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Summarize the pertinent information oblained by applying the graphing strategy and sketch the graph of \( y=f(x) \). \( f(x)= \) In \( (x+3)-2 \) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The \( y \)-intercept of \( f \) is \( y= \) In (3) -2. B. The function \( f \) has no \( y \)-intercept Select the correct choico below and, if necessary, fill in the answer box to complete your choice A. The function \( f \) is increasing on the subinterval(s) ( -3 , oo) (Type your answer in interval notation. Use a comma to separate answers as needed.) B. The function \( f \) is never increasing. Select the correct choice below and, if necessary, fill in the answer box to complote your choice. A. The function \( f \) is decieasing on the subinterval(s) (Type your answer in interval notation. Use a comma to separate answers as needed.) B. The function \( f \) is never decreesing.

Ask by Stewart Kirk. in the United States
Mar 11,2025

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The \( y \)-intercept of \( f \) is \( y = \ln(3) - 2 \). The function \( f \) is increasing on the interval \((-3, \infty)\) and is never decreasing.

Solución

1. **Determine the Domain and Transformation:** The function is \[ f(x)=\ln(x+3)-2. \] Since the natural log is defined only for positive arguments, we have \[ x+3>0 \quad \Longrightarrow \quad x>-3. \] This means the domain of \( f \) is \((-3,\infty)\). The graph is a horizontal translation of the basic logarithm \( \ln(x) \) left by 3 units and a vertical translation downward by 2 units. 2. **Find the \( y \)-Intercept:** The \( y \)-intercept occurs when \( x=0 \). Evaluating: \[ f(0)=\ln(0+3)-2=\ln(3)-2. \] This shows that the \( y \)-intercept is \[ (0,\ln(3)-2). \] **Correct Choice:** A. The \( y \)-intercept of \( f \) is \( y=\ln(3)-2 \). 3. **Determine Where the Function is Increasing:** Compute the derivative: \[ f'(x)=\frac{d}{dx}[\ln(x+3)-2]=\frac{1}{x+3}. \] Since \( x>-3 \), the denominator \( x+3 \) is always positive, and hence \[ f'(x)>0 \quad \text{for } x\in(-3,\infty). \] This means the function \( f \) is increasing on \((-3,\infty)\). **Correct Choice:** A. The function \( f \) is increasing on the subinterval \((-3, \infty)\). 4. **Determine Where the Function is Decreasing:** As noted, the derivative \( f'(x) \) is positive on the entire domain \((-3,\infty)\), so the function is never decreasing. **Correct Choice:** B. The function \( f \) is never decreasing. 5. **Sketching the Graph:** - **Domain:** \( x>-3 \). - **Vertical Asymptote:** \( x=-3 \) (since \( \ln(x+3) \) is undefined at \( x=-3 \)). - **\( y \)-Intercept:** \( (0,\ln(3)-2) \). - **Behavior:** - The graph increases on the interval \((-3,\infty)\). - It is a shifted version of \( \ln(x) \): shifted left 3 units and down 2 units. - **Key Point:** - \( x \)-intercept can be found by solving \( \ln(x+3)-2=0 \): \[ \ln(x+3)=2 \quad \Longrightarrow \quad x+3=e^2 \quad \Longrightarrow \quad x=e^2-3. \] The summarized answers are: - \( y \)-intercept: \( y=\ln(3)-2 \). - Increasing on: \((-3,\infty)\). - Never decreasing.

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Beyond the Answer

To sketch the graph of \( f(x) = \ln(x + 3) - 2 \), we first identify the \( y \)-intercept. Plugging in \( x = 0 \), we get \( f(0) = \ln(3) - 2 \), confirming the \( y \)-intercept is \( y = \ln(3) - 2 \). Thus, the first choice is A. The function is defined for \( x > -3 \), making the interval for increasing behavior from \( (-3, \infty) \), so the correct answer is A. The function does not decrease anywhere in this interval, making the last choice A. The function \( f(x) \) reflects a vertical shift of the basic logarithmic function down by 2 units. The asymptote occurs at \( x = -3 \), as \( \ln(x + 3) \) approaches negative infinity, giving the graph a leftward edge. Exploring its practical use, this type of logarithmic function can model real-world situations, such as population growth or radioactive decay, where values are constrained but exhibit rapid increases over time. This means you can use it in various fields, like economics or environmental science, to predict behaviors or outcomes based on initial conditions.

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