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Chapter 9: Problem 21

$$ y=\frac{x+1}{x+5} $$

Short Answer

Expert verified
The y-intercept is at \(y = \frac{1}{5}\), the x-intercept is at \(x = -1\), the vertical asymptote is \(x= -5\) and the horizontal asymptote is \(y=1\).

Step by step solution

01

Find the y-intercept

The y-intercept occurs when x = 0. By plugging this in, the equation becomes \(y = \frac{0 + 1}{0 + 5} = \frac{1}{5}\). Therefore, the y-intercept is at \(y = \frac{1}{5}\).
02

Find the x-intercept

The x-intercept is found by setting the numerator equal to zero. If \(x + 1 = 0\), then \(x = -1\). Therefore, the x-intercept is at \(x = -1\).
03

Find the Vertical Asymptote

The vertical asymptote of a rational function is found by setting the denominator equal to zero. If \(x + 5 = 0\), then \(x = -5\). Therefore, the vertical asymptote is \(x= -5\).
04

Find the Horizontal Asymptote

The horizontal asymptote of a rational function depends on the degrees of the numerator and denominator. Here both have degree 1. Because they are the same, we divide the leading coefficients, which gives us a horizontal asymptote of \(y=1\).

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Key Concepts

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

Y-intercept
The y-intercept of a function is the point where the graph crosses the y-axis. At this point, the value of x is zero. To find it for a rational function, like \( y = \frac{x+1}{x+5} \), you substitute zero in place of x. This will simplify the expression to \( \frac{0+1}{0+5} = \frac{1}{5} \). Thus, the y-intercept is at the point \( (0, \frac{1}{5}) \).

  • Set \( x = 0 \) in \( y = \frac{x+1}{x+5} \)
  • Calculate \( y \) to find the intercept \( \frac{1}{5} \)
  • Y-intercepts are crucial for graphing the starting longitude of the function on the y-axis
X-intercept
An x-intercept is where the graph crosses the x-axis. This happens when the output value, y, is zero. To find the x-intercept for a rational function, you need to set the numerator to zero and solve for x. For \( y = \frac{x+1}{x+5} \), we set \( x+1 = 0 \), resulting in \( x = -1 \). The x-intercept is then the point \( (-1, 0) \).

  • Equate the numerator \( x+1 \) to zero
  • Solve to find \( x = -1 \)
  • X-intercepts are pivotal for detecting where the graph touches or crosses the x-axis
Vertical Asymptote
A vertical asymptote is a line that the graph of a function approaches but never touches. These appear in rational functions where a zero exists in the denominator while the numerator is non-zero. For the function \( y = \frac{x+1}{x+5} \), set the denominator to zero: \( x+5 = 0 \). Solving this gives \( x = -5 \). Thus, the vertical asymptote is the line \( x = -5 \).

  • Set the denominator \( x+5 \) to zero
  • Solve for \( x \) to get \( x = -5 \)
  • Vertical asymptotes reveal values x cannot take to avoid division by zero
Horizontal Asymptote
Horizontal asymptotes indicate the value a function approaches as \( x \) becomes very large or very small. For rational functions, the degree of the numerator and the denominator determines the horizontal asymptote. In \( y = \frac{x+1}{x+5} \), both parts have a degree of 1. Therefore, the horizontal asymptote corresponds to the ratio of the leading coefficients. Here that ratio is \( \frac{1}{1} = 1 \). So, the horizontal asymptote is the line \( y = 1 \).

  • Compare the degrees: both are 1
  • Divide the leading coefficients: \( 1/1 \)
  • The horizontal asymptote helps understand the function's behavior at extreme x values

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Most popular questions from this chapter

\(\boldsymbol{Q}\) and \(\boldsymbol{R}\) are independent events. Find \(\boldsymbol{P}(\boldsymbol{Q} \text { and } \boldsymbol{R})\) $$ P(Q)=\frac{12}{17}, P(R)=\frac{3}{8} $$

Error Analysis A student claims that \(x=2\) is the only solution of the equation \(\frac{x}{x-2}=\frac{1}{2}+\frac{2}{x-2}\) Is the student correct? Explain.

Describe the vertical asymptotes and holes for the graph of each rational function. $$ y=\frac{(x-4)(x+5)}{(x+3)(x-4)} $$

Which expression can be simplified to \(\frac{x-1}{x-3} ?\) $$ \begin{array}{ll}{\text { A. } \frac{x^{2}-x-6}{x^{2}-x-2}} & {\text { B. } \frac{x^{2}-2 x+1}{x^{2}+2 x-3}} \\ {\text { C. } \frac{x^{2}-3 x-4}{x^{2}-7 x+12}} & {\text { D. } \frac{x^{2}-4 x+3}{x^{2}-6 x+9}}\end{array} $$

Solve each equation. $$ 2 e^{2 x}+1=5 $$
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