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 Klein Martin. in the United States

Mar 20,2025

Question

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 Klein Martin. in the United States

Mar 20,2025

UpStudy AI · Accepted Answer

It seems that I am unable to generate plots 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 $$ y^2 = x^3 + 3x^2 $$.
   - You can rewrite it as $$ y = \pm \sqrt{x^3 + 3x^2} $$ to plot both the upper and lower parts of the curve.

2. **Graph the Tangent Line**:
   - Use the equation $$ y = \frac{9}{4}x - \frac{1}{4} $$.
   - This is a straight line with a slope of $$ \frac{9}{4} $$ and a y-intercept of $$ -\frac{1}{4} $$.

### Steps to Plot

- **Using Graphing Software**:
  - Open a graphing tool (like Desmos, GeoGebra, or a graphing calculator).
  - Input the curve equation and the tangent line equation.
  - Adjust the viewing window to ensure both the curve and the tangent line are visible.

- **Using a Graphing Calculator**:
  - Enter the curve equation in the appropriate function slot.
  - Enter the tangent line equation in another function slot.
  - Set the viewing window to include the point (1, 2) and observe the intersection.

### Conclusion

You should see the Tschirnhausen cubic curve and the tangent line intersecting at the point (1, 2). The tangent line will touch the curve at this point without crossing it, illustrating the concept of a tangent line. If you have any further questions or need additional assistance, feel free to ask!
To find the tangent line at (1,2), differentiate $$ y^2 = x^3 + 3x^2 $$ to get $$ \frac{dy}{dx} = \frac{3x^2 + 6x}{2y} $$. At (1,2), the slope is $$ \frac{9}{4} $$. The tangent line is $$ y = \frac{9}{4}x - \frac{1}{4} $$. For horizontal tangents, set $$ \frac{dy}{dx} = 0 $$ to find $$ x = 0 $$ and $$ x = -2 $$. The corresponding points are (0,0), (-2,2), and (-2,-2). Graphing the curve and tangent line will show the tangent line touching the curve at (1,2) and having horizontal tangents at (-2,2), (-2,-2), and (0,0).

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