Question
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A graphing celculator is recommended.
(a) The curve with ecuation is called the Tschirnhausen cubic. Find an equation of the tangent line to this curve at the point ( 1,2 )
(t) At what points does this curve have horizontal tangents? (Order your answers from smallest to largest , then from smallest to largest )
( (b) by graphing the curve and the tangent lines on a common screen.

Ask by Simpson Fuentes. in the United States
Mar 20,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

To find the tangent line at (1,2), the equation is . The curve has horizontal tangents at , , and .

Solution

It seems that the system does not currently support plotting the graphs directly. However, I can guide you on how to graph the curve and the tangent line using graphing software or a graphing calculator.

Instructions for Graphing

  1. Graph the Curve:
    • Use the equation .
    • You can rearrange it to to plot both the positive and negative branches of the curve.
  2. Graph the Tangent Line:
    • Use the equation .
    • This is a straight line with a slope of and a y-intercept of .
  3. Set the Viewing Window:
    • For both graphs, set the x-axis from approximately -4 to 2 and the y-axis from -3 to 3 to capture the relevant features of the curve and the tangent line.

Points to Mark

  • Mark the point of tangency (1, 2) on the graph.
  • Also, mark the points where the curve has horizontal tangents: , , and .

Conclusion

By following these instructions, you should be able to visualize the Tschirnhausen cubic and its tangent line effectively. If you have access to graphing software like Desmos, GeoGebra, or a graphing calculator, you can input the equations directly to see the graphs.

Answered by UpStudy AI and reviewed by a Professional Tutor

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The Deep Dive

To find the equation of the tangent line to the curve at the point , we first need to differentiate the equation implicitly. By applying implicit differentiation, we find that . Then, substituting the point , we have , which simplifies to or . The tangent line has the equation , or in slope-intercept form, .
For horizontal tangents, we set . This requires , which results in , giving and . To find the corresponding values, substitute back into the original equation. For : . For : . Thus, the points with horizontal tangents are: , , and .
Now, graphing the curve with the tangent line will visually demonstrate how the curve behaves at the specified point and where it intersects at horizontal tangents!

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