Let's solve one or more of the selected problems (marked #43, #45, #47). The goal is to find the equation of the tangent line at the given point for the function $f(x)$. I'll start with #43 and proceed to the others if needed.

Problem #43:

$f(x) = (x - 1)^2; \quad (-2, 9)$

Steps:

1. Find $f'(x)$ (the derivative): $f'(x) = 2(x - 1)$

2. Evaluate $f'(x)$ at $x = -2$: $f'(-2) = 2(-2 - 1) = 2(-3) = -6$

   The slope of the tangent line at $x = -2$ is $-6$.

3. Point-slope form of the tangent line: The point is $(-2, 9)$, and the slope is $-6$. Using the point-slope form of a line: $y - y_1 = m(x - x_1)$ Substitute $y_1 = 9$, $m = -6$, and $x_1 = -2$: $y - 9 = -6(x + 2)$

4. Simplify the equation: $y - 9 = -6x - 12$ $y = -6x - 3$

   The equation of the tangent line is: $y = -6x - 3$

Would you like me to proceed with #45, #47, or verify results graphically? Let me know if you'd like detailed steps for any of these! Here are related questions to consider:

1. What is the geometric meaning of the derivative in tangent line problems?
2. How can we verify the tangent line equation graphically?
3. How does the slope vary for different functions at different points?
4. Why is point-slope form convenient for tangent line equations?
5. Can these techniques be applied to parametric or implicit functions?

Tip: Always cross-check your tangent line by substituting the given point into the equation to ensure it satisfies the line.