which function has the same range as (f(x)=-2sqrt{x - 3}+8)?\n(g(x)=sqrt{x - 3}-8)\n(g(x)=sqrt{x…

Explanation:

Step1: Analyze the range of $f(x)$

For the square - root function $y = \sqrt{x - 3}$, the domain is $x\geq3$ and $\sqrt{x - 3}\geq0$. Then $- 2\sqrt{x - 3}\leq0$, and $f(x)=-2\sqrt{x - 3}+8\leq8$. So the range of $f(x)$ is $(-\infty,8]$.

Step2: Analyze each option

Option 1: $g(x)=\sqrt{x - 3}-8$

Since $\sqrt{x - 3}\geq0$, then $g(x)=\sqrt{x - 3}-8\geq - 8$. The range is $[-8,\infty)$.

Option 2: $g(x)=\sqrt{x - 3}+8$

Since $\sqrt{x - 3}\geq0$, then $g(x)=\sqrt{x - 3}+8\geq8$. The range is $[8,\infty)$.

Option 3: $g(x)=-\sqrt{x + 3}+8$

The domain is $x\geq - 3$, and $\sqrt{x + 3}\geq0$, so $-\sqrt{x + 3}\leq0$ and $g(x)=-\sqrt{x + 3}+8\leq8$. The range is $(-\infty,8]$.

Option 4: $g(x)=-\sqrt{x - 3}-8$

Since $\sqrt{x - 3}\geq0$, then $-\sqrt{x - 3}\leq0$ and $g(x)=-\sqrt{x - 3}-8\leq - 8$. The range is $(-\infty,-8]$.

Answer:

$g(x)=-\sqrt{x + 3}+8$