1. **Problem statement:** We have a quadratic function $f$ with vertex at $(-2,-5)$ and y-intercept at point $P$. We need to find: (a) the axis of symmetry, (b) values of $h$ and $k$ in $f(x)=\frac{1}{4}(x-h)^2+k$, (c) the y-coordinate of $P$, (d) the equation of line $L$ passing through $P$ with gradient $-1$, (e) the distance between points $P$ and $Q$ where $Q$ is the second intersection of $L$ and $f$. 2. **Axis of symmetry:** The axis of symmetry of a parabola with vertex $(h,k)$ is the vertical line $x=h$. Since vertex is $(-2,-5)$, axis of symmetry is: $$x=-2$$ 3. **Values of $h$ and $k$:** Given $f(x)=\frac{1}{4}(x-h)^2+k$ and vertex $(h,k)=(-2,-5)$, so: $$h=-2, \quad k=-5$$ 4. **Y-coordinate of $P$ (y-intercept):** The y-intercept occurs at $x=0$, so: $$f(0)=\frac{1}{4}(0+2)^2 -5 = \frac{1}{4}(2)^2 -5 = \frac{1}{4} \times 4 -5 = 1 -5 = -4$$ So, $P=(0,-4)$. 5. **Equation of line $L$ with gradient $-1$ passing through $P(0,-4)$:** Using point-slope form: $$y - y_1 = m(x - x_1)$$ $$y + 4 = -1(x - 0)$$ $$y = -x - 4$$ 6. **Find point $Q$ where line $L$ intersects $f$ again:** Set $f(x) = y$ and line $L$ equation equal: $$\frac{1}{4}(x+2)^2 -5 = -x -4$$ Multiply both sides by 4: $$(x+2)^2 - 20 = -4x -16$$ Expand: $$x^2 + 4x + 4 - 20 = -4x -16$$ Simplify left: $$x^2 + 4x -16 = -4x -16$$ Bring all terms to one side: $$x^2 + 4x -16 + 4x +16 = 0$$ $$x^2 + 8x = 0$$ Factor: $$x(x + 8) = 0$$ Solutions: $$x=0 \quad \text{or} \quad x=-8$$ $x=0$ corresponds to point $P$, so $Q$ is at $x=-8$. Find $y$ coordinate of $Q$ using line $L$: $$y = -(-8) -4 = 8 -4 = 4$$ So, $Q=(-8,4)$. 7. **Distance between $P(0,-4)$ and $Q(-8,4)$:** Use distance formula: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(-8 - 0)^2 + (4 - (-4))^2} = \sqrt{(-8)^2 + (8)^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2}$$ **Final answers:** (a) Axis of symmetry: $x=-2$ (b) $h=-2$, $k=-5$ (c) $y$-coordinate of $P$ is $-4$ (d) Equation of line $L$: $y = -x -4$ (e) Distance between $P$ and $Q$ is $8\sqrt{2}$