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Chapter 3: Problem 42

Graph the exponential function using transformations. State the \(y\) -intercept, two additional points, the domain, the range, and the horizontal asymptote. $$f(x)=e^{x-1}+2$$

Short Answer

Expert verified
The y-intercept is \((0, \frac{1}{e} + 2)\), two points are \((1, 3)\) and \((2, e + 2)\), the domain is \(x \in \mathbb{R}\), the range is \(y > 2\), and the horizontal asymptote is \(y = 2\).

Step by step solution

01

Identify the base function

The function given is of the form \(f(x) = e^{x} + c\). The base function is \(g(x) = e^x\), which is the standard exponential function with a base of \(e\).
02

Determine the transformations

The function \(f(x) = e^{x-1} + 2\) involves transforming the base function \(g(x) = e^x\). The term \(x-1\) indicates a horizontal shift to the right by 1 unit. The "+2" indicates a vertical shift upwards by 2 units.
03

Find the y-intercept

To find the y-intercept, substitute \(x = 0\) into the function \(f(x) = e^{x-1} + 2\). This gives \(f(0) = e^{0-1} + 2 = \frac{1}{e} + 2\).
04

Find two additional points

Choose values for \(x\) to find additional points. For \(x = 1\), \(f(1) = e^{1-1} + 2 = e^0 + 2 = 3\). For \(x = 2\), \(f(2) = e^{2-1} + 2 = e + 2\). Thus, the additional points are \((1, 3)\) and \((2, e + 2)\).
05

State the domain

The domain of the function \(f(x) = e^{x-1} + 2\) is the set of all real numbers, \(x \in \mathbb{R}\), as there are no restrictions on the input \(x\).
06

State the range

Due to the vertical shift, the range of the function is \(y > 2\). This is because the minimum value of \(e^{x-1}\) is 0, and adding 2 gives a minimum y-value greater than 2.
07

Determine the horizontal asymptote

The horizontal asymptote of the function is determined by the vertical shift. For \(f(x) = e^{x-1} + 2\), the horizontal asymptote is \(y = 2\), due to the +2 shift upwards.

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Key Concepts

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

Transformations
When dealing with exponential functions like \( f(x) = e^{x-1} + 2 \), understanding transformations is crucial. Transformations change the position and shape of the base graph \( g(x) = e^x \). There are mainly two types of transformations here: horizontal and vertical shifts.
  • Horizontal Shift: The term \( x-1 \) in the exponent indicates a horizontal shift towards the right by one unit. This means every point on the graph of \( e^x \) is moved one unit to the right.
  • Vertical Shift: The constant \(+2\) affects the graph by shifting it upwards by two units. This means every point from the shifted graph of \( e^{x-1} \) is now moved 2 units up.

By applying these transformations, you can plot the new function \( f(x) = e^{x-1} + 2 \) from the base function \( g(x) = e^x \) easily.
Horizontal Asymptote
In the context of exponential functions, a horizontal asymptote is a horizontal line that the graph of the function approaches but never touches. For the function \( f(x) = e^{x-1} + 2 \), the horizontal asymptote is determined by the vertical shift.When you shift the graph of \( e^x \) upwards by two units, the horizontal asymptote also shifts up by the same amount.
  • The standard exponential function \( g(x) = e^x \) has a horizontal asymptote at \( y = 0 \).
  • For \( f(x) \), the horizontal asymptote is now at \( y = 2 \) because of the \(+2\) vertical shift.

The graph will never cross this horizontal asymptote and instead will get closer and closer to it as \( x \) goes to negative infinity.
Domain and Range
Understanding the domain and range of an exponential function helps grasp its overall behavior. The domain refers to all possible \( x \) values, while the range refers to all possible \( y \) values the function can take.
  • Domain: For \( f(x) = e^{x-1} + 2 \), the domain is all real numbers, denoted by \( x \in \mathbb{R} \). There are no restrictions on the input values for \( x \).
  • Range: The range of \( f(x) = e^{x-1} + 2 \) is determined by its lowest possible \( y \) value. Since the smallest value \( e^{x-1} \) can take is 0, the smallest value of \( f(x) \) is 2. Thus, the range is \( y > 2 \).

These concepts ensure that even though the function grows infinitely large towards positive infinity, it still has a lower boundary at \( y = 2 \).

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