Problem 43 - Precalculus with Limits

Question

Find the domain, \(x\) -intercept, and vertical asymptote of the logarithmic function and sketch its graph. \(y=-\log _{3} x+2\)

Step-by-Step Solution

Answer

The domain of the function \(y=-\log _{3} x+2\) is \(x>0\). The function has its \(x\)-intercept at \(x=9\), and the vertical asymptote is at \(x=0\). The graph is a reflection of the standard logarithmic function about the x-axis.

Step 1: Find the Domain

For the function \(y=-\log _{3} x+2\), the logarithmic function is defined only for \(x > 0\). So, the domain is \(x > 0\).

Step 2: Calculate the x-intercept

The \(x\)-intercept is the value of \(x\) when \(y = 0\). Substitute \(y = 0\) in the given function and solve for \(x\). \n 0 = -\log _{3} x + 2 \n \(-\log _{3} x\) = -2 \n \(\log _{3} x\) = 2 \n Using definition of the logarithm, the \(x\) -intercept is \(x = 3^2 = 9\).

Step 3: Determine Vertical Asymptotes

For \(y=-\log _{3} x+2\), since logarithmic functions are not defined for \(x \leq 0\), therefore, the vertical asymptote is \(x=0\).

Step 4: Sketch the Graph

Using the calculated domain (\(x>0\)), x-intercept (\(x=9\)), and vertical asymptote (\(x=0\)), sketch the graph. Because the logarithm in the function is multiplied with a negative number, the graph will be a reflection of the standard logarithmic function about the x-axis.

Key Concepts

Domain of a Function

The domain of a function is the set of all possible input values (usually denoted as \(x\)) for which the function is defined. In the context of our logarithmic function, \(y = -\log_{3} x + 2\), the logarithmic part \(\log_{3} x\) dictates the domain. A logarithm is defined only for positive input values. Therefore, \(x > 0\) for this function. This means you can only use positive numbers for \(x\) when dealing with this expression. Simply put, the domain is the set of numbers greater than zero. Remember, if \(x\) is less than or equal to zero, the function does not exist.

Asymptotes

Asymptotes are lines that a graph approaches but never actually reaches or crosses. For our function \(y = -\log_{3} x + 2\), there is a vertical asymptote at \(x = 0\). This occurs because the logarithmic function is undefined at this point, causing the graph to approach the line without making contact. Vertical asymptotes are typical in logarithmic functions, representing bounds where the function behavior changes significantly. It's important to recognize these lines, as they help define the behavior of the graph near its edges.

X-Intercept

The x-intercept of a function is the point where the graph intersects the x-axis. To find this point for the function \(y = -\log_{3} x + 2\), we set \(y = 0\) and solve for \(x\). This process results in:

• 0 = \(-\log_{3} x + 2\)
• \(-\log_{3} x = -2\)
• \(\log_{3} x = 2\)

By applying the properties of logarithms, we find that \(x = 3^2 = 9\). Thus, the x-intercept is at \(x = 9\). This intercept is crucial because it provides a point where the function crosses the x-axis, helping us understand the graph's shape and direction.

Graphing Functions

Graphing a function provides a visual representation of its behavior and characteristics. For \(y = -\log_{3} x + 2\), we consider the domain (\(x > 0\)), the x-intercept \((x = 9)\), and the vertical asymptote (\(x = 0\)) to sketch the graph. Because the logarithm is negative, the typical logarithmic curve is reflected across the x-axis, making it descend instead of ascend.

• The graph does not exist to the left of \(x = 0\) due to the domain restriction.
• It approaches the vertical asymptote at \(x = 0\), but never actually crosses it.
• The x-intercept at \(x = 9\) marks the point where the graph crosses the x-axis.

Understanding these components allows us to accurately portray the behavior of the function. The graph starts high as \(x\) gets small but positive, descends smoothly past the x-intercept, and continues decreasing indefinitely as \(x\) increases.