1. **Problem 1: Find the equation of the tangent to the curve $y = x^3 - 5x + 1$ at $x=2$.** 2. The formula for the slope of the tangent line to a curve at a point is the derivative of the function evaluated at that point: $$m = \frac{dy}{dx} \bigg|_{x=a}$$ 3. First, find the derivative of $y = x^3 - 5x + 1$: $$\frac{dy}{dx} = 3x^2 - 5$$ 4. Evaluate the derivative at $x=2$ to find the slope of the tangent: $$m = 3(2)^2 - 5 = 3 \times 4 - 5 = 12 - 5 = 7$$ 5. Find the $y$-coordinate of the point on the curve at $x=2$: $$y = (2)^3 - 5(2) + 1 = 8 - 10 + 1 = -1$$ 6. Use the point-slope form of a line equation: $$y - y_1 = m(x - x_1)$$ where $(x_1, y_1) = (2, -1)$ and $m=7$. 7. Substitute values: $$y - (-1) = 7(x - 2)$$ $$y + 1 = 7x - 14$$ 8. Simplify to get the tangent line equation: $$y = 7x - 15$$ **Final answer:** The equation of the tangent line at $x=2$ is $$y = 7x - 15$$.