1. **State the problem:** Find the equation of the tangent line to the curve $y=3x^2+12x+12$ at the point where $x=1.1$. 2. **Recall the formula:** The slope of the tangent line to a curve at a point is given by the derivative $y'$. The equation of the tangent line at $x=a$ is: $$y = y'(a)(x - a) + y(a)$$ 3. **Find the derivative:** $$y = 3x^2 + 12x + 12$$ $$y' = \frac{d}{dx}(3x^2) + \frac{d}{dx}(12x) + \frac{d}{dx}(12) = 6x + 12 + 0 = 6x + 12$$ 4. **Evaluate the derivative at $x=1.1$ to find the slope:** $$y'(1.1) = 6(1.1) + 12 = 6.6 + 12 = 18.6$$ 5. **Evaluate the original function at $x=1.1$ to find the point on the curve:** $$y(1.1) = 3(1.1)^2 + 12(1.1) + 12 = 3(1.21) + 13.2 + 12 = 3.63 + 13.2 + 12 = 28.83$$ 6. **Write the equation of the tangent line:** $$y = 18.6(x - 1.1) + 28.83$$ 7. **Simplify the tangent line equation:** $$y = 18.6x - 18.6 \times 1.1 + 28.83 = 18.6x - 20.46 + 28.83 = 18.6x + 8.37$$ **Final answer:** The equation of the tangent line at $x=1.1$ is: $$y = 18.6x + 8.37$$