Looking at the restrictions over the variable x, we know that the domain is:
[tex]x\ge2[/tex]To find the range, notice that:
[tex]\sqrt[]{x-2}\ge0[/tex]On the other hand, the function:
[tex]y=\sqrt[]{x-2}[/tex]is an increasing function (its value grows when x grows), and can get as large as we want provided a sufficiently large value for x. Then, the range of such a function would be:
[tex]y\ge0[/tex]Which does not get altered when we multiply the square root of (x-2) by 4.
Since the function:
[tex]y=-5+4\sqrt[]{x-2}[/tex]is a 5-units shift downwards, then the variable y can take any value from -5 onwards.
Then, the range of the function is:
[tex]y\ge-5[/tex]Another way to find the range is to isolate x from the equation:
[tex]\begin{gathered} y=-5+4\sqrt[]{x-2} \\ \Rightarrow y+5=4\sqrt[]{x-2} \\ \Rightarrow\frac{y+5}{4}=\sqrt[]{x-2} \\ \Rightarrow(\frac{y+5}{4})^2=x-2 \\ \Rightarrow x-2=(\frac{y+5}{4})^2 \\ \Rightarrow x=(\frac{y+5}{4})^2+2 \end{gathered}[/tex]Since we already know that x must be greater than 2, then:
[tex]\begin{gathered} 2\le x \\ \Rightarrow2\le(\frac{y+5}{4})^2+2 \\ \Rightarrow0\le(\frac{y+5}{4})^2 \\ \Rightarrow0\le|\frac{y+5}{4}| \\ \Rightarrow0\le|y+5| \end{gathered}[/tex]From here, there are two options:
[tex]\begin{gathered} 0\le y+5 \\ \Rightarrow-5\le y \\ \text{ Or} \\ 0\le-y-5 \\ \Rightarrow y\le-5 \end{gathered}[/tex]Since we know an equation for y, then:
[tex]\begin{gathered} -5\le-5+4\sqrt[]{x-2} \\ \Rightarrow0\le4\sqrt[]{x-2} \end{gathered}[/tex]Or:
[tex]\begin{gathered} -5+4\sqrt[]{x-2}\le-5 \\ \Rightarrow4\sqrt[]{x-2}\le0 \end{gathered}[/tex]The second case is not true for every x.
Therefore:
[tex]-5\le y[/tex]Therefore:
[tex]\begin{gathered} \text{Domain: }x\ge2 \\ \text{Range: }y\ge-5 \end{gathered}[/tex]