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Complete the table of inputs and outputs for the function.

[tex]\[ f(x) = -5(x+7) \][/tex]

\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$f(x)$[/tex] \\
\hline
-9 & [tex]$\square$[/tex] \\
\hline
[tex]$\square$[/tex] & 0 \\
\hline
0 & [tex]$\square$[/tex] \\
\hline
[tex]$\square$[/tex] & -60 \\
\hline
\end{tabular}

Sagot :

Certainly! Let's fill in the table step-by-step:

1. For [tex]\( x = -9 \)[/tex]:
[tex]\[ f(-9) = -5(-9 + 7) = -5(-2) = 10 \][/tex]
So, for [tex]\( x = -9 \)[/tex], [tex]\( f(x) = 10 \)[/tex].

2. We need to find [tex]\( x \)[/tex] such that [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ 0 = -5(x + 7) \implies x + 7 = 0 \implies x = -7 \][/tex]
Therefore, for [tex]\( f(x) = 0 \)[/tex], [tex]\( x = -7 \)[/tex].

3. For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -5(0 + 7) = -5(7) = -35 \][/tex]
So, for [tex]\( x = 0 \)[/tex], [tex]\( f(x) = -35 \)[/tex].

4. We need to find [tex]\( x \)[/tex] such that [tex]\( f(x) = -60 \)[/tex]:
[tex]\[ -60 = -5(x + 7) \implies x + 7 = 12 \implies x = 5 \][/tex]
Therefore, when [tex]\( f(x) = -60 \)[/tex], [tex]\( x = 5 \)[/tex].

Now, we can complete the table:

[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $f(x)$ \\ \hline -9 & 10 \\ \hline -7 & 0 \\ \hline 0 & -35 \\ \hline 5 & -60 \\ \hline \end{tabular} \][/tex]