/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 Let \(f\) be the function whose ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 47

Let \(f\) be the function whose graph is obtained by translating the graph of \(y=\frac{1}{x}\) to the right 3 units and up 2 units. (a) Write an equation for \(f(x)\) as a quotient of two polynomials. (b) Determine the zero(s) of \(f\) (c) Identify the asymptotes of the graph of \(f(x)\)

Short Answer

Expert verified
(a) \( f(x) = \frac{2x-5}{x-3} \) (b) Zero at \( x = \frac{5}{2} \) (c) Vertical asymptote at \( x=3 \, Horizontal asymptote at \( y=2 \).

Step by step solution

01

Understanding the transformation

The original function is given as \( y = \frac{1}{x} \). To translate this graph to the right by 3 units, we replace \( x \) with \( x - 3 \). This gives us the intermediate function \( y = \frac{1}{x-3} \).
02

Translating up by 2 units

Next, we translate the intermediate function \( y = \frac{1}{x-3} \) up by 2 units by adding 2, resulting in the function \( f(x) = \frac{1}{x-3} + 2 \).
03

Expressing as a single quotient

To write \( f(x) \) as a single quotient, combine the terms over a common denominator: \[ f(x) = \frac{1}{x-3} + 2 = \frac{1 + 2(x-3)}{x-3} = \frac{2x-5}{x-3} \].
04

Determine the zero(s) of f(x)

To find the zero(s) of \( f(x) \), set the numerator equal to zero and solve: \[ 2x - 5 = 0 \Rightarrow x = \frac{5}{2} \]. So, the zero of \( f(x) \) is \( x = \frac{5}{2} \).
05

Identify the asymptotes

Vertical asymptote occurs where the denominator is zero, so \( x=3 \). The horizontal asymptote for a function where the degree of the numerator is the same as the degree of the denominator is given by the ratio of the leading coefficients. Thus, the horizontal asymptote is \( y = \frac{2}{1} = 2 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

horizontal asymptote
The horizontal asymptote of a function is a horizontal line that the graph of the function approaches as the input value, usually denoted as x, goes to positive or negative infinity. For rational functions, like the one in our exercise, the horizontal asymptote is determined by the degrees of the polynomials in the numerator and denominator.
In our given function, influenced by the transformation, the function becomes:
\[ f(x) = \frac{2x-5}{x-3} \]
The degrees of the numerator and the denominator are the same (both are 1). When this is the case, the horizontal asymptote is the ratio of the leading coefficients.
In the function \( f(x) = \frac{2x-5}{x-3} \), the leading coefficient of the numerator is 2, and the leading coefficient of the denominator is 1. Therefore, the horizontal asymptote is:
\[ y = \frac{2}{1} = 2 \] This can be interpreted as the value that the function approaches as x gets very large or very small, but never actually reaches.
vertical asymptote
A vertical asymptote is a vertical line near which the function values grow without bound. In simpler words, as we approach this vertical line from either side, the function values approach infinity or negative infinity.
For the rational function \( f(x) = \frac{2x-5}{x-3} \), vertical asymptotes occur where the denominator is zero. This happens because division by zero is undefined and this leads to the function blowing up to infinity.
To find the vertical asymptote, we set the denominator equal to zero and solve for x:
\[ x - 3 = 0 \]
\[ x = 3 \]
Therefore, the vertical asymptote of the function \( f(x) \) is at \( x = 3 \). This means that the graph of the function will get closer and closer to the line \( x = 3 \), but will never actually touch it.
finding zeros
To find the zeros of a function, we need to determine the points where the function crosses the x-axis. This happens where the output value, or y-value, of the function is zero. Essentially, to find the zeros of \( f(x) \), we solve for x when \( f(x) = 0 \).
In the function \( f(x) = \frac{2x-5}{x-3} \), we need to determine when the numerator equals zero because a fraction is zero when its numerator is zero.
Thus, we set the numerator equal to zero and solve for x:
\[ 2x - 5 = 0 \]
\[ 2x = 5 \]
\[ x = \frac{5}{2} \]
Therefore, the zero of the function \( f(x) \) is at \( x = \frac{5}{2} \). This means that the graph of the function will cross the x-axis at this point.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Work each problem.What happens to \(y\) if \(y\) varies directly as \(x,\) and \(x\) is halved?

The number of long-distance phone calls between two cities in a certain time period varies directly as the populations \(p_{1}\) and \(p_{2}\) of the cities and inversely as the distance between them. If \(10,000\) calls are made between two cities \(500 \mathrm{mi}\) apart, having populations of \(50,000\) and \(125,000,\) find the number of calls between two cities 800 mi apart, having populations of \(20,000\) and \(80,000\).

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. $$f(x)=\frac{1}{x+4}$$

Find all complex zeros of each polynomial function. Give exact values. List multiple zeros as necessary. $$f(x)=32 x^{4}-188 x^{3}+261 x^{2}+54 x-27$$

Solve each problem. AIDS Deaths in the United States The table* lists the total (cumulative) number of known deaths caused by AIDS in the United States up to 2007 (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 \(g(x)=a x^{2}+b x+c\) that models the data. (d) Plot the data together with \(g\) on the same coordinate plane. How well does \(g\) model the number of AIDS cases? (e) Use \(g\) to estimate the total number of AIDS deaths in the year 2010 . (f) Consider the last two entries in the table for the years 2006 and 2007 . Is it safe to assume that the quadratic model given for \(g(x)\) will continue for years 2008 and beyond? $$\begin{array}{c|c||c|c} \hline \text { Year } & \text { AIDS Deaths } & \text { Year } & \text { AIDS Deaths } \\ \hline 1990 & 119,821 & 1999 & 419,234 \\ \hline 1991 & 154,567 & 2000 & 436,373 \\ \hline 1992 & 191,508 & 2001 & 454,099 \\ \hline 1993 & 220,592 & 2002 & 471,417 \\ \hline 1994 & 269,992 & 2003 & 489,437 \\ \hline 1995 & 320,692 & 2004 & 507,536 \\ \hline 1996 & 359,892 & 2005 & 524,547 \\ \hline 1997 & 381,738 & 2006 & 565,927 \\ \hline 1998 & 400,743 & 2007 & 583,298 \\ \hline \end{array}$$
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.