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 Moreno Montgomery. 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 Moreno Montgomery. in the United States

Mar 20,2025

UpStudy AI · Accepted Answer

It seems that I am unable to generate plots directly at this moment. 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 positive and negative branches 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 Graph Using a Graphing Calculator:

- Input the curve equation $$ y = \sqrt{x^3 + 3x^2} $$ and $$ y = -\sqrt{x^3 + 3x^2} $$ into the graphing calculator.
- Input the tangent line equation $$ y = \frac{9}{4}x - \frac{1}{4} $$.
- Set an appropriate viewing window to see the behavior of the curve and the tangent line around the point (1, 2).

### Expected Outcome

You should see the Tschirnhausen cubic curve with the tangent line intersecting it at the point (1, 2). The tangent line will touch the curve at this point and will not cross it, indicating that it is indeed a tangent.

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 $$ \frac{dy}{dx} = 0 $$, which gives $$ x = 0 $$ and $$ x = -2 $$. Substituting back, the points are (0,0), (-2,2), and (-2,-2). Graphing the curve and tangent line will show the Tschirnhausen cubic with the tangent line touching at (1,2).

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Instructions for Graphing

Steps to Graph Using a Graphing Calculator:

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