Pregunta
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\( \begin{array}{l}\text { Describe how the graph of } g(x)=\frac{1}{3} x^{3}+5 \text { can be obtained from the } g r a p h ~ o f ~\end{array}(x)=x^{3} \). Then graph the function \( g(x) \).

Ask by Sullivan Turner. in the United States
Mar 12,2025

Solución de inteligencia artificial de Upstudy

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To graph \( g(x) = \frac{1}{3} x^3 + 5 \), start with the basic cubic function \( f(x) = x^3 \), compress it vertically by a factor of \( \frac{1}{3} \), and then shift it upwards by 5 units. Key points to plot are \( (0, 5) \), \( (1, \frac{16}{3}) \), and \( (-1, \frac{14}{3}) \).

Solución

It seems that I am unable to generate the graph directly. However, I can guide you on how to graph \( g(x) = \frac{1}{3} x^3 + 5 \) using graphing software or a graphing calculator. ### Steps to Graph \( g(x) \): 1. **Plot the Basic Function**: Start by plotting the basic cubic function \( f(x) = x^3 \). This function has the following characteristics: - It passes through the origin (0,0). - It is symmetric about the origin. - It increases steeply as \( x \) moves away from zero. 2. **Apply the Vertical Compression**: For \( g(x) = \frac{1}{3} x^3 \): - The graph will be less steep than \( f(x) \). For example, at \( x = 1 \), \( f(1) = 1 \) while \( g(1) = \frac{1}{3} \). - At \( x = -1 \), \( f(-1) = -1 \) while \( g(-1) = -\frac{1}{3} \). 3. **Shift Upwards**: Finally, shift the entire graph of \( \frac{1}{3} x^3 \) upwards by 5 units: - The new points will be \( (0, 5) \), \( (1, \frac{1}{3} + 5) = (1, \frac{16}{3}) \), and \( (-1, -\frac{1}{3} + 5) = (-1, \frac{14}{3}) \). ### Key Points to Plot: - \( (0, 5) \) - \( (1, \frac{16}{3}) \approx (1, 5.33) \) - \( (-1, \frac{14}{3}) \approx (-1, 4.67) \) ### Conclusion: By following these steps, you can accurately graph \( g(x) = \frac{1}{3} x^3 + 5 \). If you have access to graphing software, you can input the function directly to visualize it.

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Mind Expander

To obtain the graph of \( g(x) = \frac{1}{3} x^{3} + 5 \) from the graph of \( f(x) = x^3 \), you can start by vertically stretching the original cubic function \( f(x) \) by a factor of \( \frac{1}{3} \) and then shifting the entire graph upward by 5 units. This means the steepness of \( f(x) \) will be reduced, producing a wider curve, and placing all points of the graph 5 units higher on the vertical axis. Graphing \( g(x) \) would show a cubic curve that starts from the lower left, gradually curls through the origin point at (0, 5), and rises through the upper right. The inflection point remains at the origin due to the nature of cubic equations, maintaining the characteristic "S" shape of cubic functions but wider and shifted up. So, the critical features are preserved, just adjusted in scale and position!

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