Problem 39 - Algebra for College Students

Question

For Problems \(39-44\), graph each of the exponential functions. See answer section. $$ f(x)=e^{x}+1 $$

Step-by-Step Solution

Answer

Graph it by plotting the y-intercept at (0, 2), additional points like (1, 3.718) and (-1, 1.368), and the asymptote at \( y = 1 \).

Step 1: Identify the Function Type

The given function is an exponential function of the form \( f(x) = e^x + 1 \). Here, \( e \) is the base of the natural logarithm, approximately equal to 2.718.

Step 2: Determine the Horizontal Asymptote

Exponential functions are known to have a horizontal asymptote. For \( f(x) = e^x + 1 \), as \( x \to -\infty \), the value of \( e^x \) approaches 0. Therefore, the asymptote is \( y = 1 \).

Step 3: Plot the Y-intercept

Set \( x = 0 \) to find the y-intercept. Thus, \( f(0) = e^0 + 1 = 1 + 1 = 2 \). The point (0, 2) is the y-intercept.

Step 4: Find Additional Points

Choose other values for \( x \) to find corresponding \( y \) values. For example:- \( x = 1 \), \( f(1) = e^1 + 1 \approx 2.718 + 1 = 3.718 \).- \( x = -1 \), \( f(-1) = e^{-1} + 1 \approx 0.368 + 1 = 1.368 \).These points are (1, 3.718) and (-1, 1.368).

Step 5: Sketch the Graph

Using the y-intercept and the additional points, sketch the graph. Draw the curve passing through the points (0, 2), (1, 3.718), (-1, 1.368), and approaching the horizontal asymptote at \( y = 1 \). The curve will extend upward to the right and approach the asymptote from above to the left.

Key Concepts

Graphing Exponential Functions

An exponential function, like the one given in the exercise, takes on the form \(f(x) = e^x + 1\). The base \(e\), approximately 2.718, ensures that this function grows rapidly as \(x\) increases. When graphing exponential functions, it's essential to start by identifying key points and a general shape to understand the behavior of the curve.
To graph \( f(x) = e^x + 1 \):

• Start by finding the y-intercept. This is done by setting \(x = 0\), giving \(f(0) = 2\).
• Select more x-values: \( x = 1\), resulting in \( f(1) \approx 3.718 \), and \( x = -1 \), leading to \( f(-1) \approx 1.368 \).
• Plot these points: (0, 2), (1, 3.718), and (-1, 1.368).
• Connect them smoothly, noting the rapid increase from left to right.

The graph will show a curve growing steeply to the right and flattening towards a line as it moves towards the left. This visual pattern is characteristic of exponential growth functions.

Horizontal Asymptotes

A horizontal asymptote is a horizontal line that the graph of a function approaches but never actually reaches. For an exponential function like \( f(x) = e^x + 1 \), observing the behavior as \( x \) goes towards infinity (both positive and negative) can help identify it.
As \( x \to -\infty \), \( e^x \to 0 \), which means the function approaches \( y = 1 \). Therefore, the horizontal asymptote here is \( y = 1 \). This means that no matter how far left on the x-axis the function moves, it will never dip below this y-value.

• The graph will hug the line \( y = 1 \) closely on the left side.
• It mirrors the concept of a ceiling that the function cannot drop below as \( x \to -\infty \).

Understanding horizontal asymptotes helps to anticipate the behavior of the function at extreme values of \( x \), providing insights into its limiting behavior.

Function Transformations

Function transformations involve shifts, stretches, or reflections that alter the graph of a function in predictable ways. For the function \( f(x) = e^x + 1 \), the "+1" represents a vertical shift. Here, the whole graph of \( e^x \) moves upward by 1 unit across the y-axis.

1. Vertical shift: The addition of 1 moves every point on the original graph of \( e^x \) up by 1 unit, including its horizontal asymptote, making it \( y = 1 \).
2. Reflection and stretching: Although not applicable here, it's helpful to remember that multiplying function values can stretch or reflect them along the y-axis.
3. Horizontal transformations: similarly involve shifting the graph left or right.

Function transformations are invaluable for quickly adjusting and predicting changes in behavior across different types of functions. They enable a deep understanding of how functions react to different manipulations.