Problem 29 - Precalculus with Calculus Previews

Question

In Problems \(25-32\), find the domain of the given function \(f .\) Find the \(x\) -intercept and the vertical asymptote of the graph. Use transformations to graph the given function \(f\). $$ f(x)=3-\log _{2}(x+3) $$

Step-by-Step Solution

Answer

Domain: \((-3, \infty)\); x-intercept: \((5, 0)\); Vertical asymptote: \(x = -3\).

Step 1: Determine the domain of the function

The function given is \( f(x) = 3 - \log_2(x+3) \). The domain of the logarithmic function \( \log_b(y) \) is \( y > 0 \). Thus, we need \( x+3 > 0 \). Solving this inequality, we get:\( x > -3 \).Therefore, the domain of the function is \( (-3, \infty) \).

Step 2: Find the x-intercept of the function

To find the x-intercept, set the function equal to zero and solve for \( x \):\[ 3 - \log_2(x+3) = 0 \].Rearrange to find:\[ \log_2(x+3) = 3 \].Solving for \( x+3 \) using the exponential form, we have:\[ x+3 = 2^3 \].\[ x+3 = 8 \].Subtract 3 from both sides:\[ x = 5 \].Thus, the x-intercept is at \( (5, 0) \).

Step 3: Identify the vertical asymptote of the function

A logarithmic function has a vertical asymptote where the argument of the logarithm is zero. Thus, set \( x+3 = 0 \):\[ x = -3 \].Therefore, the vertical asymptote is the line \( x = -3 \).

Step 4: Describe how to graph the function using transformations

The base function is \( \log_2(x) \). The given function can be rewritten as \( f(x) = 3 - \log_2(x+3) \), which involves the following transformations:1. Horizontal shift left by 3 units (\( x+3 \)).2. Reflection across the x-axis (due to negative sign before \( \log \)).3. Vertical shift upward by 3 units (adding 3).Thus, graph the base logarithmic function, apply the described transformations to obtain \( f(x) \).

Key Concepts

Domain of a Function

The domain of a function refers to all the possible input values (or "x-values") that a function can accept without resulting in undefined or problematic outputs. For logarithmic functions like \(f(x) = 3 - \log_2(x+3)\), it is crucial to remember that the argument of the logarithm (here, \(x+3\)) must be greater than zero. This is because the logarithm of a non-positive number is not defined within the real number system.
To find the domain in this case, you solve for when the argument is positive: \(x+3 > 0\). Simplifying this inequality, we find that \(x > -3\).
Thus, the domain of the function is \((-3, \infty)\), meaning that any \(x\) greater than \(-3\) can be input into the function successfully.

X-Intercept

The x-intercept of a function is the point where the graph of the function crosses the x-axis. This occurs when the output of the function is zero. To find this point for our function, \(f(x) = 3 - \log_2(x+3)\), we set \(f(x) = 0\) and solve for \(x\).
The equation simplifies to \(3 - \log_2(x+3) = 0\), which can be rewritten to \(\log_2(x+3) = 3\). Using properties of logarithms, this means \(x+3 = 2^3\).
Since \(2^3 = 8\), we further solve \(x+3 = 8\), which leads us to subtract 3 from both sides, resulting in \(x = 5\).
Therefore, the x-intercept is at the point \((5, 0)\). This means that at \(x = 5\), the graph touches or crosses the x-axis.

Vertical Asymptote

A vertical asymptote is a line that the graph of a function approaches but never touches or crosses. For logarithmic functions, this usually occurs where the argument of the logarithm equals zero.
In our function \(f(x) = 3 - \log_2(x+3)\), the vertical asymptote is determined by setting the argument of the logarithm equal to zero: \(x+3 = 0\). Solving for \(x\), we find \(x = -3\).
This means there is a vertical asymptote at the line \(x = -3\). As \(x\) approaches \(-3\) from the right, the logarithmic portion of the function becomes extremely negative, and the overall function becomes undefined, illustrating why the graph approaches but never quite reaches this line.

Graph Transformations

Graph transformations modify the appearance of a graph through changes such as shifting, reflecting, and stretching. For our function \(f(x) = 3 - \log_2(x+3)\), these transformations must be considered in sequence.

• Horizontal Shift: The \(x+3\) inside the logarithm indicates a horizontal shift to the left by 3 units, as we replace \(x\) with \(x+3\).
• Reflection: The negative sign in front of the logarithmic term \(-\log_2(x+3)\) causes a reflection across the x-axis, flipping the graph upside down.
• Vertical Shift: Adding 3 to the entire function results in a vertical shift upward by 3 units.

First, start by graphing the basic \(\log_2(x)\) function. Then, apply the horizontal shift by moving it left by 3 units. Reflect it across the x-axis by flipping it upside down. Finally, shift the entire graph upwards by 3 units to obtain the graph of \(f(x)\). This sequence of transformations allows us to accurately plot the function on a graph.