Chapter 2: Problem 48
Find the horizontal and vertical asymptotes for the graphs of the indicated functions. Then sketch their graphs. $$ g(x)=\frac{2 x}{\sqrt{x^{2}+5}} $$
Short Answer
Expert verified
No vertical asymptotes; horizontal asymptote at \( y = 2 \).
Step by step solution
01
Identify the Function Type
The function given is rational with a square root expression in the denominator: \( g(x) = \frac{2x}{\sqrt{x^2 + 5}} \). We need to find both horizontal and vertical asymptotes, which are common for rational functions.
02
Find Vertical Asymptotes
Vertical asymptotes occur where the denominator is zero, provided the numerator isn't also zero at those points. In this function, \( \sqrt{x^2 + 5} = 0 \) when \( x^2 + 5 = 0 \). Solving gives no real solutions since \( x^2 + 5 > 0 \) for all real \( x \). Thus, there are no vertical asymptotes.
03
Find Horizontal Asymptotes
To find horizontal asymptotes, consider the limits as \( x \) approaches infinity and negative infinity. Simplify the function by dividing the numerator and denominator by \( x \):\[ g(x) = \frac{2x/x}{\sqrt{(x^2 + 5)/x^2}} = \frac{2}{\sqrt{1 + 5/x^2}}. \]As \( x \to \infty \) or \( x \to -\infty \), \( 5/x^2 \to 0 \), hence the function approaches \( \frac{2}{\sqrt{1}} = 2 \). So, there is a horizontal asymptote at \( y = 2 \).
04
Sketch the Graph
The graph of \( g(x) = \frac{2x}{\sqrt{x^2 + 5}} \) will approach the horizontal line \( y = 2 \) as \( x \to \infty \) and \( x \to -\infty \). There are no vertical asymptotes. The graph will be a continuous curve that flattens and approaches the line \( y = 2 \) without touching it on both ends.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
horizontal asymptotes
Horizontal asymptotes are lines that a graph approaches as the input, or \( x \), becomes very large or very small. These asymptotes give us an idea of the behavior of a function at the extremes, even if they don't clearly intersect with the function. In the function \( g(x) = \frac{2x}{\sqrt{x^2 + 5}} \), the horizontal asymptote is found by examining the limits as \( x \to \infty \) and \( x \to -\infty \).
By simplifying \( g(x) \), we divide both the numerator and the denominator by \( x \). This gives us \( \frac{2}{\sqrt{1 + \frac{5}{x^2}}} \). As \( x \) approaches infinity, the term \( \frac{5}{x^2} \) approaches zero. This simplification shows that \( g(x) \) approaches \( 2 \).
Thus, the horizontal asymptote is \( y = 2 \). This line will be approached by the graph of the function in both directions but will never be crossed. It effectively serves as a boundary for the behavior of the graph at its far ends.
By simplifying \( g(x) \), we divide both the numerator and the denominator by \( x \). This gives us \( \frac{2}{\sqrt{1 + \frac{5}{x^2}}} \). As \( x \) approaches infinity, the term \( \frac{5}{x^2} \) approaches zero. This simplification shows that \( g(x) \) approaches \( 2 \).
Thus, the horizontal asymptote is \( y = 2 \). This line will be approached by the graph of the function in both directions but will never be crossed. It effectively serves as a boundary for the behavior of the graph at its far ends.
vertical asymptotes
Vertical asymptotes occur where the output of a function grows infinitely large in magnitude as the input approaches a certain point. These are usually found where the denominator of a rational function equals zero, given that the numerator doesn't zero out at the same point. For the function \( g(x) = \frac{2x}{\sqrt{x^2 + 5}} \), there are no vertical asymptotes.
To determine this, we set the denominator \( \sqrt{x^2 + 5} \) equal to zero and solve \( x^2 + 5 = 0 \). This results in \( x^2 = -5 \), which has no real solutions, since a square term cannot be negative with real numbers.
This means that there are no real \( x \) values making the denominator zero. So, for this specific function, there are no vertical asymptotes to consider.
To determine this, we set the denominator \( \sqrt{x^2 + 5} \) equal to zero and solve \( x^2 + 5 = 0 \). This results in \( x^2 = -5 \), which has no real solutions, since a square term cannot be negative with real numbers.
This means that there are no real \( x \) values making the denominator zero. So, for this specific function, there are no vertical asymptotes to consider.
rational functions
A rational function is a quotient of two polynomials, represented as \( f(x) = \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomial functions. In these functions, the behavior near vertical asymptotes is of particular interest, as well as the end behavior, captured by horizontal asymptotes. However, not all rational functions contain vertical asymptotes.For the function \( g(x) = \frac{2x}{\sqrt{x^2 + 5}} \), although it includes a square root in the denominator, it can still be treated as a type of rational function. It demonstrates a smooth behavior without interruptions, as evidenced by the lack of vertical asymptotes.
These types of functions are common in modeling scenarios involving ratios and can appear in several practical applications, including physics and economics. Recognizing their form and behavior helps in predicting and understanding the broader trends they represent.
These types of functions are common in modeling scenarios involving ratios and can appear in several practical applications, including physics and economics. Recognizing their form and behavior helps in predicting and understanding the broader trends they represent.