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Complete the missing parts of the table for the following function.

\begin{tabular}{c|cccccc}
\multicolumn{7}{c}{[tex]$y=\left(\frac{1}{4}\right)^x$[/tex]} \\
[tex]$x$[/tex] & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline
[tex]$y$[/tex] & [tex]$16$[/tex] & [tex]$4$[/tex] & [tex]$1$[/tex] & [tex]$\frac{1}{4}$[/tex] & [tex]$\frac{1}{16}$[/tex] & [tex]$\frac{1}{64}$[/tex]
\end{tabular}


Sagot :

Sure, let's determine the missing parts of the table for the given function \( y = \left(\frac{1}{4}\right)^x \).

Let's complete the values one by one:
- For \( x = -2 \):
[tex]\[ y = \left(\frac{1}{4}\right)^{-2} = \left(\frac{4}{1}\right)^2 = 16 \][/tex]
The missing value for \( y \) when \( x = -2 \) is \( 16 \).

- For \( x = 0 \):
[tex]\[ y = \left(\frac{1}{4}\right)^0 = 1 \][/tex]
The missing value for \( y \) when \( x = 0 \) is \( 1 \).

- For \( x = 3 \):
[tex]\[ y = \left(\frac{1}{4}\right)^3 = \frac{1}{64} \][/tex]
The missing value for \( y \) when \( x = 3 \) is \( \frac{1}{64} \).

Now, we can complete the table:
[tex]\[ \begin{tabular}{c|cccccc} \multicolumn{7}{c}{[tex]$y=\left(\frac{1}{4}\right)^x$[/tex]} \\
[tex]$x$[/tex] & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline
[tex]$y$[/tex] & 16 & 4 & 1 & \frac{1}{4} & \frac{1}{16} & \frac{1}{64} \\
\end{tabular}
\][/tex]