Pregunta
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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 Simpson Fuentes. in the United States
Mar 20,2025

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To find the tangent line at (1,2), the equation is \( y = \frac{9}{4}x - \frac{1}{4} \). The curve has horizontal tangents at \( (-2, -2) \), \( (-2, 2) \), and \( (0, 0) \).

Solución

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 \( y^2 = x^3 + 3x^2 \). - You can rearrange it to \( y = \pm \sqrt{x^3 + 3x^2} \) 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 \( -\frac{1}{4} \). 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: \( (-2, -2) \), \( (-2, 2) \), and \( (0, 0) \). ### 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.

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The Deep Dive

To find the equation of the tangent line to the curve \( y^{2} = x^{3} + 3x^{2} \) at the point \( (1, 2) \), we first need to differentiate the equation implicitly. By applying implicit differentiation, we find that \( 2y \frac{dy}{dx} = 3x^{2} + 6x \). Then, substituting the point \( (1, 2) \), we have \( 2(2)\frac{dy}{dx} = 3(1)^{2} + 6(1) \), which simplifies to \( 4\frac{dy}{dx} = 9 \) or \( \frac{dy}{dx} = \frac{9}{4} \). The tangent line has the equation \( y - 2 = \frac{9}{4}(x - 1) \), or in slope-intercept form, \( y = \frac{9}{4}x - \frac{5}{4} \). For horizontal tangents, we set \( \frac{dy}{dx} = 0 \). This requires \( 3x^{2} + 6x = 0 \), which results in \( x(x + 2) = 0 \), giving \( x = 0 \) and \( x = -2 \). To find the corresponding \( y \) values, substitute back into the original equation. For \( x = 0 \): \( y^{2} = 0 \) → \( y = 0 \). For \( x = -2 \): \( y^{2} = (-2)^{3} + 3(-2)^{2} = -8 + 12 = 4 \) → \( y = \pm 2 \). Thus, the points with horizontal tangents are: \( (0, 0) \), \( (-2, 2) \), and \( (-2, -2) \). Now, graphing the curve \( y^{2} = x^{3} + 3x^{2} \) with the tangent line \( y = \frac{9}{4}x - \frac{5}{4} \) will visually demonstrate how the curve behaves at the specified point and where it intersects at horizontal tangents!

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