Chapter 3: Problem 46
Simplify \(f(x),\) and sketch the graph of \(f\). $$f(x)=\frac{x+2}{x^{2}-4}$$
Short Answer
Expert verified
Simplified: \( g(x) = \frac{1}{x-2} \); Hole at \( x = -2 \), asymptote at \( x = 2 \).
Step by step solution
01
Factor the Denominator
The function given is \( f(x) = \frac{x+2}{x^2-4} \). The denominator \( x^2 - 4 \) is a difference of squares. This can be factored into \((x - 2)(x + 2)\).
02
Cancel Common Factors
In the function \( f(x) = \frac{x+2}{(x-2)(x+2)} \), notice that \( x + 2 \) appears in both the numerator and denominator. We can cancel the common factor \( x + 2 \), simplifying the function to \( g(x) = \frac{1}{x-2} \), except when \( x = -2 \).
03
Find Domain and Discontinuities
The original function \( f(x) \) is undefined at \( x = -2 \) and \( x = 2 \) because the denominator becomes zero. After simplification, \( g(x) = \frac{1}{x-2} \) still has these points as discontinuities, but \( x = -2 \) is a removable discontinuity and \( x = 2 \) is a vertical asymptote.
04
Determine Key Features
Since \( g(x) = \frac{1}{x-2} \), \( x = 2 \) is a vertical asymptote and the horizontal asymptote is \( y = 0 \) as \( x \to \infty \) or \( x \to -\infty \). The removable discontinuity at \( x = -2 \) occurs since there was cancellation.
05
Sketch the Graph
To sketch the graph of \( f(x) \), plot the vertical asymptote at \( x = 2 \) and the horizontal asymptote at \( y = 0 \). Note the hole (removable discontinuity) at \( x = -2 \). The general shape is a hyperbola, approaching these asymptotes without crossing them, and consider the hole in the graph at \( x = -2 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Domain and Discontinuities
Understanding the domain of a rational function is crucial because it determines where the function is defined. For the function \( f(x) = \frac{x+2}{x^2-4} \), the domain consists of all real numbers except where the denominator is zero.
To find this, factor the denominator: \[ x^2 - 4 = (x-2)(x+2) \].
This tells us that \( x = 2 \) and \( x = -2 \) are points where the denominator becomes zero, hence these are not in the domain.
Recall, removable discontinuities occur when a factor is cancelled from numerator and denominator. Therefore, canceling \( x+2 \) results in a hole at \( x = -2 \) while \( x = 2 \) becomes a vertical asymptote.
To find this, factor the denominator: \[ x^2 - 4 = (x-2)(x+2) \].
This tells us that \( x = 2 \) and \( x = -2 \) are points where the denominator becomes zero, hence these are not in the domain.
- Domain: all real numbers except \( x = 2 \) and \( x = -2 \).
- Discontinuities: these points are discontinuities in the graph; \( x = 2 \) is a vertical asymptote, and \( x = -2 \) is a removable discontinuity.
Recall, removable discontinuities occur when a factor is cancelled from numerator and denominator. Therefore, canceling \( x+2 \) results in a hole at \( x = -2 \) while \( x = 2 \) becomes a vertical asymptote.
Vertical and Horizontal Asymptotes
Asymptotes provide a guide to the behavior of rational functions. For the simplified function \( g(x) = \frac{1}{x-2} \), analyzing these asymptotes allows us to predict the function's behavior at extreme values.
The vertical asymptote occurs where the denominator equals zero after cancellation of common factors – here, at \( x = 2 \). This is because the function can approach infinity as \( x \) gets closer to this value, but will never actually reach it.
As for the horizontal asymptote, we assess the behavior as \( x \) approaches infinity. Since \( g(x) = \frac{1}{x-2} \) has a constant numerator and the denominator infinitely increases, the value approaches zero.
Understanding these asymptotes provides a structural frame of reference that will assist in accurately sketching the graph.
The vertical asymptote occurs where the denominator equals zero after cancellation of common factors – here, at \( x = 2 \). This is because the function can approach infinity as \( x \) gets closer to this value, but will never actually reach it.
As for the horizontal asymptote, we assess the behavior as \( x \) approaches infinity. Since \( g(x) = \frac{1}{x-2} \) has a constant numerator and the denominator infinitely increases, the value approaches zero.
- Vertical Asymptote: \( x = 2 \)
- Horizontal Asymptote: \( y = 0 \)
Understanding these asymptotes provides a structural frame of reference that will assist in accurately sketching the graph.
Graph Sketching
Sketching the graph of a rational function involves plotting key features such as asymptotes and discontinuities. For \( g(x) = \frac{1}{x-2} \), begin by marking the vertical asymptote at \( x = 2 \). This is a line the graph approaches but never touches.
Next, the horizontal asymptote at \( y = 0 \) must be sketched, showing the function's value trend as \( x \) increases or decreases towards infinity.
Don’t forget the removable discontinuity at \( x = -2 \) from the original function \( f(x) = \frac{x+2}{x^2-4} \). This is represented in the graph by a hole, or an open circle at \( x = -2 \) on the line.
The curve of \( g(x) \) typically takes the form of a hyperbola, residing in two separate regions split by the vertical asymptote at \( x = 2 \). As it approaches each asymptote, the curve nears but never crosses it.
Through these steps, a complete and clear representation of the function’s graph is produced, aiding visualization and understanding.
Next, the horizontal asymptote at \( y = 0 \) must be sketched, showing the function's value trend as \( x \) increases or decreases towards infinity.
Don’t forget the removable discontinuity at \( x = -2 \) from the original function \( f(x) = \frac{x+2}{x^2-4} \). This is represented in the graph by a hole, or an open circle at \( x = -2 \) on the line.
The curve of \( g(x) \) typically takes the form of a hyperbola, residing in two separate regions split by the vertical asymptote at \( x = 2 \). As it approaches each asymptote, the curve nears but never crosses it.
- Vertical Asymptote: guide for curve placement.
- Horizontal Asymptote: indicative of end behavior.
- Removable Discontinuity: mark with an open circle.
Through these steps, a complete and clear representation of the function’s graph is produced, aiding visualization and understanding.