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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 :

To solve this problem, we need to analyze the given table which presents values of \( x \) and their corresponding \( f(x) \).

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

### Finding the \( x \)-Intercepts
The \( x \)-intercepts are the points where the function \( f(x) \) crosses the \( x \)-axis. At these points, \( f(x) = 0 \).

From the table:
- When \( x = -2 \), \( f(x) = 0 \).
- When \( x = 2 \), \( f(x) = 0 \).

Thus, the \( x \)-intercepts are \( -2 \) and \( 2 \).

### Finding the \( y \)-Intercept
The \( y \)-intercept is the point where the function \( f(x) \) crosses the \( y \)-axis. This occurs when \( x = 0 \).

From the table:
- When \( x = 0 \), \( f(x) = -4 \).

Thus, the \( y \)-intercept is \( -4 \).

### Completing the Statements
The \( x \)-intercepts shown in the table are \( -2 \) and \( 2 \).

The [tex]\( y \)[/tex]-intercept shown in the table is [tex]\( -4 \)[/tex] [tex]\(\checkmark\)[/tex].
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