1. **State the problem:** We have the function $y=4\sqrt{-8x+16}-11$ and want to find its vertex, domain, and range, and rewrite it in vertex form. 2. **Rewrite the function inside the square root:** $$-8x+16 = -8(x-2)$$ 3. **Express the function as:** $$y=4\sqrt{-8(x-2)}-11$$ 4. **Simplify inside the square root:** Since $\sqrt{a b} = \sqrt{a} \sqrt{b}$, $$y=4\sqrt{-8}\sqrt{x-2}-11$$ However, $\sqrt{-8}$ is imaginary, so instead, factor out $-1$ inside the root: $$y=4\sqrt{-1 \cdot 8(x-2)}-11=4\sqrt{-1}\sqrt{8(x-2)}-11$$ Since $\sqrt{-1}=i$ (imaginary), the function is real only if $-8x+16 \geq 0$ or $x \leq 2$. 5. **Rewrite the function to isolate the vertex:** Let’s set $$u = -8x +16$$ The function is $$y=4\sqrt{u} -11$$ The square root function $\sqrt{u}$ has its minimum at $u=0$, so vertex occurs when $$-8x +16 = 0 \Rightarrow x=2$$ At $x=2$, $$y=4\sqrt{0} -11 = -11$$ So the vertex is at $(2, -11)$. 6. **Domain:** Since the expression inside the root must be non-negative, $$-8x +16 \geq 0 \Rightarrow x \leq 2$$ 7. **Range:** The smallest value of $y$ is at the vertex $-11$, and since $\sqrt{u} \geq 0$, $y$ increases from $-11$ upwards. So, $$\text{Range} = [-11, \infty)$$ 8. **General form:** The function is already in vertex form for a square root function: $$y = a \sqrt{x - h} + k$$ Here, $$a = 4\sqrt{-8}$$ is complex, so instead, keep the function as $$y = 4\sqrt{-8(x-2)} - 11$$ or equivalently, $$y = 4\sqrt{16 - 8x} - 11$$ with vertex at $(2, -11)$. **Final answers:** - Vertex: $(2, -11)$ - Domain: $(-\infty, 2]$ - Range: $[-11, \infty)$