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The task in this image asks for finding the equation of the tangent line to the parabola \( y = x^2 - 6x + 7 \) at the point \( (2, -1) \).

### Steps to solve:

1. **Find the derivative of the given parabola:**

   The function is \( y = x^2 - 6x + 7 \). Its derivative gives us the slope of the tangent line at any point \( x = a \).

   \[
   f'(x) = 2x - 6
   \]

2. **Evaluate the derivative at \( x = 2 \) to find the slope of the tangent line at the point \( (2, -1) \):**

   \[
   f'(2) = 2(2) - 6 = 4 - 6 = -2
   \]

   Therefore, the slope of the tangent line at the point \( (2, -1) \) is \( -2 \).

3. **Use the point-slope form of the equation of a line:**

   The point-slope form is:
   \[
   y - y_1 = m(x - x_1)
   \]
   where \( m \) is the slope, and \( (x_1, y_1) \) is the point of tangency.

   Plugging in \( m = -2 \), \( x_1 = 2 \), and \( y_1 = -1 \):
   \[
   y - (-1) = -2(x - 2)
   \]
   Simplifying:
   \[
   y + 1 = -2(x - 2)
   \]
   Distribute:
   \[
   y + 1 = -2x + 4
   \]
   Subtract 1 from both sides:
   \[
   y = -2x + 3
   \]

Thus, the equation of the tangent line is \( y = -2x + 3 \).

Would you like a detailed explanation or more examples? Here are some related questions:

1. How do you find the derivative of more complex functions?
2. What is the geometric meaning of a tangent line?
3. Can you always use the point-slope form to find the equation of a tangent line?
4. How does the second derivative affect the curvature at the point of tangency?
5. What happens when the slope of the tangent line is zero?

**Tip:** Always check the slope at the point of tangency; a zero slope implies a horizontal tangent line.

Find an equation of the tangent line to the parabola y = x^2 - 6x + 7 at the point (2, -1).

Learn how to find the equation of a tangent line to the parabola y = x^2 - 6x + 7 at the point (2, -1). This problem involves differentiation and uses the point-slope form of the equation of a line. Suitable for students in Grades 11-12.

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