1. **Problem statement:** We have a fire hose spraying water modeled by the quadratic function $$f(x) = -0.16x^2 + 2x$$ where $x$ is the horizontal distance in metres and $f(x)$ is the height of the water in metres. 2. **Restrictions on variables:** Since $x$ represents horizontal distance from the hose on the ground, it must be non-negative: $$x \geq 0$$. Also, the height $f(x)$ cannot be negative because water cannot go below ground level, so $$f(x) \geq 0$$. 3. **Finding the domain from $f(x) \geq 0$:** Solve $$-0.16x^2 + 2x \geq 0$$. Factor out $x$: $$x(-0.16x + 2) \geq 0$$ Set each factor to zero: $$x = 0$$ or $$-0.16x + 2 = 0 \Rightarrow x = \frac{2}{0.16} = 12.5$$ Since the parabola opens downward (coefficient of $x^2$ is negative), $f(x) \geq 0$ between $x=0$ and $x=12.5$. 4. **Summary of restrictions:** $$0 \leq x \leq 12.5$$ and $$f(x) \geq 0$$. 5. **Graph sketch details:** - The parabola intercepts the x-axis at $x=0$ and $x=12.5$. - The vertex is the maximum point, found by $$x = -\frac{b}{2a} = -\frac{2}{2 \times (-0.16)} = 6.25$$. - Calculate vertex height: $$f(6.25) = -0.16(6.25)^2 + 2(6.25) = -0.16(39.0625) + 12.5 = -6.25 + 12.5 = 6.25$$. - Label axes: horizontal distance ($x$) and height of water ($f(x)$). - Label intercepts: $(0,0)$ and $(12.5,0)$. - Label vertex: $(6.25, 6.25)$. 6. **Finding horizontal distance when height is 4 m:** Set $$f(x) = 4$$: $$-0.16x^2 + 2x = 4$$ Rearranged: $$-0.16x^2 + 2x - 4 = 0$$ Multiply both sides by -1 for easier calculation: $$0.16x^2 - 2x + 4 = 0$$ Calculate discriminant: $$\Delta = (-2)^2 - 4(0.16)(4) = 4 - 2.56 = 1.44$$ Find roots: $$x = \frac{2 \pm \sqrt{1.44}}{2 \times 0.16} = \frac{2 \pm 1.2}{0.32}$$ Two solutions: $$x_1 = \frac{2 - 1.2}{0.32} = \frac{0.8}{0.32} = 2.5$$ $$x_2 = \frac{2 + 1.2}{0.32} = \frac{3.2}{0.32} = 10$$ 7. **Interpretation:** The water reaches 4 m height at horizontal distances $x=2.5$ m and $x=10$ m. **Final answers:** - Restrictions: $$0 \leq x \leq 12.5$$ and $$f(x) \geq 0$$. - Vertex at $(6.25, 6.25)$. - Height 4 m at $x=2.5$ m and $x=10$ m.