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Chapter 3: Problem 29

Match the rational function in Column I with the appropriate description in Column II. Choices in Column II can be used only once. A. The \(x\) -intercept is \(-3\) B. The \(y\) -intercept is 5 C. The horizontal asymptote is \(y=4\) D. The vertical asymptote is \(x=-1\) E. There is a "hole" in its graph at \(x=-4\) F. The graph has an oblique asymptote. G. The \(x\) -axis is its horizontal asymptote, and the \(y\) -axis is not its vertical asymptote. H. The \(x\) -axis is its horizontal asymptote, and the \(y\) -axis is its vertical asymptote. $$f(x)=\frac{x+7}{x+1}$$

Short Answer

Expert verified
The vertical asymptote is at \(x = -1\). This matches option D.

Step by step solution

01

Determine the x-intercept

Set the numerator equal to zero and solve for x: \[(x + 7) = 0 \rightarrow x = -7\]So, the x-intercept is -7, which does not match any options.
02

Determine the vertical asymptote

Set the denominator equal to zero and solve for x: \[(x + 1) = 0 \rightarrow x = -1\]Thus, the vertical asymptote is at \(x = -1\). This matches option D.
03

Determine the horizontal asymptote

Compare the degrees of the numerator and the denominator: Since both the numerator and the denominator are linear (degree 1), divide the leading coefficients: \[\frac{1}{1} = 1 \rightarrow y = 1\]Thus, the horizontal asymptote is \(y = 1\). This does not match any given options.
04

Check for holes in the graph

A hole in the graph occurs where a factor is canceled out in both the numerator and the denominator, which is not the case here. Therefore, there is no hole.

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Key Concepts

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

Vertical Asymptote
To find the vertical asymptote of a rational function, we look at the values of the variable that make the denominator zero. For the function given in the exercise: \(f(x) = \frac{x+7}{x+1}\), we set the denominator equal to zero and solve for \(x\): \[ x + 1 = 0 \rightarrow x = -1 \]Therefore, there is a vertical asymptote at \(x = -1\). This means as \(x\) gets very close to \(-1\) from either side, the function will increase or decrease without bound. Remember, vertical asymptotes are lines that the graph approaches but never touches.

Vertical asymptotes often give insight into the behavior of the function near certain points and can indicate where the function is undefined.
Horizontal Asymptote
Horizontal asymptotes describe the behavior of a function as \(x\) approaches positive or negative infinity. To find the horizontal asymptote for our function, \(f(x) = \frac{x+7}{x+1}\), we compare the degrees of the numerator and the denominator. Both the numerator and the denominator are polynomials of degree 1.

To find the horizontal asymptote for cases where the degrees are the same, we divide the leading coefficients: \[ \frac{1}{1} = 1 \rightarrow y = 1 \] Therefore, the horizontal asymptote of \(f(x)\) is \(y = 1\). This shows that as \(x\) goes towards infinity, the function approaches the line \(y=1\).
X-Intercept
X-intercepts are points where the graph of the function crosses the x-axis. These occur where the function value is zero. To find the x-intercepts for the function \(f(x) = \frac{x+7}{x+1}\), we set the numerator equal to zero and solve for \(x\): \[ x + 7 = 0 \rightarrow x = -7 \] This means the x-intercept is at \(x = -7\). At this point, the function value is zero, meaning \(f(-7) = 0\). Keep in mind that the options in the exercise do not match this particular x-intercept, but understanding how to find it is vital.
Holes in Graph
A hole in the graph of a rational function occurs at values of \(x\) that cancel out in both the numerator and the denominator. Holes represent points where the function is not defined due to a common factor that was canceled when simplifying the function. For the given function \(f(x) = \frac{x+7}{x+1}\), there is no common factor that cancels out between the numerator and the denominator, thus no hole in the graph at any point.

It's important to understand holes as they represent breaks in the graph, but in this example, since no factors cancel out, no holes exist.

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Most popular questions from this chapter

Explain how the graph of each function can be obtained from the graph of \(y=\frac{1}{x}\) or \(y=\frac{1}{x^{2}} .\) Then graph \(f\) and give the (a) domain and (b) range. Determine the intervals of the domain for which the function is ( \(c\) ) increasing or (d) decreasing. See Examples \(1-3\). $$f(x)=\frac{1}{(x-3)^{2}}$$

Find all complex zeros of each polynomial function. Give exact values. List multiple zeros as necessary. $$f(x)=4 x^{3}+3 x^{2}+8 x+6$$

Find all complex zeros of each polynomial function. Give exact values. List multiple zeros as necessary. $$f(x)=x^{4}+x^{3}-9 x^{2}+11 x-4$$

Find all complex zeros of each polynomial function. Give exact values. List multiple zeros as necessary. $$f(x)=x^{4}+2 x^{3}-3 x^{2}+24 x-180$$

Solve each problem. AIDS Cases in the United States The table* lists the total (cumulative) number of AIDS cases diagnosed in the United States through \(2007 .\) For example, a total of \(361,509\) AIDS cases were diagnosed through 1993 (a) Plot the data. Let \(x=0\) correspond to the year 1990 . (b) Would a linear or a quadratic function model the data better? Explain. (c) Find a quadratic function defined by \(f(x)=a x^{2}+b x+c\) that models the data. (d) Plot the data together with \(f\) on the same coordinate plane. How well does \(f\) model the number of AIDS cases? (e) Use \(f\) to estimate the total number of AIDS cases diagnosed in the years 2009 and 2010 (f) According to the model, how many new cases were diagnosed in the year \(2010 ?\) $$\begin{array}{c|c||c|c} \text { Year } & \text { AIDS Cases } & \text { Year } & \text { AIDS Cases } \\\ \hline 1990 & 193,245 & 1999 & 718,676 \\ \hline 1991 & 248,023 & 2000 & 759,434 \\ \hline 1992 & 315,329 & 2001 & 801,302 \\ \hline 1993 & 361,509 & 2002 & 844,047 \\ \hline 1994 & 441,406 & 2003 & 888,279 \\ \hline 1995 & 515,586 & 2004 & 932,387 \\ \hline 1996 & 584,394 & 2005 & 978,056 \\ \hline 1997 & 632,249 & 2006 & 982,498 \\ \hline 1998 & 673,572 & 2007 & 1,018,428 \\ \hline \end{array}$$
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