2. Evaluate [tex]y=3(0.5)^x[/tex] for the given values of [tex]x[/tex].

\begin{tabular}{|r|r|r|r|r|r|}
\hline
[tex]$x$[/tex] & -2 & -1 & 0 & 1 & 2 \\
\hline
[tex]$y$[/tex] & & & & & \\
\hline
\end{tabular}



Answer :

Sure, let's work through the solution step-by-step to fill out the table given the function [tex]\( f(x) = 3 \cdot (0.5)^x \)[/tex].

1. Understanding the Function:
The function given is [tex]\( f(x) = 3 \cdot (0.5)^x \)[/tex]. This means for any value of [tex]\( x \)[/tex], you raise [tex]\( 0.5 \)[/tex] to the power of [tex]\( x \)[/tex] and then multiply the result by 3.

2. Calculating [tex]\( f(x) \)[/tex] for Different Values of [tex]\( x \)[/tex]:

- For [tex]\( x = -2 \)[/tex]:

[tex]\[ f(-2) = 3 \cdot (0.5)^{-2} \][/tex]

Recall that raised to a negative power, [tex]\( (a^{-n} = \frac{1}{a^n}) \)[/tex]:

[tex]\[ (0.5)^{-2} = \frac{1}{(0.5)^2} = \frac{1}{0.25} = 4 \][/tex]

Therefore,

[tex]\[ f(-2) = 3 \cdot 4 = 12 \][/tex]

- For [tex]\( x = -1 \)[/tex]:

[tex]\[ f(-1) = 3 \cdot (0.5)^{-1} \][/tex]

[tex]\[ (0.5)^{-1} = \frac{1}{0.5} = 2 \][/tex]

Therefore,

[tex]\[ f(-1) = 3 \cdot 2 = 6 \][/tex]

- For [tex]\( x = 0 \)[/tex]:

[tex]\[ f(0) = 3 \cdot (0.5)^0 \][/tex]

Any number raised to the power of 0 is 1:

[tex]\[ (0.5)^0 = 1 \][/tex]

Therefore,

[tex]\[ f(0) = 3 \cdot 1 = 3 \][/tex]

- For [tex]\( x = 1 \)[/tex]:

[tex]\[ f(1) = 3 \cdot (0.5)^1 \][/tex]

[tex]\[ (0.5)^1 = 0.5 \][/tex]

Therefore,

[tex]\[ f(1) = 3 \cdot 0.5 = 1.5 \][/tex]

- For [tex]\( x = 2 \)[/tex]:

[tex]\[ f(2) = 3 \cdot (0.5)^2 \][/tex]

[tex]\[ (0.5)^2 = 0.25 \][/tex]

Therefore,

[tex]\[ f(2) = 3 \cdot 0.25 = 0.75 \][/tex]

3. Filling in the Table:
Now that we have computed [tex]\( f(x) \)[/tex] for each value of [tex]\( x \)[/tex], we can fill in the table:

[tex]\[ \begin{tabular}{|r|r|r|r|r|r|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline f(x) & 12 & 6 & 3 & 1.5 & 0.75 \\ \hline \end{tabular} \][/tex]

So your completed table should look exactly like this:

[tex]\[ \begin{tabular}{|r|r|r|r|r|r|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline f(x) & 12 & 6 & 3 & 1.5 & 0.75 \\ \hline \end{tabular} \][/tex]