solve the euqation of line L=ax+b and it is normal to the graph of f at point P which is 0,-4. the equation for f is 1/4(x-h)^2+k where h=-2 and k=-5

We are given the following information:

- The line \( L = ax + b \) is normal to the graph of the function \( f(x) = \frac{1}{4}(x - h)^2 + k \) at the point \( P(0, -4) \).
- The function \( f(x) = \frac{1}{4}(x - (-2))^2 + (-5) \), or simplified:
  \[
  f(x) = \frac{1}{4}(x + 2)^2 - 5.
  \]

### Steps to solve:
1. **Find the derivative of \( f(x) \):**

   The derivative \( f'(x) \) will give us the slope of the tangent to the graph at any point \( x \). Since \( f(x) = \frac{1}{4}(x + 2)^2 - 5 \), we use the chain rule to differentiate:
   \[
   f'(x) = \frac{1}{4} \cdot 2(x + 2) = \frac{1}{2}(x + 2).
   \]

2. **Find the slope of the tangent at \( P(0, -4) \):**

   To find the slope of the tangent at \( P(0, -4) \), substitute \( x = 0 \) into \( f'(x) \):
   \[
   f'(0) = \frac{1}{2}(0 + 2) = 1.
   \]
   So, the slope of the tangent at \( P(0, -4) \) is \( 1 \).

3. **Find the slope of the normal line:**

   The slope of the normal line is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent is \( 1 \), the slope of the normal line is:
   \[
   a = -\frac{1}{1} = -1.
   \]

4. **Find the equation of the normal line:**

   The normal line passes through the point \( P(0, -4) \) and has a slope of \( -1 \). Using the point-slope form of the equation of a line:
   \[
   y - y_1 = m(x - x_1),
   \]
   where \( (x_1, y_1) = (0, -4) \) and \( m = -1 \), we get:
   \[
   y - (-4) = -1(x - 0),
   \]
   \[
   y + 4 = -x,
   \]
   \[
   y = -x - 4.
   \]
   Therefore, the equation of the line \( L \) is:
   \[
   L: y = -x - 4.
   \]

### Final Answer:
The equation of the line \( L \) is \( \boxed{y = -x - 4} \).

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Do you have any questions, or would you like more details?

Here are 5 related questions to deepen your understanding:
1. How do you find the equation of a tangent line to a curve?
2. What is the difference between the tangent and normal lines to a curve?
3. How do you determine the slope of a curve at a specific point?
4. How does the chain rule help in differentiating composite functions?
5. How do you find the derivative of a quadratic function?

**Tip:** When dealing with normal lines, remember that their slope is always the negative reciprocal of the slope of the tangent line at the point of intersection.

Solve the equation of line L = ax + b, and it is normal to the graph of f at point P(0, -4). The equation for f is (1/4)(x - h)^2 + k where h = -2 and k = -5.

Find the equation of a line normal to the parabola f(x) = (1/4)(x + 2)^2 - 5 at the point P(0, -4). This problem involves finding the derivative of the function to determine the tangent and normal slopes, followed by applying the point-slope form of a line. Suitable for students in Grades 10-12.

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