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Desi Given the furction \( \quad f(x)=(x+2) e^{-2} \). ancoer the following questions recending its properties and behavour, and then shetch its eraph Damsin lintercegts with thex axis and the y-axs Irrervals in which Be function is poultie or negative Limits Equations of horironta//vertical/slant asymptotes First derivative Intervals in which the derivative is positive or negative.

Ask by Young Riley. in Italy
Feb 19,2025

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Answer

The function \( f(x) = (x + 2)e^{-2} \) has an x-intercept at \( (-2, 0) \) and a y-intercept at \( (0, 2e^{-2}) \). It is positive for \( x > -2 \) and negative for \( x < -2 \). The function increases for all \( x \) and has no horizontal, vertical, or slant asymptotes. The first derivative is a constant \( e^{-2} \), indicating the function is always increasing.

Solution

It seems that I am unable to create a plot at the moment. However, I can describe how to sketch the graph based on the analysis: 1. **Draw the Axes:** Start by drawing the x-axis and y-axis. 2. **Plot the Intercepts:** - Mark the x-intercept at \( (-2, 0) \). - Mark the y-intercept at \( (0, 2e^{-2}) \) (approximately \( (0, 0.27) \)). 3. **Behavior of the Function:** - The function is negative for \( x < -2 \) and positive for \( x > -2 \). - As \( x \) approaches \( -\infty \), the function goes to \( -\infty \). - As \( x \) approaches \( +\infty \), the function goes to \( +\infty \). 4. **Draw the Line:** - Start from the left, drawing a curve that approaches the x-axis but never touches it (as it goes to \( -\infty \)). - At \( x = -2 \), the curve crosses the x-axis. - Continue the curve upwards, passing through the y-intercept and increasing without bound as \( x \) increases. 5. **Label the Graph:** - Label the x-intercept and y-intercept. - Indicate that the function is increasing for all \( x \). This sketch will represent the function \( f(x) = (x + 2)e^{-2} \) accurately based on the properties we analyzed. If you have graphing software or graph paper, you can use the points and behavior described to create a precise graph.

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Bonus Knowledge

To tackle the function \( f(x) = (x + 2)e^{-2} \), we can identify some interesting properties! First, let’s determine intercepts. The y-intercept occurs when \( x = 0 \) leading to \( f(0) = 2e^{-2} \). For the x-intercept, we set \( f(x) = 0 \), giving us \( x + 2 = 0 \), hence \( x = -2 \). Now, let’s look at its behavior: the function is linear since it can be expressed as \( f(x) = e^{-2} x + 2e^{-2} \). Being a linear function, it will have constant positive or negative behavior depending on the value of the slope \( e^{-2} \), which is always positive. Thus, \( f(x) \) is positive for \( x > -2 \) and negative for \( x < -2 \). On the limit fronts, as \( x \to \infty \), \( f(x) \to \infty \) and as \( x \to -\infty \), \( f(x) \to -\infty \). For asymptotes, there are none since it's a linear function; horizontal and vertical asymptotes apply to different function types. The first derivative \( f'(x) = e^{-2} \), indicating it doesn’t change—it’s positive everywhere. Therefore, the function is always increasing, confirming it would rise towards infinity. For further exploration, consider derivatives in connection to rates of change and motion—concepts beautifully illustrated in Graph Theory or Calculus. Delving into those areas can help illuminate how functions behave in real-world contexts, such as predicting population growth or analyzing financial trends.

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