Find an equation of the tangent line to the curve at the given point. \[ y=3 x^{2}-6 x+1, \quad(3,10) \] \( y=\square \)

To find the equation of the tangent line to the curve \( y = 3x^2 - 6x + 1 \) at the point \( (3, 10) \), follow these steps: 1. **Find the derivative \( y' \):** The derivative of \( y \) with respect to \( x \) gives the slope of the tangent line at any point \( x \). \[ y = 3x^2 - 6x + 1 \] \[ y' = \frac{dy}{dx} = 6x - 6 \] 2. **Evaluate the derivative at \( x = 3 \):** This gives the slope of the tangent line at the specific point. \[ y'(3) = 6(3) - 6 = 18 - 6 = 12 \] So, the slope \( m \) of the tangent line is **12**. 3. **Use the point-slope form to write the equation of the tangent line:** The point-slope form is \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is the point of tangency. \[ y - 10 = 12(x - 3) \] 4. **Simplify the equation:** \[ y - 10 = 12x - 36 \] \[ y = 12x - 26 \] **Final Answer:** \[ y = 12\,x\ -\ 26 \]