a.

Question

Graph the function f left parenthesis x right parenthesis equals e Superscript x on the window ​[minus​5,5] by ​[minus​10,30].
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.

Answer 1

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).