\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$f(x)$[/tex] \\
\hline
-4 & -10 \\
\hline
-3 & 0 \\
\hline
-2 & 0 \\
\hline
-1 & -4 \\
\hline
0 & -6 \\
\hline
1 & 0 \\
\hline
\end{tabular}

Which is the [tex]$y$[/tex]-intercept of the continuous function in the table?

A. [tex]$(0, -6)$[/tex]
B. [tex]$(-2, 0)$[/tex]
C. [tex]$(-6, 0)$[/tex]
D. [tex]$(0, -2)$[/tex]



Answer :

To determine the y-intercept of the function given in the table, we need to find the point where [tex]\( x = 0 \)[/tex]. The y-intercept is the value of the function [tex]\( f(x) \)[/tex] when [tex]\( x \)[/tex] is zero.

Let's check the table for the value of [tex]\( f(x) \)[/tex] when [tex]\( x = 0 \)[/tex]:

[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -4 & -10 \\ \hline -3 & 0 \\ \hline -2 & 0 \\ \hline -1 & -4 \\ \hline 0 & -6 \\ \hline 1 & 0 \\ \hline \end{array} \][/tex]

From the table, when [tex]\( x = 0 \)[/tex], the function [tex]\( f(x) \)[/tex] evaluates to [tex]\( -6 \)[/tex]. Therefore, the y-intercept is the point [tex]\((0, -6)\)[/tex].

Let's cross-check this with the given options:
1. [tex]\( (0, -6) \)[/tex]
2. [tex]\( (-2, 0) \)[/tex]
3. [tex]\( (-6, 0) \)[/tex]
4. [tex]\( (0, -2) \)[/tex]

Clearly, the y-intercept of the continuous function is [tex]\((0, -6)\)[/tex].