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A graphing celculator is recommended. (a) The curve with ecuation \( y^{2}=x^{3}+3 x^{2} \) 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 \( x \), then from smallest to largest \( y \) ) \( \qquad(x, y)= \) ( \( x, y)= \) (b) by graphing the curve and the tangent lines on a common screen.

Ask by Moreno Montgomery. in the United States
Mar 20,2025

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Answer

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).

Solution

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!

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Extra Insights

To find the tangent line at the point (1, 2) on the curve \( y^{2} = x^{3} + 3x^{2} \), we first need to derive the slope. Using implicit differentiation, we get \( 2y \frac{dy}{dx} = 3x^{2} + 6x \). Evaluating this at (1, 2) gives us \( 2(2) \frac{dy}{dx} = 3(1)^{2} + 6(1) \) leading to \( 4 \frac{dy}{dx} = 9 \) or \( \frac{dy}{dx} = \frac{9}{4} \). Thus, the equation of the tangent line at (1, 2) is \( y - 2 = \frac{9}{4}(x - 1) \). Now, to find points with horizontal tangents, we set \( \frac{dy}{dx} = 0 \) which implies \( 3x^{2} + 6x = 0 \). Factoring out gives \( 3x(x + 2) = 0 \), giving us \( x = 0 \) and \( x = -2 \). For these \( x \) values, substituting back to find \( y \) leads to \( y^{2} = 0 \) when \( x = 0 \) (so \( y = 0 \)), and \( y^{2} = 4 \) when \( x = -2 \) (thus \( y = 2 \) or \( y = -2 \)). Therefore, the horizontal tangent points are \( (x, y) = (-2, -2) \), \( (-2, 2) \), and \( (0, 0) \). Feel free to graph these points along with the curve \( y^{2} = x^{3} + 3x^{2} \) and draw the tangent line — it’s a great way to visualize how these elements interact! Don’t forget to check out online graphing tools for interactive experiences!

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