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.
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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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