{2}+4 x-5} $$

Step-by-Step Solution

Answer

Vertical asymptote at \(x=-5\); hole at \(x=1\).

Step 1: Factor the Denominator

Begin by factoring the denominator of the rational function. The denominator is \(x^2 + 4x - 5\). We want to factor it into the form of \((x+a)(x+b)\). Find two numbers that multiply to \(-5\) and add to \(4\). These numbers are \(5\) and \(-1\), so the factored form is \((x+5)(x-1)\).

Step 2: Identify Potential Asymptotes and Holes

Set each factor of the denominator equal to zero to find potential vertical asymptotes and holes. The factors are \((x+5)\) and \((x-1)\). Setting each to zero gives us \(x+5=0\) (\(x=-5\)) and \(x-1=0\) (\(x=1\)).

Step 3: Determine Holes

Check if the numerator \(x-1\) shares any common factors with the denominator. Since \((x-1)\) is a factor of the denominator and also the numerator, this indicates a hole at \(x=1\).

Step 4: Determine Vertical Asymptotes

Identify values from Step 2 that are not canceled by factors in the numerator. Since \(x=-5\) is not canceled as it does not appear in the numerator, there is a vertical asymptote at \(x=-5\).

Key Concepts

Vertical Asymptotes

Vertical asymptotes occur when the denominator of a rational function equals zero, resulting in the function going to infinity at those points. Essentially, they represent values of \(x\) where the function cannot exist.
To find vertical asymptotes, factor the denominator of the rational function, and set each factor equal to zero. For the function given:

• Denominator: \(x^2 + 4x - 5\)
• Factored form: \((x+5)(x-1)\)

Set each factor to zero to solve for \(x\):

• \(x+5=0\): \(x=-5\)
• \(x-1=0\): \(x=1\)

While \(x=1\) is a hole due to a common factor with the numerator, \(x=-5\) remains a vertical asymptote because it doesn't appear in the numerator.

Holes in Graphs

Holes in the graph of a rational function occur when a factor in the denominator is shared with the numerator. These shared factors cancel out, creating a gap or "hole" in the graph. This happens because both the numerator and denominator become zero at the same \(x\) value, rendering the function undefined.
In this problem, the numerator is \(x-1\), and the denominator factors to \((x+5)(x-1)\).
The common factor is \(x-1\), as it appears in both the numerator and denominator. Thus, there is a hole at \(x=1\). Remember that while the function is not defined at this point, holes are not part of vertical asymptotes because they get "canceled out" from the function.

Denominator Factoring

Factoring the denominator in a rational function is crucial to finding both vertical asymptotes and holes. Factoring simplifies the equation and reveals critical points where the denominator becomes zero.
For example, with the denominator \(x^2 + 4x - 5\), the goal was to factor it into \((x+a)(x+b)\).
By identifying numbers that multiply to \(-5\) and add to \(4\), we find \(5\) and \(-1\). Thus, the correct factorization is \((x+5)(x-1)\).

• This step is essential before proceeding to find asymptotes and holes.
• Correct factoring will correctly identify these critical points.

Common Factors

Common factors between the numerator and the denominator of a rational function indicate the existence of holes. These factors cancel each other out and need to be identified to understand the finer details of the function’s graph.
In the given function, both the numerator \((x-1)\) and the denominator \((x+5)(x-1)\) share the factor \(x-1\). When these common factors cancel:

• A hole is indicated at \(x=1\), where the shared factor equals zero.

Understanding common factors is key in differentiating between vertical asymptotes, which affect the graph as it approaches infinity, and holes, which create a point where the function is undefined but does not reach infinity.