Problem 46 - Precalculus

Question

Graph the exponential function using transformations. State the \(y\) -intercept, two additional points, the domain, the range, and the horizontal asymptote. $$f(x)=2-\left(\frac{1}{3}\right)^{x+1}$$

Step-by-Step Solution

Answer

The y-intercept is \((0, \frac{5}{3})\), two points are \((-1, 1)\) and \((1, \frac{17}{9})\), domain is all real numbers, range is \((-\infty, 2)\), and the horizontal asymptote is \(y=2\).

Step 1: Identify the Parent Function

The parent function is an exponential function of the form \(g(x) = b^x\). In this case, it is \(h(x) = \left(\frac{1}{3}\right)^x\).

Step 2: Analyze Transformations

Our function is \(f(x) = 2 - \left(\frac{1}{3}\right)^{x+1}\). The expression \(\left(\frac{1}{3}\right)^{x+1}\) can be re-written as \(\left(\frac{1}{3}\right)^{x} \cdot \left(\frac{1}{3}\right)\). Therefore, the transformation includes a vertical shift of 2 units up and a reflection across the x-axis since it is subtracted. Additionally, there is a horizontal shift 1 unit to the left due to the \(+1\) inside the exponent.

Step 3: Graph the Transformed Function

Start with the parent function \(\left(\frac{1}{3}\right)^x\). Apply the transformations: Reflect over the x-axis, shift left by 1 unit, and shift up 2 units to get the final graph of \(f(x)\).

Step 4: Determine the Y-Intercept

To find the y-intercept, set \(x = 0\) in the function: \(f(0) = 2 - \left(\frac{1}{3}\right)^{0+1} = 2 - \frac{1}{3} = \frac{6}{3} - \frac{1}{3} = \frac{5}{3}\). So the y-intercept is \(\left(0, \frac{5}{3}\right)\).

Step 5: Find Two Additional Points

Choose two other values for \(x\) to determine additional points. For \(x = -1\), \(f(-1) = 2 - \left(\frac{1}{3}\right)^{0} = 2 - 1 = 1\). For \(x = 1\), \(f(1) = 2 - \left(\frac{1}{3}\right)^{2} = 2 - \frac{1}{9} = \frac{18}{9} - \frac{1}{9} = \frac{17}{9}\). Thus, the points are \((-1, 1)\) and \( (1, \frac{17}{9})\).

Step 6: Determine the Domain and Range

The domain of an exponential function is always all real numbers \(x \in \mathbb{R}\). The range after the transformation is \((-abla, 2)\), as the function is reflected and shifted up 2 units.

Step 7: Find the Horizontal Asymptote

The horizontal asymptote is the line that the graph approaches as \(x\) goes to infinity. After applying the transformations, the horizontal asymptote is at \(y = 2\).

Key Concepts

Transformations

In the study of exponential functions, transformations play a vital role in reshaping the graph of the parent function. For the given function \(f(x) = 2 - \left(\frac{1}{3}\right)^{x+1}\), we need to interpret these transformations systematically.

First, identify the base function: \(\left(\frac{1}{3}\right)^x\). This base is modified in a few key ways:

• A vertical shift upward by 2 units is achieved through the "+2" outside the exponential term. This means that every point on the graph moves 2 units higher on the y-axis.
• The expression is subtracted, resulting in a reflection over the x-axis. This inversion flips the graph vertically.
• A horizontal shift is indicated by the "+1" inside the exponent, resulting in a shift of the graph one unit left.

By understanding these transformations, we can predict how the graph's shape and position are altered from the original parent function.

Graphing

Graphing the transformed exponential function \(f(x) = 2 - \left(\frac{1}{3}\right)^{x+1}\) involves several steps. Start with the parent function \(\left(\frac{1}{3}\right)^x\). Apply the transformations identified earlier.

First, flip the graph of \(\left(\frac{1}{3}\right)^x\) over the x-axis due to the negative sign. Next, shift every point to the left by one unit because of the "+1" in the exponent. Finally, move the entire graph two units up to account for the "+2" in the expression.

To get accurate points for plotting, calculate key values such as the y-intercept by setting \(x = 0\). Then, find additional points by plugging in values like \(x = 1\) and \(x = -1\). These steps help piece together the behavior of the function across the graphing plane.

Domain and Range

For the function \(f(x) = 2 - \left(\frac{1}{3}\right)^{x+1}\), understanding its domain and range is crucial. The domain of any exponential function is all real numbers. This means \(x \in \mathbb{R}\) since there are no restrictions on the values that \(x\) can take.

However, the range is influenced by the transformations applied to the function. Originally, the range of an exponential function of form \(b^x\) typically is positive values. With the transformations, particularly the vertical reflection and shift, the range changes significantly.

The flip and vertical movement mean that the function can now take values less than 2 but will never reach or exceed 2. Therefore, the final range of \(f(x)\) is \((-\infty, 2)\). Understanding these aspects are important for interpreting what the graph represents in terms of possible outputs.

Horizontal Asymptote

In the realm of functions, a horizontal asymptote indicates a y-value that the function approaches but never truly reaches as \(x\) progresses to positive or negative infinity. For \(f(x) = 2 - \left(\frac{1}{3}\right)^{x+1}\), the horizontal asymptote is not immediately obvious until we consider the transformations in detail.

Originally, \(\left(\frac{1}{3}\right)^x\) tends toward 0 as \(x\) approaches infinity. With our function flipped and vertically shifted, this behavior adjusts. Instead of approaching 0, \(f(x)\) now approaches 2 due to the vertical shift, which moves the entire graph upwards. Therefore, the horizontal asymptote of the transformed graph is \(y = 2\). This concept means that no matter how large \(x\) gets, the function values will approach 2, but will not actually equal 2.