Graphs of Exponential Relationships: Mastery Test

Problem:

Use the drawing tool(s) to form the correct answers on the provided grid.

Consider the function [tex]\( g(x) = -3\left(\frac{1}{2}\right)^x \)[/tex].

For the [tex]\( x \)[/tex]-values given in the table, determine the corresponding values of [tex]\( g(x) \)[/tex] and plot each point on the graph.

[tex]\[
\begin{tabular}{|c|c|c|c|c|}
\hline
$x$ & -2 & -1 & 0 & 1 \\
\hline
$g(x)$ & & & & \\
\hline
\end{tabular}
\][/tex]

Instructions:
1. Calculate [tex]\( g(x) \)[/tex] for each [tex]\( x \)[/tex]-value.
2. Fill in the table with the calculated values.
3. Plot each point on the graph using the provided grid.



Answer :

To determine the corresponding values of [tex]\( g(x) = -3 \left(\frac{1}{2}\right)^x \)[/tex] for the given [tex]\( x \)[/tex]-values, we'll evaluate the function at each [tex]\( x \)[/tex]-value and then plot the points on the graph.

1. Calculate [tex]\( g(x) \)[/tex] for each [tex]\( x \)[/tex]-value:

- For [tex]\( x = -2 \)[/tex]:
[tex]\[ g(-2) = -3 \left(\frac{1}{2}\right)^{-2} = -3 \left(2\right)^2 = -3 \cdot 4 = -12 \][/tex]

- For [tex]\( x = -1 \)[/tex]:
[tex]\[ g(-1) = -3 \left(\frac{1}{2}\right)^{-1} = -3 \left(2\right) = -3 \cdot 2 = -6 \][/tex]

- For [tex]\( x = 0 \)[/tex]:
[tex]\[ g(0) = -3 \left(\frac{1}{2}\right)^0 = -3 \cdot 1 = -3 \][/tex]

- For [tex]\( x = 1 \)[/tex]:
[tex]\[ g(1) = -3 \left(\frac{1}{2}\right)^1 = -3 \cdot \frac{1}{2} = -1.5 \][/tex]

2. Fill in the table with the calculated values:

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

3. Plot each point [tex]\((x, g(x))\)[/tex] on the graph:

- Plot the point [tex]\((-2, -12)\)[/tex]
- Plot the point [tex]\((-1, -6)\)[/tex]
- Plot the point [tex]\((0, -3)\)[/tex]
- Plot the point [tex]\((1, -1.5)\)[/tex]

Now you've successfully determined the values of the function [tex]\( g(x) = -3 \left(\frac{1}{2}\right)^x \)[/tex] at the given [tex]\( x \)[/tex]-values and plotted them on the graph!