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…

Answer:

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

Explanation:

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

For $y = f(x)=-2\sqrt{x - 3}+8$, since $\sqrt{x - 3}\geq0$, then $-2\sqrt{x - 3}\leq0$, and $y=-2\sqrt{x - 3}+8\leq8$.

Step2: Analyze option A

For $g(x)=\sqrt{x - 3}-8$, as $\sqrt{x - 3}\geq0$, $g(x)=\sqrt{x - 3}-8\geq - 8$.

Step3: Analyze option B

For $g(x)=\sqrt{x - 3}+8$, since $\sqrt{x - 3}\geq0$, $g(x)=\sqrt{x - 3}+8\geq8$.

Step4: Analyze option C

For $g(x)=-\sqrt{x + 3}+8$, because $\sqrt{x + 3}\geq0$, then $-\sqrt{x + 3}\leq0$, and $g(x)=-\sqrt{x + 3}+8\leq8$, which has the same range as $f(x)$.

Step5: Analyze option D

For $g(x)=-\sqrt{x - 3}-8$, as $\sqrt{x - 3}\geq0$, $-\sqrt{x - 3}\leq0$, and $g(x)=-\sqrt{x - 3}-8\leq - 8$.