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 - 3}+8…

Explanation:

Step1: Find the range of ( f(x)=-2\sqrt{x - 3}+8 )

The square - root function ( y=\sqrt{x-3} ) has a range of ( y\geq0 ). Multiply by (- 2) (which reflects the graph over the (x) - axis and vertically stretches it), so ( -2\sqrt{x - 3}\leq0 ). Then add (8), so the range of ( f(x)) is ( y\leq8 ).

Step2: Analyze the range of ( g(x)=-\sqrt{x - 3}+8 )

For the square - root function ( y = \sqrt{x-3}), (y\geq0). Multiply by (-1) (reflects the graph over the (x) - axis), so (-\sqrt{x - 3}\leq0). Then add (8), so the range of (g(x)=-\sqrt{x - 3}+8) is (y\leq8).

Step3: Analyze the range of ( g(x)=\sqrt{x - 3}-8 )

Since (y = \sqrt{x-3}\geq0), then (y=\sqrt{x - 3}-8\geq - 8).

Step4: Analyze the range of ( g(x)=\sqrt{x - 3}+8 )

Since (y=\sqrt{x - 3}\geq0), then (y=\sqrt{x - 3}+8\geq8).

Step5: Analyze the range of ( g(x)=-\sqrt{x + 3}-8 )

Since (y=\sqrt{x + 3}\geq0), then (y=-\sqrt{x + 3}\leq0) and (y=-\sqrt{x + 3}-8\leq - 8).

Answer:

(g(x)=-\sqrt{x - 3}+8) (the third option)