Problem 49 - Algebra 2

Question

Describe the vertical asymptotes and holes for the graph of each rational function. $$ y=\frac{x-1}{(3 x+2)(x+1)} $$

Step-by-Step Solution

Answer

The vertical asymptotes of the function \(y=\frac{x-1}{(3x+2)(x+1)}\) are at \(x = -\frac{2}{3}\) and \(x = -1\). There are no holes in the graph of this function.

Step 1: Identify the Denominator Factors

First, look at the denominator of the given function. It has two factors: \(3x + 2\) and \(x + 1\).

Step 2: Determine the x-values for Vertical Asymptotes

Next, set each factor in the denominator equal to zero to find the x-values that make the function undefined, which will be the locations of the vertical asymptotes. \n Solve \(3x + 2 = 0\) and \(x + 1 = 0\). Get \(x = -\frac{2}{3}\) and \(x = -1\) respectively.

Step 3: Identify Holes

In order for the graph to have a hole, a factor must cancel out from both the numerator and the denominator. There are no common factors in the numerator and denominator, so there are no holes.

Key Concepts

Vertical Asymptotes

In a graph of a rational function, vertical asymptotes are lines where the graph shoots off to infinity as the values of the function become undefined. Essentially, these are straight lines that the graph never touches or crosses.

To find vertical asymptotes, look at the denominator of the rational expression. When you set each factor in the denominator to zero, the solutions are where the vertical asymptotes occur. This is because a zero in the denominator causes the function to be undefined at those points.

• For the function \( y=\frac{x-1}{(3x+2)(x+1)} \), set each factor to zero: \(3x + 2\) and \(x + 1\).
• Solve for each: \(3x + 2 = 0\) gives \(x = -\frac{2}{3}\) and \(x + 1 = 0\) gives \(x = -1\).

These solutions, \(x = -\frac{2}{3}\) and \(x = -1\), are the vertical asymptotes of the graph.

Holes in Graphs

A hole in the graph occurs where a factor in the numerator and denominator cancels out. At these points, the function is not defined, resulting in a gap, or hole, in the graph.

For a hole to exist:

• There must be a common factor in both the numerator and the denominator of the rational expression.
• Once this factor is factored out or canceled, the graph will have a hole at the x-value that makes this factor zero.

In our given function, \(y=\frac{x-1}{(3x+2)(x+1)}\), the numerator \(x-1\) and the factors of the denominator \((3x+2)(x+1)\) share no common factors. Thus, there are no cancellations, implying this function has no holes.

Denominator Factors

Understanding the factors of a denominator in a rational function is crucial to identifying possible vertical asymptotes and holes.

The denominator shows where the function has restrictions. These restrictions occur because division by zero is undefined. Thus, solving each factor in the denominator set equal to zero identifies critical points for the function's graph.

• The given function's denominator is \((3x+2)(x+1)\).
• By factoring out each part, we solve \(3x + 2 = 0\) and \(x + 1 = 0\), leading to potential asymptotes or holes.

Reducing Rational Expressions

Reducing rational expressions involves simplifying them to the lowest terms by canceling out common factors.

This process makes calculations easier and reveals the graph's characteristics, such as holes, because these occur where factors in the numerator and denominator cancel each other out, creating undefined points.

To successfully reduce a rational expression:

• Factor both the numerator and the denominator completely.
• Identify and cancel any common factors.
• Express the simplified form, if possible.

In our case with the function \(y=\frac{x-1}{(3x+2)(x+1)}\), there are no common factors to cancel, so the function is already in its simplest form.