To analyze the curve given by the equation $y^2 = x^3 - x + 2$, we need to find points with vertical and horizontal tangent lines. Let’s go through the steps to determine both types of points.

Step 1: Find the Derivative

To find tangent lines, we’ll differentiate implicitly with respect to $x$ and solve for $\frac{dy}{dx}$ to find the slope of the tangent.

Given: $y^2 = x^3 - x + 2$

Differentiating both sides with respect to $x$: $2y \frac{dy}{dx} = 3x^2 - 1$

Solving for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{3x^2 - 1}{2y}$

Step 2: Conditions for Vertical and Horizontal Tangents

• Horizontal Tangent: Occurs when $\frac{dy}{dx} = 0$, which requires $3x^2 - 1 = 0$.
• Vertical Tangent: Occurs when the denominator $2y = 0$, implying $y = 0$.

Step 3: Find Points with Horizontal Tangents

1. Solve $3x^2 - 1 = 0$: $3x^2 = 1 \implies x^2 = \frac{1}{3} \implies x = \pm \frac{1}{\sqrt{3}}$

2. Substitute these $x$-values into the original equation to find the corresponding $y$-values.

   • For $x = \frac{1}{\sqrt{3}}$: $y^2 = \left(\frac{1}{\sqrt{3}}\right)^3 - \frac{1}{\sqrt{3}} + 2 = \frac{1}{3\sqrt{3}} - \frac{1}{\sqrt{3}} + 2 = 2 - \frac{2}{\sqrt{3}}$
   • Similarly, for $x = -\frac{1}{\sqrt{3}}$.

3. Find corresponding $y$-values for both cases.

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