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A Rational Function with a Slant Asymptote In Exercises \(49-62,\) (a) state the domain of the function, (b) identify all intercepts, (c) find any vertical or slant asymptotes, and (d) plot additional solution points as needed to sketch the graph of the rational function. $$f(x)=\frac{1-x^{2}}{x}$$

Short Answer

Expert verified
The domain of the function \( f(x)=\frac{1-x^{2}}{x} \) is \( x ≠ 0 \), the x-intercepts are at \( x = -1 \) and \( x = 1 \). And there's no y-intercept. The graph of the function has a vertical asymptote at \( x = 0 \) and a slant asymptote at \( y = -x \).

Step by step solution

01

Find the Domain

The domain of a function is the set of all possible input values or x-values which will result in defined output or y-values. For the rational function \( f(x) = \frac{1 - x^{2}}{x} \), the domain would be all real numbers except where the denominator \( x = 0 \). Therefore the domain of the function is except \(x=-∞\) to \(x=0\) and \(x=0\) to \(x=∞\) or \(x ≠ 0\).
02

Identify Intercepts

Intercepts are points where the graph of the function crosses the x-axis or y-axis. To find the x-intercept(s), we first set \( y, or f(x) = 0 \), then solve for \( x \). Similarly, for the y-intercept, we set \( x = 0 \), then solve for \( y \). For our function, setting \( f(x) = 0 \) results in \( x = 1 \) or \( x = -1 \). Thus, the x-intercepts of the function are at \( x = -1 \) and \( x = 1 \). The function is not defined at \( x = 0 \) (which you can see from the domain), so there's no y-intercept.
03

Find Asymptotes

Asymptotes refer to lines that the graph of the function approaches as \( x \) goes toward \( ±∞ \). A vertical asymptote indicates the values of \( x \) that make the function undefined, which in this case is \( x = 0 \). The slant asymptote can be found by performing long division of the numerator by the denominator. As the degree of the numerator is one more than the degree of the denominator, there is a slant asymptote. In this case, the slant asymptote is \( y = -x \).
04

Plot Points and Graph

Since we have obtained the x-intercepts, the vertical asymptote, and the slant asymptote, we can plot these on a graph. Along with this, we can plot some points either side of \( x = 0 \) to give us a rough idea of the shape of the graph. For example, plotting points for \( x = -2, -1, 1, \) and \( 2 \), results in the points \( (-2, -1.5), (-1, 2), (1, -2), \) and \( (2, 1.5) \). This provides enough information for us to sketch the graph of the function.

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Key Concepts

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

Rational Function Domain
Understanding the domain of a rational function is essential when graphing. A rational function has the form \( f(x) = \frac{P(x)}{Q(x)} \) where \( P(x) \) and \( Q(x) \) are polynomials. The domain consists of all the real numbers except for where the denominator, \( Q(x) \) equals zero, since division by zero is undefined.

Looking at our example, \( f(x) = \frac{1 - x^2}{x} \) has a denominator of \( x \). Therefore, the domain excludes \( x = 0 \) because it would make the denominator zero. Hence, the domain is all real numbers except zero, which we denote as \( x eq 0 \), or \( (-\infty, 0) \cup (0, \infty) \).
X-intercepts and Y-intercepts
The intercepts of a function are points where the graph crosses the x-axis and y-axis. For the x-intercepts, we set the output of the function, \( f(x) \) to zero and solve for \( x \). This is because the points on the x-axis have a y-coordinate of zero. Conversely, to find the y-intercept, we set \( x \) to zero and solve for \( f(x) \) since the points on the y-axis have an x-coordinate of zero.

In our function \( f(x) = \frac{1 - x^2}{x} \), setting \( f(x) = 0 \) yields \( x \) values of 1 and -1, leading to two x-intercepts: \( (-1, 0) \) and \( (1, 0) \). As for the y-intercept, since setting \( x = 0 \) is not possible (the function is undefined at \( x = 0 \)), our graph has no y-intercept.
Slant Asymptotes
Slant or oblique asymptotes occur in rational functions when the degree of the numerator is one more than the degree of the denominator. They describe the end behavior of the function and appear as lines that the graph tends towards as \( x \) approaches \( -\infty \) or \( \infty \).

To find a slant asymptote, you can perform polynomial long division or synthetic division of the numerator by the denominator. In our function, \( f(x) = \frac{1 - x^2}{x} \), the numerator has a degree of 2 (since \( x^2 \) is the highest degree term) and the denominator has a degree of 1. Dividing \( 1 - x^2 \) by \( x \) yields a quotient of \( -x \) with a remainder. Thus, the slant asymptote is the line \( y = -x \) since the remainder becomes insignificant as \( x \) grows in magnitude.
Vertical Asymptotes
Vertical asymptotes are lines that the graph of a function approaches but never touches. They indicate where the function heads towards infinity or negative infinity. For rational functions, vertical asymptotes occur at values of \( x \) that cause the denominator to be zero (as long as the numerator isn't also zero at these points, as that would suggest a hole instead).

For our function, the denominator is \( 0 \) when \( x = 0 \). Hence, there is a vertical asymptote at \( x = 0 \) because the function is undefined at this value. When graphing, you would draw a dashed line along \( x = 0 \) to represent this asymptote and show that the function increases or decreases without bound as it approaches this line.

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