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Complete the function table for the given domain, and plot the points on the graph.

[tex]f(x) = (x-5)^2 + 1[/tex]

\begin{tabular}{|c|c|c|c|c|c|}
\hline
[tex]$x$[/tex] & 2 & 3 & 4 & 5 & 6 \\
\hline
[tex]$f(x)$[/tex] & & & & & \\
\hline
\end{tabular}


Sagot :

To complete the function table and plot the points on the graph, we need to calculate [tex]\( f(x) \)[/tex] for each given [tex]\( x \)[/tex] in the domain using the function [tex]\( f(x) = (x-5)^2 + 1 \)[/tex]. Here are the steps:

1. Calculate [tex]\( f(2) \)[/tex]:
[tex]\[ f(2) = (2-5)^2 + 1 = (-3)^2 + 1 = 9 + 1 = 10 \][/tex]
So, [tex]\( f(2) = 10 \)[/tex].

2. Calculate [tex]\( f(3) \)[/tex]:
[tex]\[ f(3) = (3-5)^2 + 1 = (-2)^2 + 1 = 4 + 1 = 5 \][/tex]
So, [tex]\( f(3) = 5 \)[/tex].

3. Calculate [tex]\( f(4) \)[/tex]:
[tex]\[ f(4) = (4-5)^2 + 1 = (-1)^2 + 1 = 1 + 1 = 2 \][/tex]
So, [tex]\( f(4) = 2 \)[/tex].

4. Calculate [tex]\( f(5) \)[/tex]:
[tex]\[ f(5) = (5-5)^2 + 1 = (0)^2 + 1 = 0 + 1 = 1 \][/tex]
So, [tex]\( f(5) = 1 \)[/tex].

5. Calculate [tex]\( f(6) \)[/tex]:
[tex]\[ f(6) = (6-5)^2 + 1 = (1)^2 + 1 = 1 + 1 = 2 \][/tex]
So, [tex]\( f(6) = 2 \)[/tex].

Now the completed function table looks like this:

\begin{tabular}{|c|c|c|c|c|c|}
\hline [tex]$x$[/tex] & 2 & 3 & 4 & 5 & 6 \\
\hline [tex]$f(x)$[/tex] & 10 & 5 & 2 & 1 & 2 \\
\hline
\end{tabular}

Next, plot the points [tex]\((x, f(x))\)[/tex] on the graph. The points are:

- [tex]\((2, 10)\)[/tex]
- [tex]\((3, 5)\)[/tex]
- [tex]\((4, 2)\)[/tex]
- [tex]\((5, 1)\)[/tex]
- [tex]\((6, 2)\)[/tex]