Question
a.
Graph the function f left parenthesis x right parenthesis equals e Superscript x on the window [minus5,5] by [minus10,30].
b.
Find f(1), f(negative 1), and f(4), rounded to three decimal places.
c.
What is the horizontal asymptote of the graph?
d.
What is the y-intercept?
Question content area bottom
Part 1
a. Choose the correct graph below.
A.
A coordinate system has a horizontal x-axis labeled from negative 5 to 5 in increments of 1 and a vertical f of x-axis labeled from negative 10 to 30 in increments of 2. A line rises from left to right in quadrant 3, passes through the origin, and continues to rise in quadrant 1.
B.
A coordinate system has a horizontal x-axis labeled from negative 5 to 5 in increments of 1 and a vertical f of x-axis labeled from negative 10 to 30 in increments of 2. A smooth curve falls steeply in quadrant two, passes through the point (0,1), and then falls shallowly in quadrant 1 approaching and above the x-axis.
C.
Graph the function f left parenthesis x right parenthesis equals e Superscript x on the window [minus5,5] by [minus10,30].
b.
Find f(1), f(negative 1), and f(4), rounded to three decimal places.
c.
What is the horizontal asymptote of the graph?
d.
What is the y-intercept?
Question content area bottom
Part 1
a. Choose the correct graph below.
A.
A coordinate system has a horizontal x-axis labeled from negative 5 to 5 in increments of 1 and a vertical f of x-axis labeled from negative 10 to 30 in increments of 2. A line rises from left to right in quadrant 3, passes through the origin, and continues to rise in quadrant 1.
B.
A coordinate system has a horizontal x-axis labeled from negative 5 to 5 in increments of 1 and a vertical f of x-axis labeled from negative 10 to 30 in increments of 2. A smooth curve falls steeply in quadrant two, passes through the point (0,1), and then falls shallowly in quadrant 1 approaching and above the x-axis.
C.
Answers
Answered by
GPT-4o mini
To address your questions step by step:
### Part a: Graph of the Function
The function given is \( f(x) = e^x \). This is an exponential function that rises steeply as \( x \) becomes positive and approaches 0 as \( x \) becomes negative.
Given the specified window of \([-5, 5]\) by \([-10, 30]\), we should see that:
- For \( x \) values significantly negative (like -5), \( e^{-5} \) will yield a small positive value (approximately \( 0.0067 \)).
- For \( x = 0 \), \( f(0) = e^0 = 1 \).
- As \( x \) increases to positive values (like 5), \( e^5 \) is approximately \( 148.413 \).
Based on this, the correct graph should show a curve that rises steeply from the left (past the x-axis) and approaches y = 0, while going very high for positive values of \( x \).
You should choose a graph that reflects these characteristics. The description you provided of graph A seems to describe the behavior of \( f(x) = e^x \) correctly.
### Part b: Finding Values of \( f(x) \)
1. \( f(1) = e^1 = e \approx 2.718 \) (rounded to three decimal places: **2.718**)
2. \( f(-1) = e^{-1} = \frac{1}{e} \approx 0.368 \) (rounded to three decimal places: **0.368**)
3. \( f(4) = e^4 \approx 54.598 \) (rounded to three decimal places: **54.598**)
### Part c: Horizontal Asymptote
The horizontal asymptote of the function \( f(x) = e^x \) occurs as \( x \) approaches negative infinity. As \( x \to -\infty\), \( f(x) \to 0 \). Therefore, the horizontal asymptote is:
**y = 0**
### Part d: Y-Intercept
The y-intercept of a function is found by evaluating \( f(0) \):
\( f(0) = e^0 = 1 \)
So, the y-intercept is:
**(0, 1)**
### Summary
- a: Choose graph A.
- b: \( f(1) \approx 2.718 \), \( f(-1) \approx 0.368 \), \( f(4) \approx 54.598 \)
- c: Horizontal asymptote is **y = 0**.
- d: Y-intercept is **(0, 1)**.
### Part a: Graph of the Function
The function given is \( f(x) = e^x \). This is an exponential function that rises steeply as \( x \) becomes positive and approaches 0 as \( x \) becomes negative.
Given the specified window of \([-5, 5]\) by \([-10, 30]\), we should see that:
- For \( x \) values significantly negative (like -5), \( e^{-5} \) will yield a small positive value (approximately \( 0.0067 \)).
- For \( x = 0 \), \( f(0) = e^0 = 1 \).
- As \( x \) increases to positive values (like 5), \( e^5 \) is approximately \( 148.413 \).
Based on this, the correct graph should show a curve that rises steeply from the left (past the x-axis) and approaches y = 0, while going very high for positive values of \( x \).
You should choose a graph that reflects these characteristics. The description you provided of graph A seems to describe the behavior of \( f(x) = e^x \) correctly.
### Part b: Finding Values of \( f(x) \)
1. \( f(1) = e^1 = e \approx 2.718 \) (rounded to three decimal places: **2.718**)
2. \( f(-1) = e^{-1} = \frac{1}{e} \approx 0.368 \) (rounded to three decimal places: **0.368**)
3. \( f(4) = e^4 \approx 54.598 \) (rounded to three decimal places: **54.598**)
### Part c: Horizontal Asymptote
The horizontal asymptote of the function \( f(x) = e^x \) occurs as \( x \) approaches negative infinity. As \( x \to -\infty\), \( f(x) \to 0 \). Therefore, the horizontal asymptote is:
**y = 0**
### Part d: Y-Intercept
The y-intercept of a function is found by evaluating \( f(0) \):
\( f(0) = e^0 = 1 \)
So, the y-intercept is:
**(0, 1)**
### Summary
- a: Choose graph A.
- b: \( f(1) \approx 2.718 \), \( f(-1) \approx 0.368 \), \( f(4) \approx 54.598 \)
- c: Horizontal asymptote is **y = 0**.
- d: Y-intercept is **(0, 1)**.