To determine which table correctly lists the values of [tex]\( h(x) = x^2 - 3 \)[/tex] for given [tex]\( x \)[/tex] values, let's calculate [tex]\( h(x) \)[/tex] at [tex]\( x = 1 \)[/tex], [tex]\( x = 2 \)[/tex], and [tex]\( x = 3 \)[/tex] step-by-step.
First, for [tex]\( x = 1 \)[/tex]:
[tex]\[ h(1) = 1^2 - 3 = 1 - 3 = -2 \][/tex]
Next, for [tex]\( x = 2 \)[/tex]:
[tex]\[ h(2) = 2^2 - 3 = 4 - 3 = 1 \][/tex]
Finally, for [tex]\( x = 3 \)[/tex]:
[tex]\[ h(3) = 3^2 - 3 = 9 - 3 = 6 \][/tex]
So, the set of values [tex]\( h(x) \)[/tex] for [tex]\( x = 1 \)[/tex], [tex]\( x = 2 \)[/tex], and [tex]\( x = 3 \)[/tex] are [tex]\([-2, 1, 6]\)[/tex].
Now let's compare these calculated values to each of the given tables:
Option (A):
\begin{tabular}{|c|c|c|c|}
\hline[tex]$x$[/tex] & 1 & 2 & 3 \\
\hline[tex]$h(x)$[/tex] & 4 & 5 & 6 \\
\hline
\end{tabular}
Here, [tex]\( h(1) \neq -2 \)[/tex], [tex]\( h(2) \neq 1 \)[/tex], [tex]\( h(3) \neq 6 \)[/tex].
Option (B):
\begin{tabular}{|c|c|c|c|}
\hline[tex]$x$[/tex] & 1 & 2 & 3 \\
\hline[tex]$h(x)$[/tex] & -2 & 1 & 6 \\
\hline
\end{tabular}
Here, [tex]\( h(1) = -2 \)[/tex], [tex]\( h(2) = 1 \)[/tex], [tex]\( h(3) = 6 \)[/tex]. This matches the calculated values.
Option (C):
\begin{tabular}{|c|c|c|c|}
\hline[tex]$x$[/tex] & 1 & 2 & 3 \\
\hline[tex]$h(x)$[/tex] & -1 & 1 & 3 \\
\hline
\end{tabular}
Here, [tex]\( h(1) \neq -2 \)[/tex], [tex]\( h(2) = 1 \)[/tex], [tex]\( h(3) \neq 6 \)[/tex].
Option (D):
\begin{tabular}{|c|c|c|c|}
\hline[tex]$x$[/tex] & 1 & 2 & 3 \\
\hline[tex]$h(x)$[/tex] & -2 & 1 & 3 \\
\hline
\end{tabular}
Here, [tex]\( h(1) = -2 \)[/tex], [tex]\( h(2) = 1 \)[/tex], [tex]\( h(3) \neq 6 \)[/tex].
Thus, the correct table that lists the values [tex]\( x = 1, 2, 3 \)[/tex] and their corresponding [tex]\( h(x) = -2, 1, 6 \)[/tex] is:
Option (B):
\begin{tabular}{|c|c|c|c|}
\hline[tex]$x$[/tex] & 1 & 2 & 3 \\
\hline[tex]$h(x)$[/tex] & -2 & 1 & 6 \\
\hline
\end{tabular}
