The question asks to find the equation of the tangent line to various curves at specific points. Let's go step-by-step for each part:

Part a: $y = -x^2 - 6x + 3$ at $x = -2$

To find the equation of the tangent line, we need the derivative of the function to get the slope at the point and then use the point-slope form of the line equation.

1. Find the derivative: $\frac{dy}{dx} = -2x - 6$

2. Evaluate the derivative at $x = -2$ to get the slope: $\frac{dy}{dx} \bigg|_{x = -2} = -2(-2) - 6 = 4 - 6 = -2$ So the slope at $x = -2$ is $-2$.

3. Find the corresponding $y$-coordinate at $x = -2$: $y = -(-2)^2 - 6(-2) + 3 = -4 + 12 + 3 = 11$ The point is $(-2, 11)$.

4. Use the point-slope form: $y - y_1 = m(x - x_1)$ $y - 11 = -2(x + 2)$ $y - 11 = -2x - 4$ $y = -2x + 7$ The equation of the tangent line is $y = -2x + 7$.

Part b: $y = x^2 + 4x - 5$ at $(2, -16)$

1. Find the derivative: $\frac{dy}{dx} = 2x + 4$

2. Evaluate the derivative at $x = 2$: $\frac{dy}{dx} \bigg|_{x = 2} = 2(2) + 4 = 4 + 4 = 8$ The slope at $x = 2$ is $8$.

3. The point is already given as $(2, -16)$.

4. Use the point-slope form: $y - y_1 = m(x - x_1)$ $y + 16 = 8(x - 2)$ $y + 16 = 8x - 16$ $y = 8x - 32$ The equation of the tangent line is $y = 8x - 32$.

Part c: $y = \frac{4}{x}$ at $(2, 2)$

1. Find the derivative: $\frac{dy}{dx} = -\frac{4}{x^2}$

2. Evaluate the derivative at $x = 2$: $\frac{dy}{dx} \bigg|_{x = 2} = -\frac{4}{2^2} = -\frac{4}{4} = -1$ The slope at $x = 2$ is $-1$.

3. The point is $(2, 2)$.

4. Use the point-slope form: $y - 2 = -1(x - 2)$ $y - 2 = -x + 2$ $y = -x + 4$ The equation of the tangent line is $y = -x + 4$.

Would you like further details on any of these steps, or do you have any specific questions?

Related questions:

1. What is the general process for finding the equation of a tangent line?
2. How do you differentiate a rational function like $y = \frac{4}{x}$?
3. Why is the slope of a tangent line important in calculus?
4. How does the point-slope form differ from the slope-intercept form of a line?
5. What is the significance of second derivatives in understanding curves?

Tip: The point-slope form $y - y_1 = m(x - x_1)$ is very useful when you have both a point and a slope—it allows quick derivation of the tangent line equation!