Problem 45 - Precalculus

Question

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

Step-by-Step Solution

Answer

y-intercept: (0, 5), two points: (2, 2), (3, 1.5), domain: \\( (-\infty, \infty) \\\), range: \\( (1, \infty) \\\), horizontal asymptote: \\( y = 1 \\\).

Step 1: Identify the Parent Function

The given function is an exponential function based on \( \left(\frac{1}{2}\right)^{x} \), but the exponent is shifted and modified. The parent exponential function is \( y = \left(\frac{1}{2}\right)^x \).

Step 2: Determine Transformations

The given function can be written as \( f(x) = \left(\frac{1}{2}\right)^{x-2} + 1 \). The transformations involved are a horizontal shift and a vertical shift: a horizontal shift to the right by 2 units and a vertical shift up by 1 unit.

Step 3: Graph the Function Using Transformations

Start by graphing \( y = \left(\frac{1}{2}\right)^x \), then shift each point 2 units to the right and 1 unit up to account for the transformation, resulting in the graph of \( f(x) = \left(\frac{1}{2}\right)^{x-2} + 1 \).

Step 4: Identify the y-intercept

Substitute \( x = 0 \) into the function to find the y-intercept: \( f(0) = 1 + \left(\frac{1}{2}\right)^{0-2} = 1 + 4 = 5 \). Therefore, the y-intercept is (0, 5).

Step 5: Find Two Additional Points

Calculate two more points by substituting other values of x:- For \(x = 2\), \(f(2) = 1 + \left(\frac{1}{2}\right)^{2-2} = 1 + 1 = 2 \), so one point is (2, 2).- For \(x = 3\), \(f(3) = 1 + \left(\frac{1}{2}\right)^{3-2} = 1 + \frac{1}{2} = \frac{3}{2} \), so another point is (3, 1.5).

Step 6: State the Domain and the Range

The domain of the function is \( (-\infty, \infty) \) because exponential functions can take any real number as their input. The range is \( (1, \infty) \) due to the vertical shift of the graph by 1 unit upwards.

Step 7: Identify the Horizontal Asymptote

The horizontal asymptote is at \( y = 1 \) because the graph shifts vertically up by 1 unit, with the base exponential function approaching zero as \( x o \infty \).

Key Concepts

Transformations

Transformations of exponential functions involve altering the graph of the base or parent function to create a new function. In the exercise provided, we start with the parent function, which is a simple exponential function, given by \( y = \left(\frac{1}{2}\right)^x \). From this, we apply transformations which modify the graph. Specifically:

• **Horizontal Shift**: The expression inside the exponent, \( x-2 \), indicates a shift to the right by 2 units. This is because subtraction in the exponent moves the graph in the positive x-direction.
• **Vertical Shift**: The constant \(+1\) outside the exponential term indicates the graph is shifted 1 unit upwards.

These transformations change where the base function lies on the coordinate plane, assisting in determining the new graph position.

Domain and Range

Understanding the domain and range of an exponential function is crucial for graphing and comprehending its behavior. **Domain** refers to all the possible x-values the function can take. For exponential functions like **\( f(x) = 1 + \left(\frac{1}{2}\right)^{x-2} \)**, the domain is always all real numbers, \( (-\infty, \infty) \). This means there are no restrictions on x; you can input any real number into the function.
**Range**, however, describes all the possible y-values or outputs. Due to the upward vertical shift caused by \(+1\), the smallest value \( f(x) \) can approach is slightly above 1. Therefore, its range is \( (1, \infty) \), signifying that the function outputs begin just above 1 and extend upwards indefinitely.

Horizontal Asymptote

A horizontal asymptote in a function graph is a horizontal line that the graph approaches but never quite touches or intersects as \( x \) approaches \( \infty \) or \(-\infty\). For the function **\( f(x) = 1 + \left(\frac{1}{2}\right)^{x-2} \)**, the horizontal asymptote is located at \( y = 1 \).
This is due to the fact that regardless of the x-value input, as x grows larger in either direction, the expression \( \left(\frac{1}{2}\right)^{x-2} \) diminishes towards zero, leaving the constant \(+1\). As a result, the function smoothly approaches but does not reach \( y = 1 \), establishing \( y = 1 \) as the horizontal asymptote.

Graphing Functions

Graphing exponential functions like **\( f(x) = 1 + \left(\frac{1}{2}\right)^{x-2} \)** requires understanding the transformations, domain, range, and asymptotes involved. To graph this function:

• Begin by plotting the parent function \( y = \left(\frac{1}{2}\right)^x \).
• Apply the horizontal transformation by shifting the entire graph 2 units to the right.
• Then, shift the graph 1 unit upwards to reflect the vertical transformation.
• Identify the y-intercept by evaluating the function at \( x = 0 \), obtaining the point (0, 5).
• Calculate two additional points, for instance, \( (2, 2) \) and \( (3, 1.5) \), to aid in sketching the graph's trajectory.
• Finally, emphasize the horizontal asymptote at \( y = 1 \), which the graph will approach but never reach.

Mastering these steps ensures the accurate plotting of exponential graphs with various transformations.