Answers: Graphs of Logarithmic and Exponential Functions

Questions on identifying equations given graphs, with answers included, to practice graphs of logarithmic and exponential functions. Review the function transformations carefully to understand how the correct graph is identified.

Table of Contents

Transformation Cheat Sheet

Before reviewing the answers, here is a quick reference for the transformations used in these questions:

Transformation Effect on Parent Graph
\( f(x) + k \)Vertical shift: up if \( k \gt 0 \), down if \( k \lt 0 \)
\( f(x - h) \)Horizontal shift: right if \( h \gt 0 \), left if \( h \lt 0 \)
\( -f(x) \)Reflection across the x-axis
\( f(-x) \)Reflection across the y-axis
\( a f(x) \)Vertical stretch if \( |a| \gt 1 \), compression if \( 0 \lt |a| \lt 1 \)

Questions and Answers

Question 1

Identify the graph of the function:

\[ f(x) = \ln(-x) \]

Multiple choice graphs for identifying the graph of f(x) = ln(-x)

Answer: c

Explanation

The parent function is \( y = \ln(x) \) with domain \( x \gt 0 \) and a vertical asymptote at \( x = 0 \).

Replacing \( x \) with \( -x \) reflects the graph across the y-axis. The new domain is \( x \lt 0 \) (i.e., \( (-\infty, 0) \)), the vertical asymptote remains at \( x = 0 \), and the graph passes through \( (-1, 0) \).

Question 2

Identify the graph of the function:

\[ f(x) = e^{x + 2} \]

Multiple choice graphs for identifying the graph of f(x) = e^(x+2)

Answer: b

Explanation

The parent function is \( y = e^x \) with horizontal asymptote \( y = 0 \) and y-intercept \( (0, 1) \).

Replacing \( x \) with \( x + 2 \) shifts the graph 2 units to the left. The new y-intercept is \( (0, e^2) \approx (0, 7.39) \), and the horizontal asymptote remains \( y = 0 \).

Question 3

Identify the graph of the function:

\[ f(x) = \ln(x) - 1 \]

Multiple choice graphs for identifying the graph of f(x) = ln(x) - 1

Answer: d

Explanation

The parent function is \( y = \ln(x) \) with x-intercept \( (1, 0) \) and vertical asymptote \( x = 0 \).

Subtracting 1 shifts the graph down 1 unit. The new x-intercept occurs when \( \ln(x) - 1 = 0 \), i.e., \( x = e \approx 2.718 \). The vertical asymptote remains \( x = 0 \).

Question 4

Identify the graph of the function:

\[ f(x) = e^{-x} + 1 \]

Multiple choice graphs for identifying the graph of f(x) = e^(-x) + 1

Answer: a

Explanation

Start with \( y = e^x \), reflect across the y-axis to get \( e^{-x} \) (decreasing curve), then shift up 1 unit.

The horizontal asymptote becomes \( y = 1 \). The y-intercept is \( (0, e^0 + 1) = (0, 2) \). The graph decreases from \( +\infty \) as \( x \to -\infty \) toward \( y = 1 \) as \( x \to +\infty \).

Question 5

Identify the graph of the function:

\[ f(x) = -\ln(x) - 1 \]

Multiple choice graphs for identifying the graph of f(x) = -ln(x) - 1

Answer: d

Explanation

Start with \( y = \ln(x) \), reflect across the x-axis to get \( -\ln(x) \) (decreasing curve passing through \( (1, 0) \)), then shift down 1 unit.

The vertical asymptote remains \( x = 0 \). The x-intercept occurs when \( -\ln(x) - 1 = 0 \), i.e., \( \ln(x) = -1 \), so \( x = e^{-1} \approx 0.368 \). The graph decreases from \( +\infty \) near the asymptote toward \( -\infty \) as \( x \to +\infty \).

Question 6

Identify the graph of the function:

\[ f(x) = \ln(x - 1) + 1 \]

Multiple choice graphs for identifying the graph of f(x) = ln(x-1) + 1

Answer: c

Explanation

The parent function is \( y = \ln(x) \). Shifting right 1 unit gives \( \ln(x - 1) \), and shifting up 1 unit gives \( \ln(x - 1) + 1 \).

The vertical asymptote moves to \( x = 1 \). The domain is \( (1, +\infty) \). The x-intercept occurs when \( \ln(x - 1) = -1 \), i.e., \( x - 1 = e^{-1} \), so \( x = 1 + e^{-1} \approx 1.368 \).

Question 7

Identify the graph of the function:

\[ f(x) = -e^{-x} \]

Multiple choice graphs for identifying the graph of f(x) = -e^(-x)

Answer: c

Explanation

Start with \( y = e^x \), reflect across the y-axis to get \( e^{-x} \), then reflect across the x-axis to get \( -e^{-x} \).

The horizontal asymptote remains \( y = 0 \). The y-intercept is \( (0, -1) \). The graph increases from \( y = 0 \) toward \( -\infty \) as \( x \to +\infty \), and approaches \( y = 0 \) from below as \( x \to -\infty \).

Question 8

Identify the graph of the function:

\[ f(x) = -\ln(-x) - 1 \]

Multiple choice graphs for identifying the graph of f(x) = -ln(-x) - 1

Answer: b

Explanation

Start with \( y = \ln(x) \). Reflect across the y-axis to get \( \ln(-x) \) (domain \( x \lt 0 \)), then reflect across the x-axis to get \( -\ln(-x) \), then shift down 1 unit.

The vertical asymptote remains \( x = 0 \). The graph passes through \( (-1, -1) \) since \( -\ln(1) - 1 = -1 \). Domain: \( (-\infty, 0) \).

Question 9

Identify the graph of the function:

\[ f(x) = -e^{-x - 1} + 1 \]

Multiple choice graphs for identifying the graph of f(x) = -e^(-x-1) + 1

Answer: c

Explanation

Rewrite: \( f(x) = -e^{-(x + 1)} + 1 \). Start with \( y = e^x \), reflect across the y-axis to get \( e^{-x} \), shift left 1 unit to get \( e^{-(x+1)} \), reflect across the x-axis to get \( -e^{-(x+1)} \), then shift up 1 unit.

The horizontal asymptote becomes \( y = 1 \). The y-intercept is \( (0, -e^{-1} + 1) \approx (0, 0.632) \). The graph decreases from \( y = 1 \) as \( x \to -\infty \) toward \( -\infty \) as \( x \to +\infty \).

Question 10

Identify the graph of the function:

\[ f(x) = -\ln(-x + 1) \]

Multiple choice graphs for identifying the graph of f(x) = -ln(-x+1)

Answer: d

Explanation

Rewrite: \( f(x) = -\ln(1 - x) \). Start with \( y = \ln(x) \). Reflect across the y-axis to get \( \ln(-x) \), shift right 1 unit to get \( \ln(-(x - 1)) = \ln(1 - x) \), then reflect across the x-axis to get \( -\ln(1 - x) \).

The domain is \( 1 - x \gt 0 \), i.e., \( x \lt 1 \) or \( (-\infty, 1) \). The vertical asymptote is at \( x = 1 \). The graph passes through \( (0, 0) \) since \( -\ln(1) = 0 \).

Summary of Answers

Question Function Key Transformation(s) Answer
1\( \ln(-x) \)Reflect across y-axisc
2\( e^{x+2} \)Shift left 2b
3\( \ln(x) - 1 \)Shift down 1d
4\( e^{-x} + 1 \)Reflect y-axis, shift up 1a
5\( -\ln(x) - 1 \)Reflect x-axis, shift down 1d
6\( \ln(x-1) + 1 \)Shift right 1, up 1c
7\( -e^{-x} \)Reflect y-axis, reflect x-axisc
8\( -\ln(-x) - 1 \)Reflect y-axis, reflect x-axis, shift down 1b
9\( -e^{-x-1} + 1 \)Reflect y-axis, shift left 1, reflect x-axis, shift up 1c
10\( -\ln(-x+1) \)Reflect y-axis, shift right 1, reflect x-axisd

Frequently Asked Questions

How does the transformation \( f(x) = \ln(-x) \) affect the graph of \( \ln(x) \)?

The transformation \( f(x) = \ln(-x) \) reflects the graph of \( y = \ln(x) \) across the y-axis. The domain changes from \( (0, +\infty) \) to \( (-\infty, 0) \), and the vertical asymptote remains at \( x = 0 \) but the curve now extends to the left.

What is the effect of \( f(x) = e^{x+2} \) on the graph of \( e^x \)?

The transformation \( f(x) = e^{x+2} \) shifts the graph of \( y = e^x \) horizontally 2 units to the left. The horizontal asymptote remains at \( y = 0 \), and the y-intercept moves from \( (0, 1) \) to \( (0, e^2) \).

What does the transformation \( f(x) = \ln(x) - 1 \) do to the graph of \( \ln(x) \)?

The transformation \( f(x) = \ln(x) - 1 \) shifts the graph of \( y = \ln(x) \) vertically down 1 unit. The vertical asymptote remains at \( x = 0 \), and the x-intercept moves from \( (1, 0) \) to \( (e, 0) \).

How do you identify the graph of \( f(x) = -e^{-x} \)?

Start with \( y = e^x \), reflect across the y-axis to get \( e^{-x} \), then reflect across the x-axis to get \( -e^{-x} \). The resulting graph is decreasing, lies entirely below the x-axis, passes through \( (0, -1) \), and has a horizontal asymptote at \( y = 0 \).

What is the domain and range of \( f(x) = \ln(x - 1) + 1 \)?

The domain is \( (1, +\infty) \) because the argument \( x - 1 \) must be positive. The range is \( (-\infty, +\infty) \). The vertical asymptote is at \( x = 1 \), and the graph passes through \( (2, 1) \).

How does the transformation \( f(x) = -\ln(-x + 1) \) differ from \( \ln(x) \)?

The transformation \( f(x) = -\ln(-x + 1) = -\ln(1 - x) \) involves: reflecting \( \ln(x) \) across the y-axis to get \( \ln(-x) \), shifting right 1 unit to get \( \ln(1 - x) \), then reflecting across the x-axis to get \( -\ln(1 - x) \). The domain is \( (-\infty, 1) \), and the vertical asymptote is at \( x = 1 \).

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