To find the equation of the tangent line to the function $y = 3(x + 1)^2$ at the point $(2, 27)$, we follow these steps:

Step 1: Find the derivative of the function

The derivative of a function represents the slope of the tangent line at any point. To find the derivative of $y = 3(x + 1)^2$, we apply the chain rule:

$\frac{dy}{dx} = 3 \cdot 2(x + 1) \cdot \frac{d}{dx}(x + 1)$ Since $\frac{d}{dx}(x + 1) = 1$, the derivative simplifies to: $\frac{dy}{dx} = 6(x + 1)$

Step 2: Evaluate the derivative at $x = 2$

The slope of the tangent line at $x = 2$ is found by plugging $x = 2$ into the derivative: $\frac{dy}{dx} \bigg|_{x = 2} = 6(2 + 1) = 6 \times 3 = 18$

Thus, the slope of the tangent line at $(2, 27)$ is 18.

Step 3: Use the point-slope form of the tangent line

The point-slope form of a line is given by: $y - y_1 = m(x - x_1)$ where $m$ is the slope and $(x_1, y_1)$ is the point of tangency. In this case, $m = 18$ and $(x_1, y_1) = (2, 27)$.

Substituting these values into the point-slope form: $y - 27 = 18(x - 2)$ Simplifying the equation: $y - 27 = 18x - 36$ $y = 18x - 36 + 27$ $y = 18x - 9$

Final Answer:

The equation of the tangent line is: $y = 18x - 9$

Would you like further explanation or details on any of the steps?

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Here are 5 related questions for further exploration:

1. How do you find the equation of a tangent line for implicit functions?
2. What is the geometric interpretation of a derivative at a point?
3. How do you apply the chain rule to more complex functions?
4. What is the difference between a secant line and a tangent line?
5. How would the equation change if the function was cubic?

Tip: When finding the tangent line, always double-check if the point given lies on the curve by substituting it into the original function.