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Chapter 8: Problem 106

Sketch the graph of the function. Identify any asymptotes. $$f(x)=\frac{4 x}{5 x^{2}+2}$$

Short Answer

Expert verified
The function \(f(x) = \frac{4x}{5x^2+2}\) has a horizontal asymptote at y = 0. It doesn't have a vertical asymptote. The graph can be plotted using key points and following the features of rational function graphs.

Step by step solution

01

Determine the Asymptotes

For rational functions, horizontal or oblique asymptotes exist depending on whether the degree of the polynomial in the numerator is less than, equal to, or greater than the degree of the polynomial in the denominator. Here, the degree in the numerator (1) is less than that in the denominator (2), hence the x-axis (y=0) is the horizontal asymptote. To find the vertical asymptotes, set the denominator equal to zero and solve for 'x'. But in this case, the equation \(5x^2 + 2 = 0\) has no real solutions, hence there is no vertical asymptote.
02

Sketch the Graph

Plot some key points of the function and connect these points smoothly bearing in mind that the graph approaches the asymptotes as x tends towards positive or negative infinity. For the given function, there seem to be no x or y-intercepts, as setting the function equal to zero gives non-real solutions. Choose some convenient values for x and calculate the corresponding function value to gain some points on the graph. Then, use these points to draw the graph, with the curve approaching the horizontal asymptote of y=0 as x tends towards positive or negative infinity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Asymptotes
In the world of graphing rational functions, asymptotes are your guiding lines. But what are they exactly? Asymptotes are lines that the graph of a function gets infinitely close to but never actually touches. They provide an essential visual guide to understanding how a function behaves as it approaches infinity. There are different types of asymptotes: vertical, horizontal, and oblique. Each has distinct properties, depending on the nature of the rational function you're working with. For this particular case, we need to explore both horizontal and vertical asymptotes to fully understand the behavior of the function.
Horizontal Asymptote
A horizontal asymptote represents a y-value that the graph approaches as the x-values become very large (positive or negative). Determining horizontal asymptotes involves comparing the degrees of the numerator and the denominator polynomials.
  • If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0.
  • If the degrees of the numerator and denominator are equal, the horizontal asymptote is the quotient of the leading coefficients.
  • If the degree of the numerator is greater, there's no horizontal asymptote, but potentially, an oblique asymptote instead.

In the function given, \(f(x)=\frac{4x}{5x^2+2}\), the degree of the numerator (1) is less than that of the denominator (2). Therefore, the horizontal asymptote is the x-axis or \(y=0\). This implies that as \(x\) approaches infinity or negative infinity, the function \(f(x)\) will get increasingly close to zero.
Degree of Polynomial in Rational Functions
Rational functions, like our example, have polynomials in both the numerator and the denominator. The degree of a polynomial is the highest power of \(x\) present in the polynomial. This is what helps us determine key features like asymptotes. Here is how these degrees impact the function:
  • The numerator degree impacts the position and type of horizontal asymptote.
  • The denominator degree significantly affects the potential for vertical asymptotes.

In \(f(x)=\frac{4x}{5x^2+2}\), the numerator has a degree of 1, and the denominator has a degree of 2. This guides us to identify the horizontal asymptote at \(y=0\) and indicates there shouldn't be vertical asymptotes, as the denominator equation \(5x^2 + 2 = 0\) has no real solutions.
Vertical Asymptote
Vertical asymptotes occur where the function is undefined due to division by zero. To find potential vertical asymptotes, set the denominator equal to zero and solve for \(x\). If there are real solutions, those \(x\) values indicate the presence of vertical asymptotes. However, in our function, the denominator \(5x^2 + 2\) can't be zero for any real numbers.
  • The equation \(5x^2 + 2 = 0\) simplifies to \(x^2 = -\frac{2}{5}\).
  • This has no real solutions since a squared number can't be negative.

Therefore, this function has no vertical asymptotes, meaning there's no point at which the graph will shoot up to infinity or down to negative infinity within the real number system.

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