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Rational functions are defined as the ratio of two polynomials. Just like a fraction, where one number is divided by another, a rational function divides one algebraic expression by another. Here, the numerator is a polynomial, and the denominator is also a polynomial, which cannot be zero. For example, in the function \(g(x) = \frac{x+1}{2x^2 - x - 3}\), \(x + 1\) is the numerator and \(2x^2 - x - 3\) is the denominator.

A key point about rational functions is determining their domain. The domain consists of all real numbers except those which make the denominator zero, as division by zero is undefined. Hence, to find the domain of a rational function, we set the denominator not equal to zero and solve to find the restricted values of \(x\). Excluding these values from the set of all real numbers gives us the domain of the rational function.

Quadratic equations are polynomial equations of degree 2, generally expressed in the standard form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). The equation \(2x^2 - x - 3 = 0\) showcased in the exercise is a typical quadratic equation.

Quadratic equations can be solved using various methods, such as:

• Factoring: Expressing the quadratic as a product of two binomials.
• Completing the square: Rearranging the equation to form a perfect square trinomial.
• Quadratic formula: Using the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the solutions.
• Graphing: Finding the points where the parabola intersects the x-axis.

In rational functions, finding the zeroes of the denominator involves solving the quadratic equation to determine any restrictions on the domain.

Factoring quadratics is a process of breaking down a quadratic equation into products of linear factors. This is a useful technique to find the zeroes of a quadratic polynomial and is often the first step in solving quadratic equations. Let's explore how this works using \(2x^2 - x - 3 = 0\) from the exercise.

The polynomial \(2x^2 - x - 3\) is factored as \((2x + 1)(x - 3) = 0\). This means that the original quadratic can be expressed as the product of the two binomials \(2x + 1\) and \(x - 3\). To factor a quadratic:

• Identify two numbers that multiply to give the product of \(a\) and \(c\) (the first and last coefficients), and sum to \(b\) (the middle coefficient).
• Use these numbers to decompose the middle term, rewriting the quadratic and then factor by grouping.
• Alternatively, recognize patterns or use trial and error to quickly factor it if it is straightforward like in the exercise.

After factoring, set each factor equal to zero to solve, determining the roots of the equation, which help in identifying restrictions for rational functions.