The vertex of the parabola is at \((-2, -5)\). The axis of symmetry is the vertical line passing through the vertex.

Answer: \( x = -2 \)

The function is given in the form \( f(x) = \frac{1}{4}(x – h)^2 + k \). The vertex is at \((-2, -5)\), so \( h \) and \( k \) correspond to the x- and y-coordinates of the vertex.

Answer: \( h = -2 \), \( k = -5 \)

The \( y \)-intercept occurs at \( x = 0 \). Using the function \( f(x) = \frac{1}{4}(x + 2)^2 – 5 \):

\[ f(0) = \frac{1}{4}(0 + 2)^2 – 5 = \frac{1}{4} \cdot 4 – 5 = 1 – 5 = -4 \]

Answer: The \( y \)-coordinate of \( P \) is \( -4 \).

Line \( L \) is normal to \( f(x) \) at \( P(0, -4) \). First, find the slope of the tangent at \( x = 0 \):

\[ f(x) = \frac{1}{4}(x + 2)^2 – 5 \]

\[ f'(x) = \frac{1}{4} \cdot 2(x + 2) = \frac{1}{2}(x + 2) \]

\[ f'(0) = \frac{1}{2}(0 + 2) = 1 \]

The slope of the normal is the negative reciprocal: \( m = -\frac{1}{1} = -1 \).

Using point-slope form at \( P(0, -4) \):

\[ y – (-4) = -1(x – 0) \]

\[ y + 4 = -x \implies y = -x – 4 \]

Answer: \( y = -x – 4 \)

Find point \( Q \), the intersection of \( f(x) \) and line \( L \):

\[ f(x) = \frac{1}{4}(x + 2)^2 – 5 \]

\[ L: y = -x – 4 \]

Set \( f(x) = L \):

\[ \frac{1}{4}(x + 2)^2 – 5 = -x – 4 \]

Multiply by 4:

\[ (x + 2)^2 – 20 = -4x – 16 \]

\[ x^2 + 4x + 4 – 20 + 4x + 16 = 0 \]

\[ x^2 + 8x = 0 \]

\[ x(x + 8) = 0 \implies x = 0 \text{ or } x = -8 \]

\( x = 0 \) corresponds to \( P \), so \( Q \) is at \( x = -8 \).

Substitute \( x = -8 \) into \( L \):

\[ y = -(-8) – 4 = 8 – 4 = 4 \]

Thus, \( Q(-8, 4) \).

Calculate the distance between \( P(0, -4) \) and \( Q(-8, 4) \):

\[ \text{Distance} = \sqrt{(-8 – 0)^2 + (4 – (-4))^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2} \]

Answer: \( 8\sqrt{2} \)