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## Sagot :

Given the table:

[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $f(x)$ \\ \hline -2 & 20 \\ \hline -1 & 0 \\ \hline 0 & -6 \\ \hline 1 & -4 \\ \hline 2 & 0 \\ \hline 3 & 0 \\ \hline \end{tabular} \][/tex]

We need to look for the value of [tex]\( x \)[/tex] where [tex]\( f(x) = 0 \)[/tex].

Let's go through the table entries:

- At [tex]\( x = -2 \)[/tex], [tex]\( f(x) = 20 \)[/tex]

- At [tex]\( x = -1 \)[/tex], [tex]\( f(x) = 0 \)[/tex]

- At [tex]\( x = 0 \)[/tex], [tex]\( f(x) = -6 \)[/tex]

- At [tex]\( x = 1 \)[/tex], [tex]\( f(x) = -4 \)[/tex]

- At [tex]\( x = 2 \)[/tex], [tex]\( f(x) = 0 \)[/tex]

- At [tex]\( x = 3 \)[/tex], [tex]\( f(x) = 0 \)[/tex]

From this, we see that [tex]\( f(x) = 0 \)[/tex] at the following [tex]\( x \)[/tex]-values:

- [tex]\( x = -1 \)[/tex]

- [tex]\( x = 2 \)[/tex]

- [tex]\( x = 3 \)[/tex]

Therefore, the correct answer among the given options is the point where [tex]\( x = -1 \)[/tex] because all the provided points should have [tex]\( f(x) = 0 \)[/tex]. The given points are:

- [tex]\((-1, 0)\)[/tex]

- [tex]\((0, -6)\)[/tex]

- [tex]\((-6, 0)\)[/tex]

- [tex]\((0, -1)\)[/tex]

Thus, the [tex]\( x \)[/tex]-intercept of the continuous function is:

[tex]\[ (-1, 0) \][/tex]