Problem 50 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the domain, vertical asymptote, and \(x\) -intercept of the logarithmic function, and sketch its graph by hand. $$y=2+\log _{10}(x+1)$$

Step-by-Step Solution

Answer

The domain of the function \(y = 2 + \log_{10}(x+1)\) is \((-1, +\infty)\), the vertical asymptote is \(x = -1\) and the x-intercept is \(-0.99\). The function is increasing and its graph can be sketched using the provided points and asymptote.

Step 1: Find the Domain

The argument of the logarithm \(x+1\) must be greater than 0. Therefore, we solve the inequality \(x+1 > 0\) to get \(x > -1\). So, the domain of the function is \((-1, +\infty)\).

Step 2: Find the Vertical Asymptote

To find the vertical asymptote, we set the argument of the logarithm equal to zero and solve for \(x\). So \(x+1 = 0\) gives \(x = -1\). Hence, \(x = -1\) is the line of the vertical asymptote.

Step 3: Find the x-intercept

To find the x-intercept, we set \(y = 0\) in the given function and solve for \(x\). So, we have \(0 = 2 + \log_{10}(x+1)\). Rearranging this gives \(\log_{10}(x+1) = -2\). Using the definition of the logarithm, we can rewrite this as \(x+1 = 10^{-2} = 0.01\). Solving for \(x\) gives \(x = 0.01 - 1 = -0.99\). Hence, the x-intercept is \(-0.99\).

Step 4: Sketch the graph

Plot the results from steps 1 to 3: The function is defined for \(x > -1\), it has a vertical asymptote at \(x = -1\), and it crosses the x-axis at \(x= -0.99\). Note that because the base of the logarithm is greater than 1, the function is increasing. Also, remember that logarithmic functions pass through the point (1,0) when shifted. In this case, the graph will pass through the point (-1+1,0) = (0,0). This point along with the vertical asymptote and x-intercept will allow for the sketch of an accurate graph of the function.

Key Concepts

Domain of a Function

Understanding the domain of a function is a foundational concept in mathematical analysis. The domain of a function refers to all the possible input values (often represented by "x") for which the function is defined. For logarithmic functions like the one in the exercise, we need to ensure that the argument of the logarithm is greater than zero because logarithm of zero or a negative number is undefined.

In our specific case, the function is given by:

• \( y = 2 + \log_{10}(x+1) \)

To find the domain, we solve the inequality \(x+1 > 0\). This gives us:

• \( x > -1 \)

Thus, the domain of the function is all real numbers greater than \(-1\), written in interval notation as \((-1, +\infty)\). This means that any x-value larger than \(-1\) can be plugged into the function without issues.

Vertical Asymptote

A vertical asymptote is a line that a graph approaches but never touches or crosses. To find the vertical asymptote of a logarithmic function, look at the values of x that make the argument of the logarithm zero. These are the points where the function becomes undefined.

For the function \( y = 2 + \log_{10}(x+1) \), we set the argument to zero:

• \( x+1 = 0 \)

Solving this equation, we find:

• \( x = -1 \)

This reveals that \( x = -1 \) is the location of the vertical asymptote. Graphically, you'll notice the curve of the function getting infinitely close to this line, yet never actually crossing it. This asymptotic behavior represents how the logarithmic function behaves near the boundary of its domain.

X-Intercepts

The x-intercept of a graph is where the graph crosses the x-axis. At this point, the value of the function (y) equals zero. To find the x-intercept for our function, we set \( y = 0 \) and solve for \( x \).

Given the function:

• \( 0 = 2 + \log_{10}(x+1) \)

Rearranging gives us:

• \( \log_{10}(x+1) = -2 \)

Using properties of logarithms, this translates to:

• \( x+1 = 10^{-2} = 0.01 \)

Finally, solving for \( x \) we obtain:

• \( x = -0.99 \)

Thus, the x-intercept is \(-0.99\), meaning the graph crosses the x-axis very close to \(-1\). This is an important feature to note while sketching the graph, as it gives a specific point that the curve of the function will pass through.

Graphing Functions

Graphing functions allows us to visualize the behavior of the function over its domain. Sketching a logarithmic function involves understanding its asymptotes, intercepts, and the nature of the logarithmic curve.

For the function \( y = 2 + \log_{10}(x+1) \):

• The domain is \((-1, +\infty)\), telling us where the graph starts on the x-axis.
• There is a vertical asymptote at \( x = -1 \), which the graph approaches but never crosses.
• The x-intercept occurs at \( x = -0.99 \), providing a specific point on the graph.

Since the base of the logarithm is greater than 1, the function itself is increasing as x increases. It also shifts upward by 2 units because of the +2 added to the logarithm.

When graphing, mark the vertical asymptote clearly, plot the x-intercept, and ensure the curve approaches the asymptote as x nears \(-1\). Notice how it crosses the x-axis, curves upwards, and continues to rise steadily.