Answered

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\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$f(x)$[/tex] \\
\hline
-1 & -9 \\
\hline
0 & -8 \\
\hline
1 & -7 \\
\hline
2 & 0 \\
\hline
3 & 19 \\
\hline
4 & 56 \\
\hline
\end{tabular}

What is the [tex]$x$[/tex]-intercept of the continuous function shown in the table?

A. [tex]$(0, -8)$[/tex]

B. [tex]$(0, 2)$[/tex]

C. [tex]$(-8, 0)$[/tex]

D. [tex]$(2, 0)$[/tex]

Sagot :

GinnyAnswer
To determine the [tex]\(x\)[/tex]-intercept of the continuous function represented by the given table, we need to identify the point where the function [tex]\(f(x)\)[/tex] crosses the x-axis. The [tex]\(x\)[/tex]-intercept occurs when [tex]\(f(x) = 0\)[/tex].

Examining the table:

[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -1 & -9 \\ \hline 0 & -8 \\ \hline 1 & -7 \\ \hline 2 & 0 \\ \hline 3 & 19 \\ \hline 4 & 56 \\ \hline \end{array} \][/tex]

We see that at [tex]\(x = 2\)[/tex], [tex]\(f(x) = 0\)[/tex]. This means the function crosses the x-axis at the point [tex]\((2, 0)\)[/tex].

Therefore, the [tex]\(x\)[/tex]-intercept of the function is [tex]\((2, 0)\)[/tex].
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