Answer :
Sure! Let's break down the function [tex]\( f(x) = \sqrt{x - 5} - 1 \)[/tex] and complete a data table for specific values of [tex]\( x \)[/tex]. We'll use the values [tex]\( x = 6, 7, 8, 9, \)[/tex] and [tex]\( 10 \)[/tex].
### Step-by-Step Calculation:
1. When [tex]\( x = 6 \)[/tex]:
[tex]\[ f(6) = \sqrt{6 - 5} - 1 = \sqrt{1} - 1 = 1 - 1 = 0 \][/tex]
This gives the point [tex]\((6, 0)\)[/tex].
2. When [tex]\( x = 7 \)[/tex]:
[tex]\[ f(7) = \sqrt{7 - 5} - 1 = \sqrt{2} - 1 \approx 1.4142135623730951 - 1 \approx 0.41421356237309515 \][/tex]
This gives the point [tex]\((7, 0.41421356237309515)\)[/tex].
3. When [tex]\( x = 8 \)[/tex]:
[tex]\[ f(8) = \sqrt{8 - 5} - 1 = \sqrt{3} - 1 \approx 1.7320508075688772 - 1 \approx 0.7320508075688772 \][/tex]
This gives the point [tex]\((8, 0.7320508075688772)\)[/tex].
4. When [tex]\( x = 9 \)[/tex]:
[tex]\[ f(9) = \sqrt{9 - 5} - 1 = \sqrt{4} - 1 = 2 - 1 = 1 \][/tex]
This gives the point [tex]\((9, 1)\)[/tex].
5. When [tex]\( x = 10 \)[/tex]:
[tex]\[ f(10) = \sqrt{10 - 5} - 1 = \sqrt{5} - 1 \approx 2.23606797749979 - 1 \approx 1.2360679774997898 \][/tex]
This gives the point [tex]\((10, 1.2360679774997898)\)[/tex].
### Data Table:
Combining all the points we've calculated:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline 6 & 0.0 \\ 7 & 0.41421356237309515 \\ 8 & 0.7320508075688772 \\ 9 & 1.0 \\ 10 & 1.2360679774997898 \\ \hline \end{array} \][/tex]
Thus, the completed data table for the function [tex]\( f(x) = \sqrt{x - 5} - 1 \)[/tex] is:
[tex]\[ \begin{array}{c c} x & f(x) \\ \hline 6 & 0.0 \\ 7 & 0.41421356237309515 \\ 8 & 0.7320508075688772 \\ 9 & 1.0 \\ 10 & 1.2360679774997898 \\ \end{array} \][/tex]
### Step-by-Step Calculation:
1. When [tex]\( x = 6 \)[/tex]:
[tex]\[ f(6) = \sqrt{6 - 5} - 1 = \sqrt{1} - 1 = 1 - 1 = 0 \][/tex]
This gives the point [tex]\((6, 0)\)[/tex].
2. When [tex]\( x = 7 \)[/tex]:
[tex]\[ f(7) = \sqrt{7 - 5} - 1 = \sqrt{2} - 1 \approx 1.4142135623730951 - 1 \approx 0.41421356237309515 \][/tex]
This gives the point [tex]\((7, 0.41421356237309515)\)[/tex].
3. When [tex]\( x = 8 \)[/tex]:
[tex]\[ f(8) = \sqrt{8 - 5} - 1 = \sqrt{3} - 1 \approx 1.7320508075688772 - 1 \approx 0.7320508075688772 \][/tex]
This gives the point [tex]\((8, 0.7320508075688772)\)[/tex].
4. When [tex]\( x = 9 \)[/tex]:
[tex]\[ f(9) = \sqrt{9 - 5} - 1 = \sqrt{4} - 1 = 2 - 1 = 1 \][/tex]
This gives the point [tex]\((9, 1)\)[/tex].
5. When [tex]\( x = 10 \)[/tex]:
[tex]\[ f(10) = \sqrt{10 - 5} - 1 = \sqrt{5} - 1 \approx 2.23606797749979 - 1 \approx 1.2360679774997898 \][/tex]
This gives the point [tex]\((10, 1.2360679774997898)\)[/tex].
### Data Table:
Combining all the points we've calculated:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline 6 & 0.0 \\ 7 & 0.41421356237309515 \\ 8 & 0.7320508075688772 \\ 9 & 1.0 \\ 10 & 1.2360679774997898 \\ \hline \end{array} \][/tex]
Thus, the completed data table for the function [tex]\( f(x) = \sqrt{x - 5} - 1 \)[/tex] is:
[tex]\[ \begin{array}{c c} x & f(x) \\ \hline 6 & 0.0 \\ 7 & 0.41421356237309515 \\ 8 & 0.7320508075688772 \\ 9 & 1.0 \\ 10 & 1.2360679774997898 \\ \end{array} \][/tex]
