Let C be the curve y^3 - 3xy + 2 = 0. If H an | vedclass

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(B) Given the curve equation: $y^3 - 3xy + 2 = 0$. Differentiating with respect to $x$: $3y^2 \frac{dy}{dx} - 3y - 3x \frac{dy}{dx} = 0$ $\frac{dy}{dx

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$H = \phi, V = \{(1, 1)\}$

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(B) Given the curve equation: $y^3 - 3xy + 2 = 0$. Differentiating with respect to $x$: $3y^2 \frac{dy}{dx} - 3y - 3x \frac{dy}{dx} = 0$ $\frac{dy}{dx}(3y^2 - 3x) = 3y$ $\frac{dy}{dx} = \frac{y}{y^2 - x}$. For the tangent to be horizontal,$\frac{dy}{dx} = 0$,which implies $y = 0$. Substituting $y = 0$ into the curve equation: $0^3 - 3x(0) + 2 = 0 \Rightarrow 2 = 0$,which is impossible. Thus,there are no points where the tangent is horizontal,so $H = \phi$. For the tangent to be vertical,$\frac{dy}{dx} = \infty$,which implies the denominator $y^2 - x = 0$,so $x = y^2$. Substituting $x = y^2$ into the curve equation: $y^3 - 3(y^2)y + 2 = 0$ $y^3 - 3y^3 + 2 = 0$ $-2y^3 + 2 = 0 \Rightarrow y^3 = 1 \Rightarrow y = 1$. Since $x = y^2$,$x = 1^2 = 1$. Thus,$V = \{(1, 1)\}$.

Let $C$ be the curve $y^3 - 3xy + 2 = 0$. If $H$ and $V$ are the sets of points on the curve $C$ where the tangent to the curve is horizontal and vertical respectively,then

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