Problem 28 - Elementary Algebra

Question

For the following problems, find the domain of each of the rational expressions. $$ \frac{y-6}{2 y^{2}-3 y-2} $$

Step-by-Step Solution

Answer

Question: Determine the domain of the given rational expression: $$ \frac{y-6}{2y^2-3y-2} $$ Answer: The domain of the given rational expression is \(y \in \mathbb{R}, y \neq 2, y \neq -1\).

Step 1: Factorise the denominator

In order to factorise the quadratic equation in the denominator, we look for two numbers which multiply to give the product of the coefficient of the squared term (2) and the constant term (-2), and also add up to give the coefficient of the linear term (-3). In this case, these numbers are -4 and 1. So, the factorisation is: $$ 2y^2 - 3y - 2 = (2y - 4)(y + 1) $$ Now, the expression becomes: $$ \frac{y-6}{(2y-4)(y+1)} $$

Step 2: Find the roots of the denominator

To find the roots of the denominator, we need to find the values of \(y\) that make it equal to 0. For the first factor, \((2y-4)\): $$ 2y - 4 = 0 \\ 2y = 4 \\ y = 2 $$ For the second factor, \((y+1)\): $$ y + 1 = 0 \\ y = -1 $$ The roots of the denominator are \(y = 2\) and \(y = -1\). As these two values make the denominator equal to 0, they must be excluded from the domain.

Step 3: Write the domain

Since the only values of \(y\) that are not allowed are 2 and -1, the domain of the given expression can be written as the set of all real numbers, excluding 2 and -1. The domain of the rational expression is: $$ y \in \mathbb{R}, y \neq 2, y \neq -1 $$

Key Concepts

Factoring Quadratics

When dealing with quadratics, factoring is a common method to simplify expressions. A quadratic expression typically takes the form: \[ ax^2 + bx + c \]The goal of factoring is to rewrite this as a product of two binomials.
To begin factoring, find two numbers that multiply to give the product of the first and last coefficients (\(a\) and \(c\)) and add to the middle coefficient (\(b\)).
This approach can help simplify the expression and solve for specific variable values.

• Place the quadratic expression in standard form.
• Identify coefficients \(a\), \(b\), and \(c\).
• Determine two numbers that multiply to \(ac\) and add to \(b\).
• Rewrite the middle term using these numbers and factor by grouping.

In our example, the expression \(2y^2 - 3y - 2\) is factored into \((2y - 4)(y + 1)\).

Rational Expressions

Rational expressions are simply fractions where the numerator and/or the denominator are polynomials. These expressions can often be simplified by factoring. Simplifying helps identify undefined values, which appear when the denominator is zero.
To handle rational expressions:

• Simplify the numerator and the denominator separately by factoring.
• Cancel out any common factors found in both the numerator and the denominator.

In rational expressions, excluding certain values from the domain is crucial. These are the values that make the denominator zero. For the expression \( \frac{y-6}{(2y-4)(y+1)} \), the important step is to identify these potential issues to understand the domain correctly.

Set Notation

To express the domain of a function efficiently, especially when restrictions are involved, set notation is highly useful. Set notation offers a concise way to list out constraints and specifies which elements are part of a set.Here's how to apply set notation to our problem:

• The domain of a rational function is the set of all real numbers, except those that make the denominator zero.
• List these exceptions using "|" to denote conditions, and "\in" to signify membership.
• Example: For the function \( \frac{y-6}{(2y-4)(y+1)} \), use: \( \{y \in \mathbb{R} \mid y eq 2, y eq -1\} \).

This notation makes it easy to see the domain visually and understand what numbers are excluded.

Excluded Values in Domain

In rational expressions, determining the domain comes down to recognizing values to be excluded. These are values where the denominator equals zero—commonly referred to as excluded values.Why exclude them?

• When the denominator is zero, the expression becomes undefined. Division by zero is not possible.
• Identify these values by setting the denominator equal to zero and solving for the variable.

For the given expression \( \frac{y-6}{(2y-4)(y+1)} \), we set each factor of the denominator equal to zero: - \((2y-4) = 0\) gives \(y=2\).- \((y+1)=0\) gives \(y=-1\).These two values, \(y = 2\) and \(y = -1\), are excluded from the domain, simplifying the original expression without including undefined points.