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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!

Tentukan persamaan garis singgung pada kurva berikut: a. y = -x^2 - 6x + 3 di titik yang berabasis -2, b. y = x^2 + 4x - 5 di titik (2, -16), c. y = 4/x di titik (2, 2).

This problem involves finding the equation of tangent lines to different curves using differentiation. The solutions apply the derivative of functions and the point-slope form of a line. Suitable for students in Grades 10-12.

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