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Find the domain, vertical asymptote, and \(x\) -intercept of the logarithmic function. Then sketch its graph.\(h(x)=\log _{2}(x+4)\)

Short Answer

Expert verified
The function \(h(x) = \log_2(x+4)\) has domain \(x > -4\), a vertical asymptote at \(x = -4\), and an x-intercept at \(-3\).

Step by step solution

01

Determine the Domain of the Function

For logarithmic functions, the argument of the logarithm (in this case, \(x+4\)) needs to be greater than zero. Hence, for \(h(x) = \log_2(x+4)\) the domain is determined by the inequality \(x+4 > 0\). Solving this inequality for \(x\) gives \(x > -4\). So, the domain is all real numbers greater than -4.
02

Identifying the Vertical Asymptote

The vertical asymptote is the vertical line where the function approaches but never reaches as \(x\) gets close to a specific value. For logarithmic functions, the line \(x = a\) is a vertical asymptote if the function is defined for \(x > a\) but not for \(x = a\). Since the function \(h(x) = \log_2(x+4)\) is not defined for \(x = -4\), there is a vertical asymptote at \(x = -4\).
03

Find the X-intercept

The x-intercept of a function is the point where the graph intersects the x-axis. This means, we are looking for the value of \(x\) when \(h(x) = 0\). So for the function \(h(x) = \log_2(x+4)\), we set \(h(x) = 0\) and solve for \(x\), i.e., \(\log_2(x+4) = 0\). Convert it to exponential form we get \(x+4 = 2^0 = 1\). Solving this for \(x\) gives \(x = -3\).
04

Sketch the Graph

The graph of the function \(h(x) = \log_2(x+4)\) would start at the point \((-3, 0)\) because that is the x-intercept and it would increase slowly going to the right side. It would also approach the vertical asymptote \(x = -4\) from the right but never reach or cross this line. See a graphing calculator for a precise image.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Domain
In understanding the domain of a logarithmic function, we need to remember that the argument of the logarithm must always be greater than zero. This stems from the fact that you cannot take the logarithm of a non-positive number in elementary mathematics.

For the given function, \( h(x) = \log_2(x+4) \), the argument is \( x+4 \). Therefore, we set up the inequality: \( x+4 > 0 \). Solving this gives \( x > -4 \).

Thus, the domain of the function is all real numbers greater than \(-4\).
  • Domain: \(( -4, \infty )\)
Understanding the domain helps us know where the function is operating and also influences the graph's behavior.
Vertical Asymptote
A vertical asymptote in a logarithmic function is a straight line that the graph approaches but never touches or crosses. For the function \( h(x) = \log_2(x+4) \), the vertical asymptote occurs when the argument of the logarithm equals zero since the logarithm is undefined at this point.

Thus, we set \( x+4 = 0 \), solving for \( x \) gives \( x = -4 \).

The vertical asymptote is important because it tells us about the boundary beyond which the function does not exist. Knowing the asymptote helps in accurately drawing the graph and understanding the behavior of the function as \( x \) approaches this critical value.
  • Vertical Asymptote: \( x = -4 \)
X-intercept
The x-intercept is the point where the graph of the function crosses the x-axis, which happens when the function value is zero. In mathematical terms, we are looking for \( x \) when \( h(x) = 0 \). For the given function \( h(x) = \log_2(x+4) \), setting \( h(x) = 0 \) yields the equation \( \log_2(x+4) = 0 \).

When we translate \( \log_2(x+4) = 0 \) to exponential form, we get \( x+4 = 2^0 = 1 \). Solving for \( x \), we find \( x = -3 \). This means the x-intercept of the function is at the point \((-3, 0)\).

The x-intercept is crucial in graph sketching as it provides a definitive starting point from which the function will either rise or fall as \( x \) moves away from this point.
  • X-intercept: \((-3, 0)\)
Graph Sketching
Graph sketching involves plotting the key features of a function to understand its overall behavior visually. For \( h(x) = \log_2(x+4) \), we have determined the domain, vertical asymptote, and x-intercept, which are pivotal in creating the graph.

Begin with the x-intercept at \((-3, 0)\). Since the function rises slowly toward the right, it is a characteristic of logarithmic growth.

Keep in mind the vertical asymptote at \( x = -4 \). The graph will come very close to this line but will never touch or cross it.

As the function progresses to the right from the x-intercept, it will ascend because logarithmic functions slowly increase when their base is greater than one.
  • Start at the x-intercept and rise gradually.
  • Approach the vertical asymptote from the right, but never reach it.
In a graphing calculator, you can see how these elements come together, giving you a concise view of the function's behavior across its domain.

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Most popular questions from this chapter

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(3 \ln 5 x=10\)

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Population The populations \(P\) of the United States (in thousands) from 1990 to 2005 are shown in the table. (Source: U.S. Census Bureau)$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Population } \\ \hline 1990 & 250,132 \\ \hline 1991 & 253,493 \\ \hline 1992 & 256,894 \\ \hline 1993 & 260,255 \\ \hline 1994 & 263,436 \\ \hline 1995 & 266,557 \\ \hline 1996 & 269,667 \\ \hline 1997 & 272,912 \\ \hline \end{array} $$$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Population } \\ \hline 1998 & 276,115 \\ \hline 1999 & 279,295 \\ \hline 2000 & 282,403 \\ \hline 2001 & 285,335 \\ \hline 2002 & 288,216 \\ \hline 2003 & 291,089 \\ \hline 2004 & 293,908 \\ \hline 2005 & 296,639 \\ \hline \end{array} $$(a) Use a graphing utility to create a scatter plot of the data. Let \(t\) represent the year, with \(t=0\) corresponding to 1990 . (b) Use the regression feature of a graphing utility to find an exponential model for the data. Use the Inverse Property \(b=e^{\ln b}\) to rewrite the model as an exponential model in base \(e\). (c) Use the regression feature of a graphing utility to find a linear model and a quadratic model for the data. (d) Use a graphing utility to graph the exponential model in base \(e\) and the models in part (c) with the scatter plot. (e) Use each model to predict the populations in 2008 , 2009 , and 2010 . Do all models give reasonable predictions? Explain.

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(\log _{10} x+\log _{10}(x+1)=\log _{10}(x+3)\)

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