/*! 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 30 Graph the function and specify t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 30

Graph the function and specify the domain, range, intercept(s), and asymptote. $$y=2^{x-1}-1$$

Short Answer

Expert verified
Domain: (-鈭�, 鈭�), Range: (-1, 鈭�), X-intercept: (1, 0), Y-intercept: (0, -1/2), Asymptote: y = -1.

Step by step solution

01

Identify the Function Type

The given function is an exponential function of the form \(y=a^{b(x-c)}+d\), where \(a=2\), \(b=1\), \(c=1\), and \(d=-1\). This means it is an exponential growth function.
02

Determine the Domain

The domain of any exponential function is all real numbers, because for any real number \(x\), the expression \(2^{x-1}\) is defined. Hence, the domain is \((-\infty, \infty)\).
03

Determine the Range

The range of the function is determined by the vertical shift. Since it is shifted down by 1, the range is \((-1, \infty)\).
04

Calculate the X-intercept

To find the x-intercept, set \(y = 0\) and solve for \(x\): \(2^{x-1} - 1 = 0\). This simplifies to \(2^{x-1} = 1\). Therefore, \(x-1=0\), so \(x=1\). The x-intercept is \((1, 0)\).
05

Find the Y-intercept

To calculate the y-intercept, set \(x = 0\) and solve for \(y\): \(y = 2^{0-1} - 1 = \frac{1}{2} - 1 = -\frac{1}{2}\). So, the y-intercept is \((0, -\frac{1}{2})\).
06

Identify the Horizontal Asymptote

For an exponential function of this type, the horizontal asymptote is affected by the vertical shift \(d\). Therefore, the horizontal asymptote is at \(y = -1\).
07

Graph the Function

To graph the function, plot key points such as the intercepts and draw the curve showing exponential growth, approaching the horizontal asymptote \(y = -1\) but never reaching it, extending from the x-intercept to the right and starting below from the negative y-axis toward the x-intercept.

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.

Understanding the Domain and Range of Exponential Functions
Exponential functions are mathematical expressions in which the variable is an exponent. A key feature of these functions is their unrestricted domain. For the function given, \(y = 2^{x-1} - 1\), it means you can substitute any real number for \(x\), and the expression \(2^{x-1}\) will be defined. Hence, the domain here is all real numbers, represented as \((-\infty, \infty)\).

Next is the range of the function. The range of an exponential function often depends on any vertical shifts applied to it. In our example, \(y = 2^{x-1} - 1\), we see a downward shift by 1 unit. This vertical shift moves the entire graph down by 1 unit, meaning that the smallest value \(y\) can achieve is just above \(-1\).

Thus, the range of the function is \((-1, \infty)\), starting just above \(-1\) and continuing indefinitely upwards. This reflects the nature of the exponential growth of the function, albeit shifted downward.
Finding Intercepts of Exponential Functions
Intercepts are points where the graph of a function crosses the x-axis or the y-axis. Finding these points for the function \(y = 2^{x-1} - 1\) involves some basic algebra.

  • X-intercept: To find this, set \(y=0\). By substituting and solving we get, \(2^{x-1} - 1 = 0\), which simplifies to \(2^{x-1} = 1\). Given that \(2^0 = 1\), it implies \(x-1 = 0\), or \(x = 1\). Thus, the x-intercept is at \((1, 0)\).
  • Y-intercept: To discover the y-intercept, substitute \(x = 0\) into the function and solve for \(y\). This gives us \(y = 2^{0-1} - 1 = \frac{1}{2} - 1 = -\frac{1}{2}\). Hence, the y-intercept is at \((0, -\frac{1}{2})\).
These intercepts are crucial because they provide points through which the graph will pass, aiding in sketching the graph.
Recognizing Asymptotes in Exponential Functions
Asymptotes are lines that a graph approaches but never actually touches. For exponential functions, horizontal asymptotes are common as they reflect the end behavior of the function.

In \(y = 2^{x-1} - 1\), the horizontal asymptote is notably affected by the vertical translation \(d\), which is \(-1\) in this case. As \(x\) tends towards \(-\infty\), the value of \(y\) approaches, but does not reach, \(-1\).

This behavior illustrates the characteristic feature of exponential decay or growth functions depending on the direction of the graph. It signifies that no matter how small \(x\) may become, \(y\) will always be a tiny fraction above \(-1\), thus never actually reaching it, establishing \(y = -1\) as our horizontal asymptote.

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

Decide which of the following properties apply to each function. (More than one property may apply to a function.)A. The function is increasing for \(-\infty

Let \(\mathcal{N}=\mathcal{N}_{0 e^{k t}} .\) In this exercise we show that if \(\Delta t\) is very small, then \(\Delta \mathcal{N} / \Delta t \approx k \mathcal{N} .\) In other words, over very small intervals of time, the average rate of change of \(\mathcal{N}\) is proportional to \(\mathcal{N}\) itself. (a) Show that the average rate of change of the function \(\mathcal{N}=\mathcal{N}_{0} e^{t t}\) on the interval \([t, t+\Delta t]\) is given by $$\frac{\Delta \mathcal{N}}{\Delta t}=\frac{\mathcal{N}_{0} e^{k t}\left(e^{k \Delta t}-1\right)}{\Delta t}=\frac{\mathcal{N}\left(e^{k \Delta t}-1\right)}{\Delta t}$$ (b) In Exercise 26 of Section 5.2 we saw that \(e^{x} \approx x+1\) when \(x\) is close to zero. Thus, if \(\Delta t\) is sufficiently small, we have \(e^{k \Delta t} \approx k \Delta t+1 .\) Use this approximation and the result in part (a) to show that \(\Delta \mathcal{N} / \Delta t \approx k N\) when \(\Delta t\) is sufficiently close to zero.

(a) How large must \(x\) be before the graph of \(y=\ln x\) reaches a height of \(y=100 ?\) (b) How large must \(x\) be before the graph of \(y=e^{x}\) reaches a height of (i) \(y=100 ?\) (ii) \(y=10^{6} ?\)

(a) Let \(\mathcal{N}(t)=\mathcal{N}_{0} e^{k t} .\) Show that \([\mathcal{N}(t+1)-\mathcal{N}(t)] / \mathcal{N}(t)=e^{k}-1 .\) (This is actually done in detail in the text. So, ideally, you should look back only if you get stuck or want to check your answer.) (b) Assume as given the following approximation, which was introduced in Exercise 26 of Section 5.2 \(e^{x} \approx x+1 \quad\) provided \(x\) is close to zero Use this approximation to explain why \(e^{k}-1 \approx k\) provided that \(k\) is close to zero. Remark: Combining this result with that in part (a), we conclude that the relative growth rate for the function \(\mathcal{N}(t)=\mathcal{N}_{0} e^{k t}\) is approximately equal to the growth constant \(k .\) As explained in the text, this is one of the reasons why in applications we've not distinguished between the relative growth rate and the decay constant \(k\)

The intensity of the sounds that the human ear can detect varies over a very wide range of values. For instance, a whisper from 1 meter away has an intensity of approximately \(10^{-10}\) watts per square meter \(\left(\mathrm{W} / \mathrm{m}^{2}\right)\), whereas, from a distance of 50 meters, the intensity of a launch of the Space Shuttle is approximately \(10^{8} \mathrm{W} / \mathrm{m}^{2} .\) For a sound with intensity \(I\), the sound level \(\beta\) is defined by $$ \beta=10 \log _{10}\left(I / I_{0}\right) $$ where the constant \(I_{0}\) is the sound intensity of a barely audible sound at the threshold of hearing. The units for the sound level \(\beta\) are decibels, abbreviated dB. (a) Solve the equation \(\beta=10 \log _{10}\left(1 / I_{0}\right)\) for \(I\) by first dividing by 10 and then converting to exponential form. (b) The sound level for a power lawnmower is \(\beta=100 \mathrm{db}\). and that for a cat purring is \(\beta=10\) db. Use your result in part (a) to determine how many times more intense is the power mower sound than the cat's purring.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

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.