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\begin{tabular}{|l|l|l|l|}
\hline
\multicolumn{4}{|c|}{[tex]$h(x)=4(x-2)$[/tex]} \\
\hline
[tex]$x$[/tex] & -2 & -1 & 1 \\
\hline
[tex]$y$[/tex] & & & \\
\hline
\end{tabular}


Sagot :

To solve this problem, let's determine the corresponding [tex]\( y \)[/tex] values for the given [tex]\( x \)[/tex] values using the function [tex]\( h(x) = 4(x - 2) \)[/tex].

### Step-by-Step Solution:

1. Given Function:
[tex]\[ h(x) = 4(x - 2) \][/tex]

2. Determine [tex]\( y \)[/tex] when [tex]\( x = -2 \)[/tex]:
[tex]\[ h(-2) = 4(-2 - 2) = 4(-4) = -16 \][/tex]
So, when [tex]\( x = -2 \)[/tex], [tex]\( y = -16 \)[/tex].

3. Determine [tex]\( y \)[/tex] when [tex]\( x = -1 \)[/tex]:
[tex]\[ h(-1) = 4(-1 - 2) = 4(-3) = -12 \][/tex]
So, when [tex]\( x = -1 \)[/tex], [tex]\( y = -12 \)[/tex].

4. Determine [tex]\( y \)[/tex] when [tex]\( x = 1 \)[/tex]:
[tex]\[ h(1) = 4(1 - 2) = 4(-1) = -4 \][/tex]
So, when [tex]\( x = 1 \)[/tex], [tex]\( y = -4 \)[/tex].

### Filling in the Table:

[tex]\[ \begin{tabular}{|c|c|c|c|} \hline \multicolumn{4}{|c|}{ $h(x) = 4(x-2)$ } \\ \hline $x$ & -2 & -1 & 1 \\ \hline $y$ & -16 & -12 & -4 \\ \hline \end{tabular} \][/tex]

In summary, the corresponding [tex]\( y \)[/tex] values for [tex]\( x = -2, -1, 1 \)[/tex] are [tex]\( -16, -12, -4 \)[/tex] respectively.
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