/*! 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 25 Graph the function and specify t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Short Answer

Expert verified
Domain: all real numbers; Range: \(y < 1\); y-intercept: (0, 0); Asymptote: \(y = 1\).

Step by step solution

01

Identify the Function

The function given is \(y = -2^x + 1\). It is an exponential function with a negative base that is reflected over the x-axis and shifted upward by 1 unit.
02

Determine the Domain

The domain of an exponential function is all real numbers, \(x \in \mathbb{R}\), because there are no restrictions on the values that \(x\) can take.
03

Determine the Range

For the function \(y = -2^x + 1\), the transformation affects the range. Since the function is reflected and shifted up, the range is \(y < 1\).
04

Find the Intercept(s)

To find the y-intercept, set \(x = 0\): \[y = -2^0 + 1 = -1 + 1 = 0\]Thus, the y-intercept is \((0, 0)\). There is no x-intercept for this function because the function never crosses the x-axis beyond the y-intercept point.
05

Identify Asymptote

The horizontal asymptote is the horizontal line that the function approaches as \(x\to \infty\). Here, this line is \(y = 1\) due to the vertical shift.

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 91Ó°ÊÓ!

Key Concepts

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

Function Graphing
Graphing the function \(y = -2^x + 1\) involves understanding its transformations. The base of the function \(2^x\) indicates exponential growth for positive values but the negative sign in front indicates a reflection over the x-axis. This flips the curve upside down. Additionally, the constant \(+1\) indicates a vertical shift upwards by 1 unit. By plotting key points and understanding these transformations, you can sketch a graph representing the exponential decline from the positive side along the x-axis.When you start plotting, remember:
  • The graph curves steeply downwards.
  • The basic shape of exponential growth becomes decline due to the negative sign.
  • The curve will never touch or cross a horizontal line, where \(y = 1\).
Grasping these concepts will help you visually represent the function accurately.
Domain and Range
The domain of an exponential function like \(y = -2^x + 1\) is the set of all real numbers. This means no matter the value of \(x\), you will get a valid result when plugging it into the function.The range, however, is affected by the vertical transformation. Typically, for a basic exponential function \(2^x\), the range is \(y > 0\). Here, due to the transformation of reflection and vertical shift, the range changes. The function outputs values less than 1, thus, the range becomes \(y < 1\). This means the graph will stay below the horizontal asymptote, which is at \(y = 1\).
Intercepts
Intercepts are points where the graph crosses the axes. For \(y = -2^x + 1\), the intercepts can be calculated to identify where it intersects with the y-axis and x-axis.- **Y-intercept:** To find the y-intercept, set \(x = 0\): \(y = -2^0 + 1 = 0\) Thus, the y-intercept is at the point \((0, 0)\).- **X-intercept:** Usually, you would solve \(y = 0\) to find the x-intercept. But here, after calculation, you find that the function never crosses the x-axis once you get past the y-intercept. Hence there are no additional x-intercepts.Understanding intercepts helps in quickly identifying important points on the graph.
Asymptotes
An asymptote is a line the graph approaches but never touches. In the context of \(y = -2^x + 1\), we deal with a horizontal asymptote. For transformed exponential functions, this horizontal line often defines the boundary for the range.For this function, due to the vertical shift, the horizontal asymptote is \(y = 1\). This means that as \(x\) becomes increasingly positive, the value of \(y\) will get closer and closer to 1, but will never actually reach or exceed it.Understanding asymptotes is crucial as they define the behavior of the graph in extreme conditions, providing insights into the limits of the function.

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

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$6\left(5-1.6^{x}\right) \geq 13$$

(a) Use a graphing utility to estimate the root(s) of the equation to the nearest one-tenth (as in Example 6). (b) Solve the given equation algebraically by first rewriting it in logarithmic form. Give two forms for each answer: an exact expression and a calculator approximation rounded to three decimal places. Check to see that each result is consistent with the graphical estimate obtained in part (a). $$e^{1-4 t}=12.405$$

Solve each equation. $$x^{1+\log _{x} 16}=4 x^{2}$$

Exercises \(55-60\) introduce a model for population growth that takes into account limitations on food and the environment. This is the logistic growth model, named and studied by the nineteenth century Belgian mathematician and sociologist Pierre Verhulst. (The word "logistic" has Latin and Greek origins meaning "calculation" and "skilled in calculation," respectively. However, that is not why Verhulst named the curve as he did. See Exercise 56 for more about this.) In the logistic model that we "I study, the initial population growth resembles exponential growth. But then, at some point owing perhaps to food or space limitations, the growth slows down and eventually levels off, and the population approaches an equilibrium level. The basic equation that we'll use for logis- tic growth is where \(\mathcal{N}\) is the population at time \(t, P\) is the equilibrium population (or the upper limit for population), and a and b are positive constants. $$\mathcal{N}=\frac{P}{1+a e^{-b t}}$$ The following figure shows the graph of the logistic function \(\mathcal{N}(t)=4 /\left(1+8 e^{-t}\right) .\) Note that in this equation the equilibrium population \(P\) is 4 and that this corresponds to the asymptote \(\mathcal{N}=4\) in the graph. (a) Use the graph and your calculator to complete the following table. For the values that you read from the graph, estimate to the nearest \(0.25 .\) For the calculator values, round to three decimal places. (b) As indicated in the graph, the line \(\mathcal{N}=4\) appears to be an asymptote for the curve. Confirm this empirically by computing \(\mathcal{N}(10), \mathcal{N}(15),\) and \(\mathcal{N}(20) .\) Round each answer to eight decimal places. (c) Use the graph to estimate, to the nearest integer, the value of \(t\) for which \(\mathcal{N}(t)=3\) (d) Find the exact value of \(t\) for which \(\mathcal{N}(t)=3 .\) Evaluate the answer using a calculator, and check that it is consistent with the result in part (c). TABLE AND GRAPH CANT COPY

Simplify each expression. (a) \(\ln e\) (b) \(\ln e^{-2}\) (c) \((\ln e)^{-2}\)

See all solutions

Recommended explanations on Math Textbooks

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.