Question: Find an equation of the tangent line to the graph of the function $$f(x) = x^2 + 3x + 3$$ at the point $(-3, 3)$.
1. **State the problem:** We need to find the equation of the tangent line to the curve defined by $$f(x) = x^2 + 3x + 3$$ at the point $$(-3, 3)$$.
2. **Recall the formula for the tangent line:** The equation of the tangent line to a function $$f(x)$$ at $$x = a$$ is given by:
$$y = f'(a)(x - a) + f(a)$$
where $$f'(a)$$ is the derivative of $$f(x)$$ evaluated at $$x = a$$.
3. **Find the derivative:**
$$f(x) = x^2 + 3x + 3$$
Using the power rule:
$$f'(x) = 2x + 3$$
4. **Evaluate the derivative at $$x = -3$$:**
$$f'(-3) = 2(-3) + 3 = -6 + 3 = -3$$
5. **Use the point and slope to write the tangent line equation:**
Given $$a = -3$$, $$f(a) = 3$$, and $$f'(a) = -3$$, substitute into the tangent line formula:
$$y = -3(x - (-3)) + 3 = -3(x + 3) + 3$$
6. **Simplify the equation:**
$$y = -3x - 9 + 3 = -3x - 6$$
7. **Final answer:** The equation of the tangent line is:
$$y = -3x - 6$$
This corresponds to **Option C**.