Problem 93 - Algebra and Trigonometry

Question

The equation for \(f\) is given by the simplified expression that results after performing the indicated operation. Write the equation for \(f\) and then graph the function. $$ \frac{1-\frac{3}{x+2}}{1+\frac{1}{x-2}} $$

Step-by-Step Solution

Answer

The equation for \(f\) is given by \[f(x) = \frac{x^2 - 3x + 2}{x^2 - 1}\]. The graph of this function shows the points of discontinuity in the line at \(x = -1\) and \(x = 1\). As \(x\) approaches to \(\pm \infty\), \(f(x)\) also approaches to 1.

Step 1: Simplify the fraction

This is a fractional operation. We need to simplify it first. Multiply the numerator and denominator by \((x+2)(x-2)\) to get rid of the fraction:\[f(x) = \frac{(x+2)(x-2)-3(x-2)}{(x+2)(x-2)+1(x+2)}\]

Step 2: Expand and simplify further

Now, expand the expression and simplify it to get\[f(x) = \frac{x^2 -4 - 3x + 6}{x^2 -1}\] Simplify more by combining terms:\[f(x) = \frac{x^2 - 3x + 2}{x^2 - 1}\]

Step 3: Graph the function

For graphing the function, calculate the function values for a range of \(x\) values and plot the corresponding points on an \(xy\) graph. Check the points where \(x = -1\) and \(x = 1\) as these are the points of discontinuity. As \(x \to \pm \infty\), the function \(f(x) \to 1\).

Key Concepts

Rational Function Graphing

Graphing a rational function involves understanding its behavior based on algebraic expressions. Rational functions can be recognized by the presence of a polynomial divided by another polynomial, just as in the exercise provided. When graphing such functions, it is important to identify key features like intercepts, asymptotes, and any discontinuities.

To graph the function, first evaluate it at various points, but pay special attention to values that could result in zero in the denominator, which are excluded from the domain. In our exercise example, plotting points around the x-values of -2 and 2 will reveal important behavior of the graph as these are close to the vertical asymptotes. Also, looking at the end behavior as x approaches positive or negative infinity gives insight into horizontal asymptotes.

Complex Fraction

Complex fractions have a fraction in the numerator, the denominator, or both. The goal when dealing with them is to simplify the expression into a single fraction that's easier to work with and understand. The step-by-step solution initially multiplies the numerator and the denominator by the least common denominator \( (x+2)(x-2) \) to eliminate the smaller fractions within.

Simplifying complex fractions is crucial for further operations such as graphing or finding discontinuities. This step can be considered analogous to finding a common ground before comparing or combining different parts—once on the same 'denominator', it becomes straightforward to integrate them into a unified expression.

Algebraic Expressions

An algebraic expression is a mathematical phrase containing numbers, variables, and operation symbols. In our given exercise, the algebraic expression represents a rational function, which is a type of algebraic expression where one polynomial is divided by another. After simplifying the complex fraction, we get an algebraic expression which can be further simplified by combining like terms or by factoring.

Algebraic expressions are the backbone of understanding and graphing functions. They must be manipulated carefully to reveal critical features such as the x-intercepts (the roots of the numerator) and y-intercept (the function value when \( x=0 \)), which are essential in the graphing step of the process.

Function Discontinuity

Discontinuity in a function occurs where the function is not continuous; essentially, there's a break in the graph. For rational functions, discontinuities commonly arise at the x-values that make the denominator equal to zero, since division by zero is undefined.

In the context of the provided exercise, the function displays discontinuity at \( x=-2 \) and \( x=2 \), which we exclude from the domain of the function. When graphing, these points will correspond with vertical asymptotes, assuming the numerator doesn't cancel out the zero in the denominator, which would then represent a hole in the graph instead of an asymptote. Identifying discontinuities is a crucial step for correct graphing and understanding the behavior of rational functions.