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 at the point (1,2), you first need to find the derivative of the curve. The given equation is \( y^2 = x^3 + 3x^2 \). To differentiate implicitly, we get \( 2y \frac{dy}{dx} = 3x^2 + 6x \). Solving for \(\frac{dy}{dx}\) gives us \(\frac{dy}{dx} = \frac{3x^2 + 6x}{2y}\). Substituting \( x = 1 \) and \( y = 2 \), we find that the slope at this point is \(\frac{3(1)^2 + 6(1)}{2(2)} = \frac{9}{4}\). Thus, the equation of the tangent line in point-slope form is \( y - 2 = \frac{9}{4}(x - 1) \). Next, for horizontal tangents, we need to set the derivative \(\frac{dy}{dx}\) to zero, which occurs when \(3x^2 + 6x = 0\). Factoring gives us \(3x(x + 2) = 0\), leading to \(x = 0\) and \(x = -2\). Now, we find the corresponding \(y\)-values by substituting \(x = 0\) and \(x = -2\) back into the original equation. For \(x = 0\), \(y^2 = 0 + 0\) which gives \(y = 0\). For \(x = -2\), \(y^2 = (-2)^3 + 3(-2)^2 = -8 + 12 = 4\) so \(y = ±2\). Thus, the horizontal tangents occur at \((-2, -2)\), \((-2, 2)\), and \((0, 0)\). To visualize the curve and tangent lines, you can use graphing software to plot \( y^2 = x^3 + 3x^2 \) along with the line \( y - 2 = \frac{9}{4}(x - 1) \) at the point (1, 2). The horizontal tangents can also be included at the identified points! Happy graphing!
