1. **Problem 4:** Determine the equation in vertex form for the parabola given the vertex and a point. 2. The vertex form of a parabola is given by: $$y = a(x - h)^2 + k$$ where $(h, k)$ is the vertex. 3. Given vertex $(-3, 2)$ and point $(3, -10)$, substitute into the vertex form to find $a$: $$-10 = a(3 - (-3))^2 + 2$$ 4. Simplify inside the parentheses: $$-10 = a(3 + 3)^2 + 2$$ $$-10 = a(6)^2 + 2$$ $$-10 = 36a + 2$$ 5. Subtract 2 from both sides: $$-10 - 2 = 36a$$ $$-12 = 36a$$ 6. Divide both sides by 36: $$\frac{-12}{36} = a$$ Show cancellation: $$a = \frac{\cancel{-12}}{\cancel{36}} = -\frac{1}{3}$$ 7. Write the final vertex form equation: $$y = -\frac{1}{3}(x + 3)^2 + 2$$ --- 1. **Problem 5:** Sketch the graph of the toy rocket's flight path given by: $$h = -(d - 4)(d - 16)$$ where $d$ is horizontal distance and $h$ is height. 2. Find the $d$-intercepts by setting $h=0$: $$0 = -(d - 4)(d - 16)$$ This implies: $$d - 4 = 0 \quad \text{or} \quad d - 16 = 0$$ So, $$d = 4 \quad \text{or} \quad d = 16$$ 3. Find the vertex (maximum point) since the parabola opens downward (negative leading coefficient). The axis of symmetry is the midpoint between the roots: $$d = \frac{4 + 16}{2} = 10$$ 4. Find the height at $d=10$: $$h = -(10 - 4)(10 - 16) = -(6)(-6) = 36$$ So the vertex is at $(10, 36)$. 5. The axis of symmetry is the vertical line: $$d = 10$$ 6. Summary: - $d$-intercepts at $(4, 0)$ and $(16, 0)$ - Vertex at $(10, 36)$ - Axis of symmetry $d=10$ The parabola opens downward, reaching a maximum height of 36 metres at $d=10$.