To find the equation of the tangent line to the graph of $f(x) = \sqrt{x + 12}$ at the point $(0, 11)$, we need to follow these steps:

Step 1: Find the derivative $f'(x)$

The derivative $f'(x)$ represents the slope of the tangent line at any point $x$.

$f(x) = \sqrt{x + 12} = (x + 12)^{1/2}$

Using the chain rule, the derivative is:

$f'(x) = \frac{1}{2}(x + 12)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{x + 12}}$

Step 2: Evaluate $f'(x)$ at $x = 0$

We need to find the slope of the tangent line at the point $(0, 11)$. Substitute $x = 0$:

$f'(0) = \frac{1}{2\sqrt{0 + 12}} = \frac{1}{2\sqrt{12}} = \frac{1}{2 \cdot 2\sqrt{3}} = \frac{1}{4\sqrt{3}}$

To simplify, we rationalize the denominator:

$f'(0) = \frac{\sqrt{3}}{12}$

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

The point-slope form of a line is:

$y - y_1 = m(x - x_1)$

where $(x_1, y_1) = (0, 11)$ and $m = \frac{\sqrt{3}}{12}$. Substituting these values:

$y - 11 = \frac{\sqrt{3}}{12}(x - 0)$

$y = \frac{\sqrt{3}}{12}x + 11$

Thus, the equation of the tangent line at the point $(0, 11)$ is:

$\boxed{y = \frac{\sqrt{3}}{12}x + 11}$

Would you like more details or have any other questions?

Here are 5 related questions you might find helpful:

1. How do you find the derivative of functions involving roots?
2. What is the chain rule in calculus, and how is it applied?
3. How do you rationalize the denominator when simplifying expressions?
4. How can you determine if a tangent line is horizontal or vertical?
5. What is the general method to find the tangent line to any curve?

Tip: Remember that finding the derivative helps in understanding the rate of change or slope at a specific point on a curve.