/*! 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 21 Explain how the graph of \(f\) c... [FREE SOLUTION] | 91Ó°ÊÓ

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Explain how the graph of \(f\) can be obtained from the graph of \(y=\frac{1}{x}\) or \(y=\frac{1}{x^{2}} .\) Draw a sketch of the graph of \(f\) by hand. Then generate an accurate depiction of the graph with a graphing calculator. Finally, give the domain and range. $$f(x)=\frac{1}{x}+1$$

Short Answer

Expert verified
The graph of \( f(x) = \frac{1}{x} + 1 \) is a vertically shifted hyperbola; domain: \((-\infty, 0) \cup (0, \infty)\), range: \((-\infty, 1) \cup (1, \infty)\).

Step by step solution

01

Identify the Parent Function

The given function is related to the parent function \( y = \frac{1}{x} \). We will use this as our base graph for transformations.
02

Determine the Transformation

The function given is \( f(x) = \frac{1}{x} + 1 \). This indicates a vertical shift of the graph of \( y = \frac{1}{x} \) upwards by 1 unit.
03

Sketch the Basic Graph

Start by sketching the graph of \( y = \frac{1}{x} \). This graph is a hyperbola with vertical and horizontal asymptotes at \( x=0 \) and \( y=0 \), respectively. The graph consists of two curves, one in the first quadrant and the other in the third quadrant.
04

Apply the Transformation

Shift the entire graph of \( y = \frac{1}{x} \) vertically upwards by 1 unit. This means that every \( y \)-value on the graph of \( y = \frac{1}{x} \) is increased by 1. As a result, the horizontal asymptote shifts from \( y=0 \) to \( y=1 \).
05

Plot Critical Points

Choose points on the graph of \( y = \frac{1}{x} \), such as \( (1, 1) \), \( (-1, -1) \), \( (2, 0.5) \), and \( (-2, -0.5) \). Increase each \( y \)-value by 1 to get points for the graph of \( f(x) \). For example, point \( (1, 1) \) becomes \( (1, 2) \); \( (-1, -1) \) becomes \( (-1, 0) \).
06

Draw the Transformed Graph

Sketch the new hyperbola based on the shifted points and asymptotes. It will have a vertical asymptote at \( x = 0 \) and a horizontal asymptote at \( y = 1 \). The graph should still be in the first and third quadrants but shifted upwards.
07

Determine Domain and Range

The domain of the function \( f(x) = \frac{1}{x} + 1 \) is all real numbers except \( x = 0 \), so it's \( (-\infty, 0) \cup (0, \infty) \). The range is all real numbers except \( y = 1 \), so it's \( (-\infty, 1) \cup (1, \infty) \).

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Key Concepts

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

Parent Function
The starting point for understanding graph transformations is recognizing the parent function. In this case, the parent function is \( y = \frac{1}{x} \). It is fundamental in graph transformations because it serves as the original graph, or "parent," that we modify to get the graph of our new function.

A parent function like \( y = \frac{1}{x} \) is a hyperbola, known for its unique shape with two branches positioned in opposite quadrants. In this graph, you see these branches in the first and third quadrants, declining toward both axes without ever touching them. Understanding the parent function helps visualize where the branches of the new graph will lie.

Keeping a clear picture of the basic structure of the parent function will let you apply transformations, like shifts or reflections, accurately and easily.
Vertical Shift
One of the transformations we often apply to a parent function is a vertical shift. In the function \( f(x) = \frac{1}{x} + 1 \), the "+1" signifies a vertical shift. Here's how it works:

- When you add a number outside of the parent function, it shifts the entire graph up by that amount. Similarly, subtracting a number would shift it down.
- For \( f(x) = \frac{1}{x} + 1 \), every point on the graph of \( y = \frac{1}{x} \) moves up by 1 unit. Thus, the horizontal asymptote of \( y = 0 \) for \( y = \frac{1}{x} \) shifts to \( y = 1 \).

This fundamental transformation does not change the vertical asymptote; it remains at \( x = 0 \). Recognizing this shift ensures that you correctly adjust the graph while maintaining its general hyperbolic shape.
Hyperbola
The graph of \( y = \frac{1}{x} \), a hyperbola, has distinct characteristics. A hyperbola consists of two disconnected curves or branches. Each branch approaches its respective axes but never crosses them, forming asymptotes.

For \( y = \frac{1}{x} \):
  • The vertical asymptote is at \( x = 0 \).
  • The horizontal asymptote is initially at \( y = 0 \), which changes with transformations like vertical shifts.

When transforming a hyperbola such as through a vertical shift, keep in mind that the shape remains the same. Only the asymptote positions and the curve points shift, maintaining the general hyperbolic form. This is crucial for accurately sketching or identifying the transformed graph in any quadrant.
Domain and Range
Understanding the domain and range is vital for describing the complete behavior of a function. The domain refers to all the possible input values \( x \), while the range is all the possible output values \( y \) for the function.

For \( f(x) = \frac{1}{x} + 1 \):
  • The domain is the set of all real numbers except where the function is undefined. Since division by zero is undefined, the function's domain excludes \( x = 0 \). Thus, the domain is \( (-\infty, 0) \cup (0, \infty) \).
  • The range is all real numbers except the horizontal asymptote. With the horizontal asymptote at \( y = 1 \), the range excludes this value, resulting in \( (-\infty, 1) \cup (1, \infty) \).

Recognizing the domain and range helps ensure you grasp where the function exists and anticipate its outputs, which is crucial in graph interpretation and analysis.

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Most popular questions from this chapter

Find all complex solutions for each equation by hand. $$\frac{x}{x-3}+\frac{4}{x+3}=\frac{18}{x^{2}-9}$$

Find all complex solutions for each equation by hand. $$\frac{1}{x+2}+\frac{3}{x+7}=\frac{5}{x^{2}+9 x+14}$$

Rational function has an oblique asymptote. Determine the equation of this asymptote. Then use a graphing calculator to graph both the function and the asymptote in the window indicated. $$f(x)=\frac{x^{2}+9}{x+3} ;[-13.2,13.2] \text { by }[-25,25]$$

Train Curves When curves are designed for trains, sometimes the outer rail is elevated or banked so that a locomotive can safely negotiate the curve at a higher speed. Suppose a circular curve is being designed for a speed of 60 mph. The rational function \(f(x)=\frac{2540}{x}\) computes the elevation \(y\) in inches of the outer track for a curve with a radius of \(x\) feet, where \(y=f(x) .\) (Image can't copy) (a) Evaluate \(f(400)\) and interpret its meaning. (b) Graph \(f\) in the window \([0,600]\) by \([0,50] .\) Discuss how the elevation of the outer rail changes with the radius \(x\) (c) Interpret the horizontal asymptote. (d) What radius is associated with an elevation of 12.7 inches?

CONCEPT CHECK In some cases, it is possible to solve a rational inequality simply by deciding what sign the numerator and the denominator must have and then using the rules for quotients of positive and negative numbers to determine the solution set. For example, consider the rational inequality $$ \frac{1}{x^{2}+1}>0 $$ The numerator of the rational expression, 1, is positive, and the denominator, \(x^{2}+1,\) must always be positive because it is the sum of a nonnegative number, \(x^{2},\) and a positive number, 1. Therefore, the rational expression is the quotient of two positive numbers, which is positive. Because the inequality requires that the rational expression be greater than \(0,\) and this will always be true, the solution set is \((-\infty, \infty)\) Use similar reasoning to solve each inequality. $$\frac{x^{4}+2}{-6} \geq 0$$

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