Find the equation of the tangent and normal to the curve $y = x^3 - 3x + 2$ at $x = 1$.

Step 1: Find the point of tangency

To find the point of tangency, we substitute the given x-coordinate, $x=1$, into the equation of the curve $y = x^3 - 3x + 2$. This will give us the corresponding y-coordinate of the point where the tangent and normal lines touch the curve.

Step 2: Calculate the slope of the tangent

The slope of the tangent to the curve at any point is given by the first derivative of the curve's equation. We differentiate $y = x^3 - 3x + 2$ with respect to x to find $\frac{dy}{dx}$. Then, we evaluate this derivative at $x=1$ to get the specific slope of the tangent line at the point of tangency.

Step 3: Find the slope of the normal

The normal line is perpendicular to the tangent line at the point of tangency. Therefore, the slope of the normal line is the negative reciprocal of the slope of the tangent line. We use the formula $m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}$.

Step 4: Equation of the tangent

Now we can find the equation of the tangent line using the point-slope form: y - $y_1 = m(x - x_1)$. We substitute the point of tangency ($x_1, y_1$) = (1, 0) and the slope of the tangent $m_{\text{tangent}} = 0$ into this formula.

Step 5: Equation of the normal

Similarly, we find the equation of the normal line using the point-slope form. We substitute the point of tangency ($x_1, y_1$) = (1, 0) and the slope of the normal $m_{\text{normal}}$ into the formula. Since the slope of the normal is undefined, this indicates a vertical line.

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