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\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$f(x)$[/tex] \\
\hline
-3 & 50 \\
\hline
-2 & 0 \\
\hline
-1 & -6 \\
\hline
0 & -4 \\
\hline
1 & -6 \\
\hline
2 & 0 \\
\hline
\end{tabular}

Use the table to complete the statements.

The [tex]$x$[/tex]-intercepts shown in the table are [tex]$\square$[/tex] and [tex]$\square$[/tex].

The [tex]$y$[/tex]-intercept shown in the table is [tex]$\square$[/tex].

Sagot :

GinnyAnsweridn
To complete the statements using the given table:

\begin{tabular}{|c|c|}
\hline
[tex]\(x\)[/tex] & [tex]\(f(x)\)[/tex] \\
\hline
-3 & 50 \\
\hline
-2 & 0 \\
\hline
-1 & -6 \\
\hline
0 & -4 \\
\hline
1 & -6 \\
\hline
2 & 0 \\
\hline
\end{tabular}

We need to identify the [tex]\(x\)[/tex]-intercepts and the [tex]\(y\)[/tex]-intercept.

1. Finding the [tex]\(x\)[/tex]-intercepts: The [tex]\(x\)[/tex]-intercepts are the values of [tex]\(x\)[/tex] where [tex]\(f(x) = 0\)[/tex]. By looking at the table, we see that [tex]\(f(x) = 0\)[/tex] at [tex]\(x = -2\)[/tex] and [tex]\(x = 2\)[/tex].

Therefore, the [tex]\(x\)[/tex]-intercepts shown in the table are [tex]\(\boxed{-2}\)[/tex] and [tex]\(\boxed{2}\)[/tex].

2. Finding the [tex]\(y\)[/tex]-intercept: The [tex]\(y\)[/tex]-intercept is the value of [tex]\(f(x)\)[/tex] when [tex]\(x = 0\)[/tex]. From the table, [tex]\(f(0) = -4\)[/tex].

Therefore, the [tex]\(y\)[/tex]-intercept shown in the table is [tex]\(\boxed{-4}\)[/tex].
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