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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 & \\
\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 based on the given table, let's analyze it step by step.

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

Therefore, the [tex]\( x \)[/tex]-intercept in the table is [tex]\( -2 \)[/tex].

2. The [tex]\( y \)[/tex]-intercept is the value of [tex]\( f(x) \)[/tex] where [tex]\( x = 0 \)[/tex]. From the table, this gives us:
- When [tex]\( x = 0 \)[/tex], [tex]\( f(x) = -4 \)[/tex].

Hence, the [tex]\( y \)[/tex]-intercept is [tex]\( -4 \)[/tex].

By extracting this information, we complete the statements as follows:

The [tex]\( x \)[/tex]-intercept shown in the table is [tex]\(-2\)[/tex] and the [tex]\( y \)[/tex]-intercept shown in the table is [tex]\(-4\)[/tex].
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