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Chapter 2: Problem 18

(a) state the domain of the function, (b) identify all intercepts, (c) find any vertical or horizontal asymptotes, and (d) plot additional solution points as needed to sketch the graph of the rational function. $$f(x)=\frac{1}{x-3}$$

Short Answer

Expert verified
The function \(f(x)=\frac{1}{x-3}\) has domain all real numbers except \(x = 3\), no x-intercepts, y-intercept at (0, -1/3), vertical asymptote \(x = 3\) and horizontal asymptote \(y = 0\). The function curve is in quadrant II and IV, which is divided by asymptotes into two parts.

Step by step solution

01

Finding the Domain

The domain of a function is all the real numbers except those that make the denominator equal to zero. Therefore, for this function, if \(x-3 = 0\), then \(x = 3\). So, the domain of the function is: \(x \in \mathbb{R}, x \neq 3\).
02

Finding the Intercepts

The x-intercepts are found when the value of the function is zero, but since the numerator of this function is constant (1), the function will never equals to zero. So, there is no x-intercept. The y-intercept is found by setting \(x = 0\) in the function, this gives \(f(0) = \frac{1}{0 - 3} = -\frac{1}{3}\). This means the y-intercept is at (0, -1/3).
03

Finding Asymptotes

The vertical asymptote is the value of \(x\) where the function becomes undefined. As already established, \(x = 3\) is a vertical asymptote. A horizontal asymptote could be determined by the behavior of the function as \(x\) approaches infinity. Given that the degree of the numerator < the degree of the denominator, the horizontal asymptote is \(y = 0\).
04

Plotting Solution Points

Plot points on either side of the vertical asymptote \(x = 3\). For example, at \(x = 2, f(x) = 1; x = 4, f(x) = -1\). This gives us some additional key points for the accurate sketching of the graph. A point below the x-axis that approaches the vertical asymptote from the left, a point above the x-axis that approaches the vertical asymptote from the right, and the y-intercept.

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