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Chapter 0: Problem 15

Sketch the graph of the equation by translating, reflecting, compressing, and stretching the graph of \(y=x^{2}, y=\sqrt{x}\), \(y=1 / x, y=|x|\), or \(y=\sqrt[3]{x}\) appropriately. Then use a graphing utility to confirm that your sketch is correct. $$ y=2-\frac{1}{x+1} $$

Short Answer

Expert verified
Graph of \( y = 2 - \frac{1}{x+1} \) is a reflected, shifted version of \( y = \frac{1}{x} \) with asymptotes at \( y = 2 \) and \( x = -1 \).

Step by step solution

01

Identify the Parent Function

The equation given is \( y = 2 - \frac{1}{x+1} \). Recognizing the structure, we see that it is based on the reciprocal function \( y = \frac{1}{x} \). Thus, the parent function is \( y = \frac{1}{x} \).
02

Apply Horizontal Shift to the Parent Function

The term \( x+1 \) in the denominator indicates a horizontal shift to the left by 1 unit. Therefore, start by shifting the graph of \( y = \frac{1}{x} \) leftward by 1 unit to get \( y = \frac{1}{x+1} \).
03

Apply Vertical Reflection and Shift

The negative sign in front of \( \frac{1}{x+1} \) reflects the graph across the x-axis. The addition of 2 outside the fraction translates the entire graph upwards by 2 units, resulting in the final transformation equation \( y = 2 - \frac{1}{x+1} \).
04

Sketch the Transformed Graph

Start by sketching the basic shape of \( y = \frac{1}{x} \), shift it left by 1 unit, reflect it across the x-axis, and finally move it up by 2 units accordingly. The graph will approach 2 as an asymptote when \( x \) approaches infinity or negative infinity, and there will be a vertical asymptote at \( x = -1 \).
05

Use a Graphing Utility to Confirm

Enter the equation \( y = 2 - \frac{1}{x+1} \) into a graphing utility. Verify that your sketch matches the computer-generated graph, ensuring the transformations were applied correctly, with horizontal asymptote at \( y = 2 \) and vertical asymptote at \( x = -1 \).

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Key Concepts

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

Reciprocal Function
The reciprocal function is a fundamental type of function where the variable is in the denominator, typically expressed as \( y = \frac{1}{x} \). This forms a hyperbola with two branches, one in each of the first and third quadrants for positive values, assuming no further transformations.

Key characteristics of this function include:
  • It has a vertical asymptote at \( x = 0 \) because the function is undefined at this point.
  • It has a horizontal asymptote at \( y = 0 \), indicating that as the absolute value of \( x \) becomes very large, the function values approach zero.
  • The function is symmetric with respect to the origin, meaning it exhibits a point symmetry about the origin \((0, 0)\).
Understanding these properties helps in identifying the function as a parent when faced with transformations, like in this exercise.
Horizontal Shift
A horizontal shift in a function occurs when every point on the graph is moved left or right. For our transformation involving \( y = 2 - \frac{1}{x+1} \), the \( +1 \) in the denominator adjusts the horizontal positioning.

Here's how it works:
  • A \( +1 \) inside the function causes the graph to shift left by one unit.
  • This is opposite to what some might expect; generally, \( y = f(x+c) \) shifts to the left by \( c \).
  • It's essential to track these shifts carefully as they can completely alter where key features like asymptotes appear in the transformed graph.
In this particular example, the shifts help position our reciprocal function's central behaviors around new vertical asymptote at \( x = -1 \). The shift stays important when combining further transformations.
Vertical Reflection
Vertical reflection involves flipping the graph over a horizontal line, often the x-axis. This is what happens when a negative sign is introduced in front of a function.

In the given equation \( y = 2 - \frac{1}{x+1} \), the negative sign in front of the reciprocal portion \( -\frac{1}{x+1} \) indicates a vertical reflection:
  • The graph of \( y = \frac{1}{x+1} \) which typically appears in the first and third quadrants, flips so that its branches switch to the second and fourth quadrants.
  • This means any upward trend becomes downward, and downward becomes upward, relative to the x-axis.
  • Such reflections can affect intercepts, the approach of asymptotes, and the general direction of slopes in the graph.
Mastering reflections is crucial for controlling how the function behaves around key points.
Asymptotes
Asymptotes are lines that a graph approaches but never touches. They are crucial in describing the behavior at the extremes.

For the function \( y = 2 - \frac{1}{x+1} \), there are two significant asymptotes:
  • Vertical Asymptote: This occurs at \( x = -1 \), due to the denominator \( x+1 = 0 \). The graph of the reciprocal function cannot exist at this point, thus creating the vertical barrier.
  • Horizontal Asymptote: This happens at \( y = 2 \), indicating that as \( x \) approaches \( \pm \) infinity, \( y \) levels off towards 2, which is influenced by the vertical shift.
Asymptotes anchor the behavior of the graph near extremes, providing a guide for sketching or identifying limits and intercepts. Knowing where they lie will help in discriminating graph orientations or stretches, especially in more complex functions.

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Most popular questions from this chapter

Express \(f\) as a composition of two functions; that is, find \(g\) and \(h\) such that \(f=g \circ h .\) [Note: Each exercise has more than one solution.] (a) \(f(x)=\sin ^{2} x\) (b) \(f(x)=\frac{3}{5+\cos x}\)

Prove: (a) \(\sin ^{-1} x=\tan ^{-1} \frac{x}{\sqrt{1-x^{2}}} \quad(|x|<1)\) (b) \(\cos ^{-1} x=\frac{\pi}{2}-\tan ^{-1} \frac{x}{\sqrt{1-x^{2}}} \quad(|x|<1)\).

Solve for \(x\) without using a calculating utility. Use the natural logarithm anywhere that logarithms are needed. $$ 5^{-2 x}=3 $$

Classify the functions whose values are given in the accompanying table as even, odd, or neither. $$ \begin{array}{|c|r|r|r|r|r|r|r|} \hline x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline f(x) & 5 & 3 & 2 & 3 & 1 & -3 & 5 \\ \hline g(x) & 4 & 1 & -2 & 0 & 2 & -1 & -4 \\ \hline h(x) & 2 & -5 & 8 & -2 & 8 & -5 & 2 \\ \hline \end{array} $$

Solve for \(x\) without using a calculating utility. Use the natural logarithm anywhere that logarithms are needed. $$ x e^{-x}+2 e^{-x}=0 $$
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