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 equation of the axis of symmetry. (b) The values of $h$ and $k$ in the vertex form $f(x) = \frac{1}{4}(x-h)^2 + k$. (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 $L$ with the graph. 2. **Axis of symmetry:** The axis of symmetry of a parabola in vertex form $f(x) = a(x-h)^2 + k$ is the vertical line $x = h$. Since the vertex is at $(-2,-5)$, the axis of symmetry is: $$x = -2$$ 3. **Values of $h$ and $k$:** From the vertex form $f(x) = \frac{1}{4}(x-h)^2 + k$, the vertex is at $(h,k)$. Given vertex $(-2,-5)$, we have: $$h = -2, \quad k = -5$$ 4. **Y-coordinate of $P$ (y-intercept):** The y-intercept occurs when $x=0$. Substitute $x=0$ into $f(x)$: $$f(0) = \frac{1}{4}(0 - (-2))^2 + (-5) = \frac{1}{4}(2)^2 - 5 = \frac{1}{4} \times 4 - 5 = 1 - 5 = -4$$ So, the y-coordinate of $P$ is $-4$. 5. **Equation of line $L$ passing through $P$ with gradient $-1$:** Point $P$ is $(0,-4)$ and slope $m = -1$. Using point-slope form: $$y - y_1 = m(x - x_1)$$ $$y - (-4) = -1(x - 0)$$ $$y + 4 = -x$$ $$y = -x - 4$$ 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 to clear fraction: $$(x+2)^2 - 20 = -4x - 16$$ Expand $(x+2)^2 = x^2 + 4x + 4$: $$x^2 + 4x + 4 - 20 = -4x - 16$$ Simplify left side: $$x^2 + 4x - 16 = -4x - 16$$ Add $4x + 16$ to both sides: $$x^2 + 4x - 16 + 4x + 16 = 0$$ $$x^2 + 8x = 0$$ Factor: $$x(x + 8) = 0$$ Solutions: $$x = 0 \quad \text{or} \quad x = -8$$ We already know $x=0$ 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) $x = -2$ (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 $8\sqrt{2}$